In this section, we compute the independence complexes of graphs $\Tilde{\Gamma}_{n,5},\Tilde{\Lambda}_{n,5}, \Tilde{A}_{n,5}, \Tilde{B}_{n,5},$ $\Tilde{C}_{n,5}$, $\Tilde{D}_{n,5}, \Tilde{E}_{n,5}$ and $\Tilde{F}_{n,5}$. In computing the independence complexes of these graphs, as discussed in <Ref>, our approach focuses on performing computations primarily on one end (the leftmost part) of the graph while holding the other end (the rightmost part) fixed. Notably, for $n\geq 1$, we observe that the left ends of the graphs $H_{n,5}$ and $\Tilde{H}_{n,5}$ coincide, where $H_{n,5}\in \{\Gamma_{n,5},\Lambda_{n,5}, A_{n, 5}, B_{n 5}, C_{n, 5}, D_{n, 5}, E_{n, 5}, F_{n, 5}\}$. This observation suggests a method for computing the independence complexes of graphs $\Tilde{\Gamma}_{n,5},\Tilde{\Lambda}_{n,5}, \Tilde{A}_{n,5}, \Tilde{B}_{n,5}, \Tilde{C}_{n,5}, \Tilde{D}_{n,5}, \Tilde{E}_{n,5}$, and $\Tilde{F}_{n,5}$ using a similar strategy to that outlined in <Ref>. Furthermore, it is notable that for $n\geq 1$, these intermediary graphs adhere to a similar recurrence relation as those described in the preceding section.  Since the base graphs of these intermediary graphs are completely different from the graphs in <Ref>, we give the complete computations for base cases. Also, we give the explicit recurrence relation for each graph and higher values of $n$.

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]180:$g_{21}$] at (21) ;
[label=[label distance=-3pt]180:$g_{31}$] at (31) ;
[label=[label distance=-3pt]180:$g_{41}$] at (41) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

[vertex] (l1) at (-2,6)  ;
[vertex] (l2) at (-2,4)  ;
[vertex] (l3) at (-2,2)  ;
[vertex] (l4) at (-2,0)  ;

[edge] (l1) – (l2) ;
[edge] (l2) – (l3) ;
[edge] (l3) – (l4) ;
[edge] (l1) – (11) ;
[edge] (l2) – (21) ;
[edge] (l3) – (31) ;
[edge] (l4) – (41) ;
[edge] (l1) – (21) ;
[edge] (l2) – (11) ;
[edge] (l4) – (31) ;
[edge] (l3) – (41) ;

[label=[label distance=-3pt]180:$l_{1}$] at (l1) ;
[label=[label distance=-3pt]180:$l_{2}$] at (l2) ;
[label=[label distance=-3pt]180:$l_{3}$] at (l3) ;
[label=[label distance=-3pt]180:$l_{4}$] at (l4) ;

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

[vertex] (a1) at (-2,6)  ;
[vertex] (a2) at (-2,0)  ;

[edge] (a1) – (11) ;
[edge] (a1) – (21) ;
[edge] (a2) – (31) ;
[edge] (a2) – (41) ;

[label=[label distance=-3pt]180:$a_{1}$] at (a1) ;
[label=[label distance=-3pt]180:$a_{2}$] at (a2) ;

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

[vertex] (b1) at (-2,6)  ;
[vertex] (b2) at (-2,4)  ;
[vertex] (b3) at (-2,2)  ;
[vertex] (b4) at (-2,0)  ;
[vertex] (b31) at (-4,2)  ;
[vertex] (b41) at (-4,0)  ;

[edge] (b1) – (11) ;
[edge] (b1) – (21) ;
[edge] (b1) – (b2) ;
[edge] (b2) – (b3) ;
[edge] (b2) – (21) ;
[edge] (b2) – (11) ;
[edge] (b3) – (31) ;
[edge] (b3) – (41) ;
[edge] (b3) – (b4) ;
[edge] (b4) – (41) ;
[edge] (b4) – (31) ;

[edge] (b31) – (b3) ;
[edge] (b31) – (b2) ;
[edge] (b31) – (b41) ;
[edge] (b41) – (b4) ;

[label=[label distance=-3pt]180:$b_{3}$] at (b1) ;
[label=[label distance=-3pt]180:$b_{4}$] at (b2) ;
[label=[label distance=-9pt]45:$b_{5}$] at (b3) ;
[label=[label distance=-9pt]135:$b_{6}$] at (b4) ;
[label=[label distance=-3pt]180:$b_{1}$] at (b31) ;
[label=[label distance=-3pt]180:$b_{2}$] at (b41) ;

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

[vertex] (c2) at (-2,0)  ;

[edge] (c2) – (31) ;
[edge] (c2) – (41) ;

[label=[label distance=-3pt]180:$c_{1}$] at (c2) ;

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

[vertex] (d1) at (-2,4)  ;
[vertex] (d2) at (-2,2)  ;

[edge] (d1) – (11) ;
[edge] (d1) – (21) ;
[edge] (d1) – (d2) ;
[edge] (d2) – (31) ;
[edge] (d2) – (41) ;

[label=[label distance=-3pt]180:$d_{1}$] at (d1) ;
[label=[label distance=-3pt]180:$d_{2}$] at (d2) ;

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

[vertex] (e1) at (-2,3)  ;

[edge] (e1) – (11) ;
[edge] (e1) – (21) ;
[edge] (e1) – (31) ;
[edge] (e1) – (41) ;

[label=[label distance=-3pt]180:$e_{1}$] at (e1) ;

in 1,2,...,4

[vertex] (1) at (2*,3*2) ;

in 1,2,...,4

[vertex] (2) at (2*,2*2) ;

in 1,2,...,4

[vertex] (3) at (2*,1*2) ;

in 1,2,...,4

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 1,2,...,3

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 1,2,3,4
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 2

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 1,3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]0:$g_{24}$] at (24) ;
[label=[label distance=-3pt]0:$g_{34}$] at (34) ;
[label=[label distance=-3pt]0:$g_{44}$] at (44) ;

[vertex] (f2) at (-2,4)  ;
[vertex] (f3) at (-2,2)  ;
[vertex] (f4) at (-2,0)  ;

[edge] (f2) – (f3) ;
[edge] (f3) – (f4) ;
[edge] (f2) – (21) ;
[edge] (f3) – (31) ;
[edge] (f4) – (41) ;
[edge] (f2) – (11) ;
[edge] (f4) – (31) ;
[edge] (f3) – (41) ;

[label=[label distance=-3pt]180:$f_{1}$] at (f2) ;
[label=[label distance=-3pt]180:$f_{2}$] at (f3) ;
[label=[label distance=-3pt]180:$f_{3}$] at (f4) ;

* $\ind(\tilde{A}_{n, 5})$:

Since $N(g_{12})\subseteq N(g_{21})$ and $N(g_{42})\subseteq N(g_{31})$ in the graphs $\Tilde{A}_{0,5}$ and $\Tilde{A}_{0,5}\setminus g_{21}$, respectively, using <Ref>, we get $\ind(\Tilde{A}_{0,5})\simeq \ind(\Tilde{A}_{0,5}\setminus\{g_{21},g_{31}\})$. Furthermore, the graph $\Tilde{A}_{0,5}\setminus\{g_{21},g_{31}\}$ is a path with 8 vertices, and hence $\ind(\Tilde{A}_{0,5})\simeq \mathbb{S}^2$ by <Ref>. For $n\geq 1$, one can use the same procedures as in <Ref> to obtain the independence complex of $\Tilde{A}_{n,5}$. Therefore, we have the following:

\begin{equation}\label{An'}
\ind(\Tilde{A}_{n,5}) \simeq \begin{cases}
\mathbb{S}^2 & \text{for $n = 0$,} \\
\bigvee_2 \Sigma^2\ind(\Tilde{C}_{n-1,5}) \bigvee \Sigma^2\ind(\Tilde{D}_{n-1,5})  & \text{for $n \geq 1$.}
\end{cases}
\end{equation}

* $\ind(\tilde{B}_{n, 5})$:

Since $N(g_{12})\subseteq N(g_{21})$, $N(g_{42})\subseteq N(g_{31})$ and $N(b_2)\subseteq N(b_5)$ in $\Tilde{B}_{0,5}$, $\Tilde{B}_{0,5}\setminus g_{21}$ and $\Tilde{B}_{0,5}\setminus\{g_{21},g_{31}\}$, respectively, we get that $\ind(\Tilde{B}_{0,5})\simeq \ind(\Tilde{B}_{0,5}\setminus\{g_{21},g_{31},b_5\})$. Note that $b_3$ is a simplicial vertex and $N(b_3)=\{g_{11},b_4\}$ in the graph $\Tilde{B}_{0,5}\setminus\{g_{21},g_{31},b_5\}$. Therefore, $\ind(\Tilde{B}_{0,5}\setminus\{g_{21},g_{31},b_5\})\simeq \Sigma \ind((\Tilde{B}_{0,5}\setminus\{g_{21},g_{31},b_5\})\setminus N[g_{11}])\vee \Sigma \ind((\Tilde{B}_{0,5}\setminus\{g_{21},g_{31},b_5\})\setminus N[{b_4}])$. However, both the graphs  $(\Tilde{B}_{0,5}\setminus\{g_{21},g_{31},b_5\})\setminus N[g_{11}]$ and $(\Tilde{B}_{0,5}\setminus\{g_{21},g_{31},b_5\})\setminus N[b_4]$ are isomorphic to $P_7$. Therefore, using <Ref>, we find that $\ind(\Tilde{B}_{0,5})$ is contractible. For $n\geq 1$, $\ind(\Tilde{B}_{n,5})$, as mentioned below, can be computed using the same argument as in <Ref>.

\begin{equation}
\ind(\Tilde{B}_{n,5}) \simeq \begin{cases}
\ast & \text{for $n = 0$,} \\
\Sigma\ind(\Tilde{C}_{n,5}) \bigvee\Sigma^2\ind(\tilde{\Lambda}_{n-1, 5}) & \text{for $n \geq 1$.}
\end{cases}
\end{equation}

in 4,5,6,7

[vertex] (1) at (2*,3*2) ;

in 4,5,6,7

[vertex] (2) at (2*,2*2) ;

in 3,...,5,6,7

[vertex] (3) at (2*,1*2) ;

in 2,3,...,5,6,7

[vertex] (4) at (2*,0*2) ;

[evaluate=as using +1] in 4,6

[edge] (1) – (1);

[evaluate=as using +1] in 4,6

[edge] (2) – (2);

[evaluate=as using +1] in 3,...,4,6

[edge] (3) – (3);

[evaluate=as using +1] in 2,3,...,4,6

[edge] (4) – (4);

in 4,5,6,7
[edge] (1) – (2);
in 4,5,6,7

[edge] (2) – (3);

in 3,4,5,6,7

[edge] (3) – (4);

[evaluate==int(1+)] in 4

[edge] (1) – (2);

[evaluate==int(1+)] in 4

[edge] (3) – (4);

[evaluate==int(-1)] in 5

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5

[edge] (3) – (4);

[evaluate==int(1+)] in 6

[edge] (2) – (3);

[evaluate==int(-1)] in 4,7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,1) – (12,1);

in 4,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

in 2,3,4,5

[label=[label distance=-3pt]-90:$g_{4\x}$] at (4) ;

[label=[label distance=-3pt]180:$g_{33}$] at (33) ;

in 2,3,4,5

[vertex] (1) at (2*,3*2) ;

in 2,3,4,5

[vertex] (2) at (2*,2*2) ;

in 1,3,4,5

[vertex] (3) at (2*,1*2) ;

in 1,2,3,4,5

[vertex] (4) at (2*,0*2) ;

[evaluate=as using +1] in 2,4

[edge] (2) – (2);

[evaluate=as using +1] in 4

[edge] (3) – (3);

[evaluate=as using +1] in 1,2,4

[edge] (4) – (4);

[edge] (12) – (13) ;
[edge] (14) – (15) ;

in 2,3,4,5

[edge] (1) – (2);

in 3,4,5

[edge] (2) – (3);

in 3,4,5

[edge] (3) – (4);

[edge] (31) – (41);

[evaluate==int(1+)] in 2

[edge] (1) – (2);

[evaluate==int(-1)] in 3

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 4

[edge] (2) – (3);

[evaluate==int(-1)] in 2,5

[edge] (2) – (3);

[dotted, thick,shorten >=4pt, shorten <=4pt] (6,5) – (8,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (6,3) – (8,3);
[dotted, thick,shorten >=4pt, shorten <=4pt] (6,1) – (8,1);

[vertex] (d1) at (-2,4)  ;
[vertex] (d2) at (-2,2)  ;

[edge] (d1) – (d2) ;

[edge] (12) – (33) ;
[edge] (12) – (43) ;
[edge] (22) – (33) ;
[edge] (22) – (43) ;

in 2,3

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

in 1,2,3

[label=[label distance=-3pt]-90:$g_{4\x}$] at (4) ;

[label=[label distance=-4pt]-180:$g_{22}$] at (22) ;
[label=[label distance=-4pt]0:$g_{23}$] at (23) ;
[label=[label distance=-4pt]0:$g_{33}$] at (33) ;
[label=[label distance=-4pt]90:$g_{31}$] at (31) ;
[label=[label distance=-3pt]180:$d_{1}$] at (d1) ;
[label=[label distance=-3pt]180:$d_{2}$] at (d2) ;

* $\ind(\tilde{C}_{n, 5})$:

Using <Ref>, we get $\ind(\Tilde{C}_{0,5})\simeq \ind(\Tilde{C}_{0,5}\setminus\{g_{21},g_{31}\})$. The graph $\Tilde{C}_{0,5}\setminus\{g_{21},g_{31}\}$ is a path on 7 vertices and  hence  $\ind(\Tilde{C}_{0,5})$ is contractible.

Since $\n(g_{11})\subseteq \n(g_{22})$, $\n(g_{14})\subseteq \n(g_{23})$ and $\n(g_{44})\subseteq \n(g_{33})$ in the graphs $\Tilde{C}_{1,5}$ (see <Ref>), $\Tilde{C}_{1, 5}\setminus g_{22}$ and $\Tilde{C}_{1, 5}\setminus \{g_{22},g_{23}\}$, respectively, we get $\ind(\Tilde{C}_{1,5})\simeq \ind(\Tilde{C}_{1, 5}\setminus\{g_{22},g_{23},g_{33}\}.$ Let $\tilde{C}_{1, 5}^1 = \Tilde{C}_{1, 5}\setminus\{g_{22},g_{23},g_{33}\} $.  Observe that $c_1$ is a simlicial vertex in $\tilde{C}_{1, 5}^1$ with $\n(c_1)=\{g_{31},g_{41}\}$. Thus, using <Ref>,
we get $\ind(\Tilde{C}_{1, 5})\simeq \Sigma\ind(\tilde{C}_{1, 5}^1 \setminus\n[g_{31}]) \vee \Sigma\ind(\tilde{C}_{1, 5}^1 \setminus\n[g_{41}])$. Since the graph $\tilde{C}_{1, 5}^1\setminus\n[g_{31}]$ is isomorphic to $P_9$ and $\tilde{C}_{1, 5}^1\setminus\n[g_{41}]$ is isomorphic to cycle graph $\mathcal{C}_{10}$, using <Ref>, we get
$\ind(\Tilde{C}_{1, 5}) \simeq \mathbb{S}^3\vee\mathbb{S}^3$.
For $n\geq 2$, we use the same argument as in <Ref> to get recurrence relations for $\ind(\Tilde{C}_{n,5})$. The homotopy type of $\ind(\Tilde{C}_{n,5})$ is given by the following:

\begin{equation}
\ind(\Tilde{C}_{n,5}) \simeq \begin{cases}
* & \text{for $n = 0$,} \\
\mathbb{S}^3\vee\mathbb{S}^3& \text{for $n = 1$,} \\
\Sigma^{3}\ind(\Tilde{A}_{n-2,5}) \bigvee \Sigma^2\ind(\Tilde{E}_{n-1,5}) \bigvee \Sigma^{3}\ind(\Tilde{F}_{n-2,5}) & \text{for $n \geq 2$.}
\end{cases}
\end{equation}

* $\ind(\tilde{D}_{n, 5})$:

Consider the graph $\Tilde{D}_{0,5}$. Using <Ref>, we obtain $\ind(\Tilde{D}_{0,5})\simeq\ind(\Tilde{D}_{0,5}\setminus\{g_{21},g_{31}\})$. The graph $\Tilde{D}_{0,5}\setminus\{g_{21},g_{31}\}$ is a cycle with 8 vertices. Therefore, using <Ref>, we get that $\ind(\Tilde{D}_{0,5})\simeq \mathbb{S}^2$.

For $n\geq 1$, we proceed in the same way as in <Ref> .  We obtain the following relations :

* $\ind(\Tilde{D}_{n,5})\simeq \ind(\Tilde{D}_{n,5}^4)\simeq \ind(\Tilde{D}_{n,5}^4 \setminus g_{21}) \vee \Sigma^2 \ind(\Tilde{E}_{n-1,5})$,

* $\ind(\Tilde{D}_{n,5}^6) \simeq  \ind(\Tilde{D}_{n,5}^6 \setminus g_{12}) \vee \Sigma\ind(\Tilde{D}_{n,5}^6 \setminus \n[g_{12}])$,

* $\ind(\Tilde{D}_{n,5}^4\setminus g_{21})\simeq \ind(\Tilde{D}_{n,5}^6).$

By combining all the above relations, we get
$\ind(\Tilde{D}_{n,5})\simeq \ind(\Tilde{D}_{n,5}^6 \setminus g_{12}) \vee \Sigma\ind(\Tilde{D}_{n,5}^6 \setminus \n[g_{12}])\vee  \Sigma^2 \ind(\Tilde{E}_{n-1,5})$.
The graph $\Tilde{D}_{n,5}^6$ (see <Ref>) is obtained from the graph $D_{n,5}^6$ (see <Ref>). In the case of $n=1$, we find that the graph $\Tilde{D}_1^6 \setminus \n[g_{12}]$ consists of three components: two edges and one with a path on $4$ vertices. Thus, using <Ref>, we get $\ind(\Tilde{D}_1^6 \setminus \n[g_{12}])$ is contractible. On the other hand, the graph $\Tilde{D}_1^6 \setminus g_{12}$ is isomorphic to a three-component graph, where two are edges and the remaining one is a graph isomorphic to $\Tilde{E}_{0,5}$. Therefore, $\ind(\Tilde{D}_{n,5}^6 \setminus g_{12})\simeq \mathbb{S}^3$. Hence, we get  $\ind(\Tilde{D}_1)\simeq \mathbb{S}^3\vee\mathbb{S}^3$. For $n\geq 2$, the independence complexes of $\Tilde{D}_{n,5}$ follow the same pattern as those in <Ref>. Thus, we have the following:

\begin{equation}
\ind(\Tilde{D}_{n,5}) \simeq \begin{cases}
\mathbb{S}^2 & \text{for $n =0$,} \\
\mathbb{S}^3\vee \mathbb{S}^3 & \text{for $n =1$,} \\
\Sigma^3\ind(\Tilde{\Lambda}_{n-2,5}) \bigvee \Sigma^2\ind(\Tilde{E}_{n-1,5}) \bigvee \Sigma^2 \ind(\Tilde{E}_{n-1,5}) & \text{for $n \geq 2$.}
\end{cases}
\end{equation}

* $\ind(\tilde{\Gamma}_{n, 5})$:

Since $N(g_{11})\subseteq N(g_{22})$ and $N(g_{41})\subseteq N(g_{32})$ in the graphs $\Tilde{\Gamma}_{0,5}$, and $\Tilde{\Gamma}_{0,5}\setminus g_{22}$ using <Ref>, we get $\ind(\Tilde{\Gamma}_{0,5})\simeq \ind(\Tilde{\Gamma}_{0,5}\setminus\{g_{22},g_{32}\})$. The graph $\Tilde{\Gamma}_{0,5}\setminus\{g_{22},g_{32}\}$ is a path with 6 vertices, and hence, using <Ref>, we get $\ind(\Tilde{\Gamma}_{0,5})\simeq \mathbb{S}^1$.

Now, consider $n\geq 1$. Application of the same procedure as that used in <Ref> provides us with a graph $\Tilde{\Gamma}^{5}_{n,5}$ such that $\ind(\Tilde{\Gamma}_{n,5}) \simeq \Sigma^{2}\ind(\Tilde{\Gamma}^{5}_{n,5})$, where the graph $\Tilde{\Gamma}^{5}_{n,5}$ (see <Ref>) is obtained from $\Gamma_{n,5}^5$ (see <Ref>, in the proof of <Ref>).

in 3,...,7

[vertex] (1) at (2*,3*2) ;

in 3,...,7

[vertex] (2) at (2*,2*2) ;

in 3,...,7

[vertex] (3) at (2*,1*2) ;

in 3,...,7

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 3,...,4,6

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);
[edge] (4) – (4);

in 3,4,5,...,7
[edge] (1) – (2);
[edge] (2) – (3);
[edge] (3) – (4);

[evaluate==int(1+)] in 4

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(-1)] in 5

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 6

[edge] (2) – (3);

[evaluate==int(-1)] in 7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,1) – (12,1);

[edge] (13) to [bend right=45] (33);
[edge] (13) to [bend right=45] (43);
[edge] (23) to [bend right=45] (43);

in 3,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;
[label=[label distance=-3pt]270:$g_{4\x}$] at (4) ;

[label=[label distance=-8pt]45:$g_{23}$] at (23) ;
[label=[label distance=-8pt]-45:$g_{33}$] at (33) ;

in 3,5,6,7

[vertex] (1) at (2*,3*2) ;

in 4,5,6,7

[vertex] (2) at (2*,2*2) ;

in 3,4,5,6,7

[vertex] (3) at (2*,1*2) ;

in 3,4,5,6,7

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 3,4,6

[edge] (3) – (3);
[edge] (4) – (4);

[evaluate=as using int(+1)] in 4,6

[edge] (2) – (2);

[edge] (16) – (17);

in 3,4,5,6,7

[edge] (3) – (4);

in 4,5,6,7
[edge] (2) – (3);
in 5,6,7

[edge] (1) – (2);

[edge] (13) to [bend right=45] (33);
[edge] (13) to [bend right=45] (43);

[evaluate==int(1+)] in 4

[edge] (3) – (4);

[evaluate==int(-1)] in 5

[edge] (1) – (2);
[edge] (3) – (4);

[evaluate==int(1+)] in 6

[edge] (2) – (3);

[evaluate==int(-1)] in 4,7

[edge] (2) – (3);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,1) – (12,1);

in 3,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;
[label=[label distance=-3pt]270:$g_{4\x}$] at (4) ;

[label=[label distance=-3pt]-90:$g_{44}$] at (44) ;
[label=[label distance=0pt]90:$g_{33}$] at (33) ;

$\Tilde{\Gamma}_{n,5}^6 \setminus g_{14}$

in 6,7,8,9

[vertex] (1) at (2*,3*2) ;

in 6,7,8,9

[vertex] (2) at (2*,2*2) ;

in 3,4,5,6,7,8,9

[vertex] (3) at (2*,1*2) ;

in 3,4,5,6,7,8,9

[vertex] (4) at (2*,0*2) ;

[evaluate=as using int(+1)] in 6,8

[edge] (1) – (1);

[evaluate=as using int(+1)] in 3,5,6,8

[edge] (3) – (3);

[evaluate=as using int(+1)] in 3,5,6,8

[edge] (4) – (4);

[evaluate=as using int(+1)] in 6,8

[edge] (2) – (2);

in 3,4,5,6,7,8,9

[edge] (3) – (4);

in 6,7,8,9
[edge] (2) – (3);

in 6,7,8,9

[edge] (1) – (2);

[evaluate==int(1+)] in 6

[edge] (1) – (2);
[evaluate==int(1+)] in 6

[edge] (3) – (4);

[evaluate==int(-1)] in 7

[edge] (1) – (2);
[evaluate==int(-1)] in 7

[edge] (3) – (4);

[evaluate==int(1+)] in 8

[edge] (2) – (3);

[evaluate==int(-1)] in 6,9

[edge] (2) – (3);

[dotted, thick,shorten >=4pt, shorten <=4pt] (16,5) – (14,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (16,3) – (14,3);
[dotted, thick,shorten >=4pt, shorten <=4pt] (16,1) – (14,1);

in 6,7

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

in 3,...,5,6,7

[label=[label distance=-3pt]270:$g_{4\x}$] at (4) ;

[label=[label distance=0pt]90:$g_{33}$] at (33) ;
[label=[label distance=0pt]90:$g_{34}$] at (34) ;

[vertex] (b1) at (-2,6)  ;
[vertex] (b2) at (-2,4)  ;
[vertex] (b3) at (-2,2)  ;
[vertex] (b4) at (-2,0)  ;
[vertex] (b31) at (-4,2)  ;
[vertex] (b41) at (-4,0)  ;

[edge] (b1) – (b2) ;
[edge] (b2) – (b3) ;
[edge] (b3) – (b4) ;

[edge] (b31) – (b3) ;
[edge] (b31) – (b2) ;
[edge] (b31) – (b41) ;
[edge] (b41) – (b4) ;

[label=[label distance=-3pt]180:$b_{3}$] at (b1) ;
[label=[label distance=-3pt]180:$b_{4}$] at (b2) ;
[label=[label distance=-9pt]45:$b_{5}$] at (b3) ;
[label=[label distance=-3pt]-90:$b_{6}$] at (b4) ;
[label=[label distance=-3pt]180:$b_{1}$] at (b31) ;
[label=[label distance=-3pt]-90:$b_{2}$] at (b41) ;

If $n =1 $, then $\n(g_{14})\subseteq \n(g_{23})$ and $\n(g_{44})\subseteq \n(g_{33})$ in the graphs $\Tilde{\Gamma}^{5}_{1,5}$ and $\Tilde{\Gamma}^{5}_{1,5}\setminus g_{23}$, respectively. Thus,  $\ind(\Tilde{\Gamma}^{5}_{1,5})\simeq \ind(\Tilde{\Gamma}^{5}_{1,5}\setminus \{g_{23},g_{33}\})$. We can observe that the resultant graph is isomorphic to a cycle on $6$ vertices. Now  <Ref> implies that $\ind(\mathcal{C}_6) \simeq \Sp^1 \vee \Sp^1$ and therefore $\ind(\Tilde{\Gamma}_{1, 5}) \simeq \Sp^3 \vee \Sp^3$.

Now, let  $n\geq 2$ and proceed with the similar methods as in <Ref>. We obtain the following relations:

* $\ind(\Tilde{\Gamma}_{n,5}) \simeq \Sigma^2\left(\ind(\Tilde{\Gamma}_{n,5}^5\setminus g_{23}) \vee \Sigma\ind(\Tilde{A}_{n-2,5}) \right),$

* $\ind(\Tilde{\Gamma}_{n,5}^6) \simeq \Sigma\ind(\Tilde{\Gamma}_{n,5}^7 )\vee \ind(\Tilde{\Gamma}_{n,5}^6 \setminus g_{14})$, where $\Tilde{\Gamma}_{n,5}^6 \coloneq \Tilde{\Gamma}_{n,5}^5\setminus g_{23},$

* $\ind(\Tilde{\Gamma}_{n,5}^6 \setminus g_{14}) \simeq \Sigma \ind((\Tilde{\Gamma}_{n,5}^6 \setminus g_{14})\setminus N[g_{33}]) \vee \Sigma \ind((\Tilde{\Gamma}_{n,5}^6 \setminus g_{14}) \setminus N[g_{43}]),$

* $\ind(\Tilde{\Gamma}_{n,5}^7) \simeq \ind(\Tilde{\Gamma}_{n,5}^8).$

By combining all these relations, we find that

$$\ind(\Tilde{\Gamma}_{n,5})\simeq \Sigma^3(\ind(\Tilde{A}_{n-2,5}))\bigvee \Sigma^3\ind(\Tilde{\Gamma}_{n,5}^8)) \bigvee \Sigma^2(\ind(\Tilde{\Gamma}_{n,5}^6 \setminus g_{14})).$$

For $n=2$, we can observe that the graph $\Tilde{\Gamma}_{2,5}^8$ consists of two connected components, one is a cycle on $4$ vertices and the other one is $\Tilde{B}_{-1}$ (see <Ref>. Since $\n(b_3)\subseteq \n(b_1)\cap\n(b_5)$ in graph $\Tilde{B}_{-1}$, using <Ref>, $\ind(\Tilde{B}_{-1})\simeq \ind(\Tilde{B}_{-1}\setminus\{b_1,b_5\})\simeq \mathbb{S}^1$. This implies that  $\ind(\Tilde{\Gamma}_{2,5}^8)\simeq \mathbb{S}^2$. On the other hand the graphs $(\Tilde{\Gamma}_{2,5}^6 \setminus g_{14})\setminus N[g_{33}]$ and $(\Tilde{\Gamma}_{2,5}^6 \setminus g_{14})\setminus N[g_{43}]$ are isomorphic to the graphs $\Tilde{C}_{0,5}$ and $\Tilde{D}_{0,5}$. Thus, we have $\ind(\Tilde{\Gamma}_{2,5})\simeq \Sigma^3(\ind(\Tilde{A}_{0, 5}))\bigvee \Sigma^3(\mathbb{S}^2) \bigvee \Sigma^3(\ind(\Tilde{C}_{0,5}) \vee \ind(\Tilde{D}_{0,5})).$ Hence, using the above computations of $\ind(\Tilde{A}_{0, 5})$, $\ind(\Tilde{C}_{0, 5})$, and $\ind(\Tilde{D}_{0, 5})$, we get that  $\ind(\Tilde{\Gamma}_{2,5})\simeq \mathbb{S}^5\vee \mathbb{S}^5\vee \mathbb{S}^5$. For $n\geq 3$, one can use the same procedures as in <Ref> to obtain the independence complex of $\Tilde{\Gamma}_{n,5}$. Therefore, we have the following:

\begin{equation}
\small
\ind(\Tilde{\Gamma}_{n,5}) \simeq \begin{cases}
\mathbb{S}^1 & \text{for $n = 0$,} \\
\mathbb{S}^3\vee \mathbb{S}^3 & \text{for $n = 1$,} \\
\mathbb{S}^5\vee \mathbb{S}^5\vee \mathbb{S}^5  & \text{for $n = 2$,} \\
\Sigma^{3}\ind(\Tilde{A}_{n-2,5}) \vee \Sigma^4 \ind(\Tilde{B}_{n-3,5}) \vee \Sigma^{3} \ind(\Tilde{C}_{n-2,5}) \vee \Sigma^{3} \ind(\Tilde{D}_{n-2,5}) & \text{for $n \geq 3$.}
\end{cases}
\end{equation}

* $\ind(\tilde{\Lambda}_{n, 5})$:

Using <Ref>, we get $\ind(\Tilde{\Lambda}_{0,5})\simeq \ind(\Tilde{\Lambda}_{0,5}\setminus\{g_{21},g_{31}\})$. Observe that $l_1$ is a simplicial vertex, and $N(l_1)=\{ g_{11},l_2\}$ in the graph $\Tilde{\Lambda}_{0,5}\setminus\{g_{21},g_{31}\}$. Therefore, <Ref> implies that  $\ind(\Tilde{\Lambda}_{0,5})\simeq \Sigma \ind((\Tilde{\Lambda}_{0,5}\setminus\{g_{21},g_{31}\})\setminus \cn{}{g_{11}})\vee  \Sigma \ind((\Tilde{\Lambda}_{0,5}\setminus\{g_{21},g_{31}\})\setminus \cn{}{l_2})$. The graph $(\Tilde{\Lambda}_{0,5}\setminus\{g_{21},g_{31}\})\setminus \cn{}{g_{11}}$ consists of two connected components; one component is an edge, and the other component is isomorphic to $K_3$. On the other hand, $(\Tilde{\Lambda}_{0,5}\setminus\{g_{21},g_{31}\})\setminus \cn{}{l_2}$ is a path with $6$ vertices. Therefore, by <Ref> and <Ref>, $\ind(\Tilde{\Lambda}_{0,5})\simeq \mathbb{S}^2\vee \mathbb{S}^2$. For $n\geq 1$, $\ind(\Tilde{\Lambda}_{n,5})$ follow the same pattern as those in <Ref>. Thus, we have the following:

\begin{equation}
\small
\ind(\Tilde{\Lambda}_{n,5}) \simeq \begin{cases}
\mathbb{S}^2\vee \mathbb{S}^2 & \text{for $n = 0$,} \\
\Sigma^2\left(\bigvee_2(\ind(\Tilde{A}_{n-1,5})) \bigvee_2 (\ind(\Tilde{C}_{n-1,5}))\bigvee \ind(\Tilde{D}_{n-1,5}) \bigvee \ind(\Tilde{F}_{n-1,5})  \right) & \text{for $n \geq 1$.}
\end{cases}
\end{equation}

* $\ind(\tilde{E}_{n, 5})$:

Using <Ref>, we get $\ind(\Tilde{E}_{0,5})\simeq \ind(\Tilde{E}_{0,5}\setminus\{g_{21},g_{31}\})$. Observe that the graph $\Tilde{E}_{0,5}\setminus\{g_{21},g_{31}\}$ is a cycle on $7$ vertices. Thus,
$\ind(\Tilde{E}_{0,5})\simeq \mathbb{S}^1$.

For $n\geq 1$, we proceed in the same way as in <Ref>. Using <Ref>, we get $\ind(\Tilde{E}_{n,5}) \simeq \Sigma^2 \left(\vee_2 \ind(\Tilde{C}_{n,5}^5)\right) \vee \Sigma\ind(\Tilde{A}_{n-1,5}) \vee \ind(\Tilde{\Gamma}_{n,5}),$ where the graph $\Tilde{C}_{n,5}^5$ (see <Ref>) is obtained from the graph $C_{n,5}^5$ (see <Ref>).
In case of $n=1$,  $\n(g_{44})\subseteq \n(g_{33})$ in $\Tilde{C}_1^5$ and therefore  $\ind(\Tilde{C}_1^5)\simeq \ind(\Tilde{C}_1^5\setminus g_{33})$. Since the graph $\Tilde{C}_{n,5}^5\setminus g_{33}$  is a graph on
$6$ vertices, $\ind(\Tilde{C}_1^5)\simeq \mathbb{S}^1$.
Now $\ind(\Tilde{A}_{0})\simeq \mathbb{S}^2$ and $\ind(\Tilde{\Gamma}_{1,5})\simeq \mathbb{S}^3\vee \mathbb{S}^3$ imply that $\ind(\Tilde{E}_1) \simeq \bigvee_5{\mathbb{S}^3}$. For $n\geq 2$, one can use the same arguments as those in <Ref> to get  $\ind(\Tilde{C}_{n,5}^5) \simeq \Sigma\ind(\Tilde{A}_{n-2,5}) \vee \Sigma\ind(\Tilde{F}_{n-2,5})$. Thus, combining all of these, we have the following:

\begin{equation}
\small
\ind(\Tilde{E}_{n,5}) \simeq \begin{cases}
\mathbb{S}^1 & \text{for $n = 0$,} \\
\bigvee_5\mathbb{S}^3 & \text{for $n = 1$,} \\
\Sigma^3\left(\bigvee_2 \ind(\Tilde{A}_{n-2,5}) \bigvee_2 \ind(\Tilde{F}_{n-2,5})\right) \vee \Sigma\ind(\Tilde{A}_{n-1,5}) \vee \ind(\Gamma'_{n,5})
& \text{for $n \geq 2$.}
\end{cases}
\end{equation}

* $\ind(\tilde{F}_{n, 5})$:

From <Ref>,  $\ind(\Tilde{F}_{0,5})\simeq \ind(\Tilde{F}_{0,5}\setminus\{g_{21},g_{31}\})$. Since $f_3$ is a simplicial vertex in $\Tilde{F}_{0,5}\setminus\{g_{21},g_{31}\}$ with $N(f_3)=\{g_{41},f_2\}$, $\ind(\Tilde{F}_{0,5})\simeq \Sigma \ind((\Tilde{F}_{0,5}\setminus\{g_{21},g_{31}\})\setminus\cn{}{g_{41}})\vee \Sigma \ind((\Tilde{F}_{0,5}\setminus\{g_{21},g_{31}\})\setminus\cn{}{f_2}) $. Both the graphs $(\Tilde{F}_{0,5}\setminus\{g_{21},g_{31}\})\setminus\cn{}{g_{41}}$ and $(\Tilde{F}_{0,5}\setminus\{g_{21},g_{31}\})\setminus\cn{}{f_2}$ are isomorphic to $P_5$. Hence,  $\ind(\Tilde{F}_{0,5})\simeq \mathbb{S}^2\vee\mathbb{S}^2$. For $n\geq 1$, the relations for $\ind(\Tilde{F}_{n,5})$ follows from a similar argument as those in <Ref>. Thus,

\begin{equation}\label{lambda_{n,5}'}
\small
\ind(\Tilde{F}_{n,5}) \simeq \begin{cases}
\mathbb{S}^2\vee\mathbb{S}^2 & \text{for $n =0$,} \\
\Sigma\ind(\Tilde{B}_{n-1,5}) \bigvee
\Sigma^2\ind(\Tilde{C}_{n-1,5}) \bigvee
\Sigma^2\ind(\Tilde{C}_{n-1,5}) \bigvee
\Sigma^2\ind(\Tilde{D}_{n-1,5})
& \text{for $n \geq 1$.}
\end{cases}
\end{equation}

For any graph $\Tilde{G}_{n,5}\in \set{\Tilde{\Gamma}_{n,5}, \Tilde{\Lambda}_{n,5}, \Tilde{A}_{n,5}, \Tilde{B}_{n,5}, \Tilde{C}_{n,5}, \Tilde{D}_{n,5}, \Tilde{E}_{n,5}, \Tilde{F}_{n,5} : n\geq 0 }$, $\ind(\Tilde{G}_{n,5})$ is homotopy equivalent to a wedge of spheres.

The result follows from An'lambda_n,5', and the induction on $n$.

§.§ Homotopy type of $\M(P_n \times P_5)$

In this section, we conclude the homotopy type of $\M(P_n \times P_5)$. We also determine the maximum and minimum dimensions of the spheres occurring in the homotopy type of the independence complexes of the graphs in
$\{\Gamma_{n,5},\Lambda_{n,5},A_{n,5},B_{n,5}$, $C_{n,5},D_{n,5},E_{n,5}, F_{n,5} \}$.

For $n\geq 2$,  $\M(P_{n}\times P_5)$ is homotopy equivalent to a wedge of spheres.

Case 1: If $n$ is even, i.e. $n=2k$ for some positive integer $k$, then  $P_n\times P_5$ comprises of two isomorphic connected components, whose line graph is isomorphic to $\Gamma_{k-1,5}$. Thus,
$$M(P_n\times P_5)\simeq \ind(\Gamma_{k-1,5})\ast \ind(\Gamma_{k-1,5}).$$
Now, the result follows from <Ref>.

Case 2: If $n$ is odd, i.e. $n=2k+1$ for some positive integer $k$, then  $P_n\times P_5$ comprises of two non-isomorphic connected components with corresponding line graphs isomorphic to $\Lambda_{k-1,5}$ and $\Tilde{\Gamma}_{k-1,5}$. Thus,
$$M(P_n\times P_5)\simeq \ind(\Lambda_{k-1,5})\ast \ind(\Tilde{\Gamma}_{k-1,5}).$$

Result follows from <Ref> and <Ref>.

For a graph $G$ whose independence complex is homotopy equivalent to a wedge of spheres, let $\dimax{G}$ and $\dimin{G}$ denote the maximum and minimum dimension of spheres appearing in the homotopy type of $\ind(G)$.

For $n\geq 0$, the maximum and minimum dimension of the spheres in the homotopy type of the independence complexes of graphs in $\{\Gamma_{n, 5}, \Lambda_{n, 5}, A_{n, 5}, B_{n, 5}, C_{n, 5}, $  $D_{n, 5}, E_{n, 5}, F_{n, 5}\}$  are given in <Ref> and <Ref> respectively.

hline2-Z = 0.5pt, vline2-Z = 0.5pt,hline1 = 1-Z0.5pt,vline1 = 1-Z0.5pt,colspec=cccccc,r,cell12 = c=5c, cell11 = r=2c
$n \equiv$     $d_{\mathit{max}}$ of independence complexes of graphs

$\Gamma_{n,5},E_{n,5}$     $\Lambda_{n,5}$     $A_{n,5}$, $F_{n,5}$     $B_{n,5}$     $C_{n,5}$, $D_{n,5}$

$0 \mod{4}$     $7(\frac{n}{4}) (n \neq 0)$     $7(\frac{n}{4})+1$     $7(\frac{n}{4})+1$     $7(\frac{n}{4})+2$     $7(\frac{n}{4})+1$

$1 \mod{4}$     $7(\frac{n-1}{4})+2$     $7(\frac{n-1}{4})+3$     $7(\frac{n-1}{4})+3$     $7(\frac{n-1}{4})+3$     $7(\frac{n-1}{4})+2$

$2 \mod{4}$     $7(\frac{n-2}{4})+4$     $7(\frac{n-2}{4})+5$     $7(\frac{n-2}{4})+4$     $7(\frac{n-2}{4})+5$     $7(\frac{n-2}{4})+4$

$3 \mod{4}$     $7(\frac{n-3}{4})+6$     $7(\frac{n-3}{4})+6$     $7(\frac{n-3}{4})+6$     $7(\frac{n+1}{4})$     $7(\frac{n-3}{4})+6$

hline2-Z = 0.5pt, vline2-Z = 0.5pt,hline1 = 1-Z0.5pt,vline1 = 1-Z0.5pt,colspec=cccccc,r,cell12 = c=5c, cell11 = r=2c
$n \equiv$     $d_{\mathit{min}}$ of independence complexes of graphs

$\Gamma_{n,5}$, $E_{n,5}$     $\Lambda_{n,5},A_{n,5}$      $F_{n,5}$, $D_{n,5}$     $B_{n,5}$     $C_{n,5}$

$0 \mod{5}$     $8(\frac{n}{5}) (n \neq 0)$         $8(\frac{n}{5})+1$     $8(\frac{n}{5})+1$     $8(\frac{n}{5})+1$     $8(\frac{n}{5})+1$

$1 \mod{5}$     $8(\frac{n-1}{5})+2$     $8(\frac{n-1}{5})+3$     $8(\frac{n-1}{5})+2$     $8(\frac{n-1}{5})+3$     $8(\frac{n-1}{5})+2$

$2 \mod{5}$     $8(\frac{n-2}{5})+4$     $8(\frac{n-2}{5})+4$     $8(\frac{n-2}{5})+4$     $8(\frac{n-2}{5})+5$     $8(\frac{n-2}{5})+4$

$3 \mod{5}$     $8(\frac{n-3}{5})+5$     $8(\frac{n-3}{5})+6$     $8(\frac{n-3}{5})+6$     $8(\frac{n-3}{5})+6$     $8(\frac{n-3}{5})+5$

$4 \mod{5}$     $8(\frac{n-4}{5})+7$     $8(\frac{n-4}{5})+7$     $8(\frac{n-4}{5})+7$     $8(\frac{n-4}{5})+8$     $8(\frac{n-4}{5})+7$

We prove the statement by using induction on the graphs in $\{\Gamma_{n,5},\Lambda_{n,5}, A_{n,5},B_{n,5}$, $C_{n,5},D_{n,5},E_{n,5},F_{n,5}\}$ for $n\geq 0$. The statement holds for $n=0$ from homA basehomL base and $\ind(\Lambda_{0, 5})$.  Assume that the statement holds for all integers less than $n$. For $\Gamma_{n,5}$, <Ref> implies that the statement holds for $n=1,2$ as well. Further, for $n\geq 3$, we have

\begin{equation}{\label{eq: gamma-n-5 rec}}
\ind(\Gamma_{n,5}) \simeq \Sigma^{3}\ind(A_{n-2,5}) \vee \Sigma^4 \ind(B_{n-3,5}) \vee \Sigma^{3} \ind(C_{n-2,5}) \vee \Sigma^{3} \ind(D_{n-2,5}).
\end{equation}

Therefore, $\dimax{\Gamma_{n,5}}$ is given by

\begin{equation*}
\begin{split}
\max\{ & 3+\dimax{A_{n-2,5}},4+\dimax{B_{n-3,5}},\\ & 3+\dimax{C_{n-2,5}}, 3+\dimax{D_{n-2,5}} \}.
\end{split}
\end{equation*}

Let $n \equiv 0 \mod{4}$. Then $n-2$ and $n-3$ are congruent to $2$ and $1$ modulo $4$, respectively. By the induction hypothesis, we have

\begin{equation*}
\begin{split}
\dimax{\Gamma_{n,5}} = &
\max\{ 3+(\sfrac{7(n-4)}{4}+4),4+(\sfrac{7(n-4)}{4}+3),\\ & \phantom{\max\{}3+(\sfrac{7(n-4)}{4}+4), 3+(\sfrac{7(n-4)}{4}+4) \} \\
= & 7\left(\frac{n}{4}\right).
\end{split}
\end{equation*}

For $n \equiv 1 \mod{4}$, $n-2$ and $n-3$ are congruent to $3$ and $2$ modulo $4$, respectively. Thus by induction hypothesis, we have

\begin{equation*}
\begin{split}
\dimax{\Gamma_{n,5}} = &
\max\{ 3+(\sfrac{7(n-5)}{4}+6),4+(\sfrac{7(n-5)}{4}+5),\\ & \phantom{\max\{}3+(\sfrac{7(n-5)}{4}+6), 3+(\sfrac{7(n-4)}{4}+6) \} \\
= & 7\left(\frac{n-5}{4}\right) + 9 = 7\left(\frac{n-1}{4}\right) + 2.
\end{split}
\end{equation*}

For $n \equiv 2 \mod{4}$, $n-2$ and $n-3$ are congruent to $0$ and $3$ modulo $4$, respectively. By induction hypothesis, we have

\begin{equation*}
\begin{split}
\dimax{\Gamma_{n,5}} = &
\max\{ 3+(\sfrac{7(n-2)}{4}+1),4+(\sfrac{7(n-2)}{4}),\\ & \phantom{\max\{}3+(\sfrac{7(n-2)}{4}+1), 3+(\sfrac{7(n-2)}{4}+1) \} \\
= & 7\left(\frac{n-2}{4}\right)+4.
\end{split}
\end{equation*}

For $n \equiv 3 \mod{4}$, $n-2$ and $n-3$ are congruent to $1$ and $0$ modulo $4$, respectively. By induction hypothesis, we have

\begin{equation*}
\begin{split}
\dimax{\Gamma_{n,5}} = &
\max\{ 3+(\sfrac{7(n-3)}{4}+3),4+(\sfrac{7(n-3)}{4})+2,\\ & \phantom{\max\{}3+(\sfrac{7(n-3)}{4}+2), 3+(\sfrac{7(n-3)}{4}+2) \} \\
= & 7\left(\frac{n-3}{4}\right)+6.
\end{split}
\end{equation*}

On the other hand, $\dimin{\Gamma_{n,5}}$ is given by

\begin{equation*}
\begin{split}
\min\{ & 3+\dimin{A_{n-2,5}},4+\dimin{B_{n-3,5}},\\ & 3+\dimin{C_{n-2,5}}, 3+\dimin{D_{n-2,5}} \}.
\end{split}
\end{equation*}

For $n\equiv 0 \mod{5}$, $n-2$ and $n-3$ are congruent to $3$ and $2$ modulo 5, respectively. Therefore, using induction hypothesis, we have

\begin{equation*}
\begin{split}
\dimin{\Gamma_{n,5}} = &
\min\{ 3+(\sfrac{8(n-5)}{5}+6),4+(\sfrac{8(n-5)}{5}+5),\\ & \phantom{\max\{} 3+(\sfrac{8(n-5)}{5}+5), 3+(\sfrac{8(n-5)}{5}+6) \} \\
= & 8\left(\frac{n-5}{5}\right)+8 = 8\left(\frac{n}{5}\right).
\end{split}
\end{equation*}

For $n\equiv 1 \mod{5}$, $n-2$ and $n-3$ are congruent to $4$ and $3$ modulo 5, respectively. Therefore, we have

\begin{equation*}
\begin{split}
\dimin{\Gamma_{n,5}} = &
\min\{ 3+(\sfrac{8(n-6)}{5}+7),4+(\sfrac{8(n-6)}{5}+6),\\ & \phantom{\max\{} 3+(\sfrac{8(n-6)}{5}+7), 3+(\sfrac{8(n-6)}{5}+7) \} \\
= & 8\left(\frac{n-6}{5}\right)+10 = 8\left(\frac{n-1}{5}\right)+2.
\end{split}
\end{equation*}

For $n\equiv 2 \mod{5}$, $n-2$ and $n-3$ are congruent to $0$ and $4$ modulo 5, respectively. Therefore, we have

\begin{equation*}
\begin{split}
\dimin{\Gamma_{n,5}} = &
\min\{ 3+(\sfrac{8(n-2)}{5}+1),4+(\sfrac{8(n-7)}{5}+8),\\ & \phantom{\max\{} 3+(\sfrac{8(n-2)}{5}+1), 3+(\sfrac{8(n-2)}{5}+1) \} \\
= & 8\left(\frac{n-2}{5}\right)+4.
\end{split}
\end{equation*}

For $n\equiv 3 \mod{5}$, $n-2$ and $n-3$ are congruent to $1$ and $0$ modulo 5, respectively. Therefore, we have

\begin{equation*}
\begin{split}
\dimin{\Gamma_{n,5}} = &
\min\{ 3+(\sfrac{8(n-3)}{5}+3),4+(\sfrac{8(n-3)}{5}+1),\\ & \phantom{\max\{} 3+(\sfrac{8(n-3)}{5}+2), 3+(\sfrac{8(n-3)}{5}+2) \} \\
= & 8\left(\frac{n-3}{5}\right)+5.
\end{split}
\end{equation*}

For $n\equiv 4 \mod{5}$, $n-2$ and $n-3$ are congruent to $2$ and $1$ modulo 5, respectively. Therefore, we have

\begin{equation*}
\begin{split}
\dimin{\Gamma_{n,5}} = &
\min\{ 3+(\sfrac{8(n-4)}{5}+4),4+(\sfrac{8(n-4)}{5}+3),\\ & \phantom{\max\{} 3+(\sfrac{8(n-4)}{5}+4), 3+(\sfrac{8(n-4)}{5}+4) \} \\
= & 8\left(\frac{n-4}{5}\right)+7.
\end{split}
\end{equation*}

The proof follows similarly for the maximum and minimum dimension of the wedge in independence complexes of $\Lambda_{n,5},A_{n,5},B_{n,5},C_{n,5},D_{n,5},E_{n,5}$ and $F_{n,5}$.

For $n\geq 0$, the maximum and minimum dimension of the spheres in the homotopy type of the independence complexes of graphs in $\{\tilde{\Gamma}_{n, 5}, \tilde{\Lambda}_{n, 5}, \tilde{A}_{n, 5}, \tilde{B}_{n, 5}, \tilde{C}_{n, 5}, $  $\tilde{D}_{n, 5}, \tilde{E}_{n, 5}, \tilde{F}_{n, 5}\}$  are given in <Ref> and <Ref> respectively.

hline2-Z = 0.5pt, vline2-Z = 0.5pt,hline1 = 1-Z0.5pt,vline1 = 1-Z0.5pt,colspec=cccccc,r,cell12 = c=5c, cell11 = r=2c
$n \equiv$     $d_{\mathit{max}}$ of independence complexes of graphs

$\Tilde{\Gamma}_{n,5},\Tilde{E}_{n,5}$     $\Tilde{\Lambda}_{n,5}$     $\Tilde{A}_{n,5}$, $\Tilde{F}_{n,5}$     $\Tilde{B}_{n,5}$     $\Tilde{C}_{n,5}$, $\Tilde{D}_{n,5}$

$0 \mod{4}$     $7(\frac{n}{4})+1$ $(n \neq 0)$    $7(\frac{n}{4})+2$     $7(\frac{n}{4})+2$     $7(\frac{n}{4})+3$     $7(\frac{n}{4})+2$

$1 \mod{4}$     $7(\frac{n-1}{4})+3$     $7(\frac{n-1}{4})+4$     $7(\frac{n-1}{4})+4$     $7(\frac{n-1}{4})+4$     $7(\frac{n-1}{4})+3$

$2 \mod{4}$     $7(\frac{n-2}{4})+5$     $7(\frac{n-2}{4})+6$     $7(\frac{n-2}{4})+5$     $7(\frac{n-2}{4})+6$     $7(\frac{n-2}{4})+5$

$3 \mod{4}$     $7(\frac{n+1}{4})$     $7(\frac{n+1}{4})$     $7(\frac{n+1}{4})$     $7(\frac{n+1}{4})+1$     $7(\frac{n+1}{4})$

hline2-Z = 0.5pt, vline2-Z = 0.5pt,hline1 = 1-Z0.5pt,vline1 = 1-Z0.5pt,colspec=cccccc,r,cell12 = c=5c, cell11 = r=2c
$n \equiv$     $d_{\mathit{min}}$ of independence complexes of graphs

$\tilde{\Gamma}_{n,5}$, $\tilde{E}_{n,5}$     $\tilde{\Lambda}_{n,5},\tilde{A}_{n,5}$      $\tilde{F}_{n,5}$     $\tilde{B}_{n,5}$     $\tilde{C}_{n,5}$, $\tilde{D}_{n,5}$

$0 \mod{5}$     $8(\frac{n}{5})+2$ $(n \neq 0)$       $8(\frac{n}{5})+2$     $8(\frac{n}{5})+2$     $8(\frac{n}{5})+3$     $8(\frac{n}{5})+2$

$1 \mod{5}$     $8(\frac{n-1}{5})+3$     $8(\frac{n-1}{5})+4$     $8(\frac{n-1}{5})+4$     $8(\frac{n-1}{5})+4$     $8(\frac{n-1}{5})+3$

$2 \mod{5}$     $8(\frac{n-2}{5})+5$     $8(\frac{n-2}{5})+5$     $8(\frac{n-2}{5})+5$     $8(\frac{n-2}{5})+6$     $8(\frac{n-2}{5})+5$

$3 \mod{5}$     $8(\frac{n-3}{5})+6$     $8(\frac{n-3}{5})+7$     $8(\frac{n-3}{5})+7$     $8(\frac{n-3}{5})+7$     $8(\frac{n-3}{5})+7$

$4 \mod{5}$     $8(\frac{n+1}{5})$     $8(\frac{n+1}{5})+1$     $8(\frac{n+1}{5})$     $8(\frac{n+1}{5})+1$     $8(\frac{n+1}{5})$

The proof is verbatim to the proof of <Ref>.

The maximum and minimum dimension of spheres in the homotopy type of $\M(P_n \times P_5)$ as given in <Ref> can be  concluded from tab: max Pn5tab: min 2nd Pn5.

§ MATCHING COMPLEXES OF  PN X P4

In this section, we aim to determine the homotopy type of the matching complex of the categorical product $P_n\times P_4$.
The categorical product $P_n\times P_4$ is characterized by being disconnected, comprising precisely two connected components. Notably, regardless of whether
$n$ is odd or even, these components are isomorphic.
For $n \geq 0$, we denote the connected components of the line graphs of $P_{2n+2}\times P_4$ and $P_{2n+3}\times P_4$ by $\Gamma_{n,4}$ (see <Ref>) and $\Lambda_{n,4}$ (see <Ref>), respectively. Therefore, we have the following:

\begin{equation*}
M(P_{2n+2}\times P_4) \simeq \ind(\Gamma_{n,4}) * \ind(\Gamma_{n,5}),
\end{equation*}

\begin{equation*}
M(P_{2n+3}\times P_4) \simeq \ind(\Lambda_{n,5}) * \ind(\Lambda_{n,5}).
\end{equation*}

During the computation of the homotopy type of the independence complexes of graphs $\Gamma_{n,4}$ and $\Lambda_{n,4}$ for $n\geq 0$, we encounter several intermediary graphs denoted as $A_{n,4}, B_{n,4}, C_{n,4}$, and $D_{n,4}$. We illustrate that the independence complexes of $\Gamma_{n,4}$ and $\Lambda_{n,4}$ can be represented as the wedge of suspensions of these intermediary graphs. By employing induction on $n$, we establish that the independence complexes of $\Gamma_{n,4}$, $\Lambda_{n,4}$, and the intermediary graphs are homotopy equivalent to a wedge of spheres. The graphs $A_{n,4}, B_{n,4}, C_{n,4}$, and $D_{n,4}$ are shown in <Ref>, respectively. In the subsequent subsections, we proceed to determine the homotopy type of the independence complexes of
$\Gamma_{n,4}$, $\Lambda_{n,4}$ and the aforementioned intermediary graphs.

§.§ Graph $\Gamma_{n,4}$

For $n\geq 0$, we define the graph $\Gamma_{n,4}$ as follows:

\begin{equation*}
\begin{split}
V(\Gamma_{n,4}) = & \set{g_{ij} : i=1,2,3 \  \text{ and }j = {1,2,\dots,2n{+}1}},\\
E(\Gamma_{n,4}) = & \ \set{g_{ij}g_{i+1\,j} : i=1,2 \text{ and } j=1,2,\dots,2n{+}1} \\
& \cup \set{g_{ij}g_{i\,j+1} : i=1,2,3 \text{ and } j=1,2,\dots,2n} \\
& \cup \set{g_{ij}g_{i+1\,j+1}, g_{ij+1}g_{i+1\,j} :  i=1,2, \; j=1,2,\dots,2n  \text{ and } i+j\equiv 1 \mod{2}}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

in 1,2,...,8

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);

in 1,2,3,4,5,...,8
[edge] (1) – (2);
[edge] (2) – (3);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

[evaluate==int(1+)] in 1,3,6

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4,7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

(31) – node[below=6pt] $1$(33);
(33) – node[below=6pt] $2$(35);
(36) – node[below=6pt] $n$(38);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]180:$g_{21}$] at (21) ;
[label=[label distance=-3pt]180:$g_{31}$] at (31) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;
[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;
[label=[label distance=-3pt]0:$g_{3\,{2n+1}}$] at (38) ;

[] at (10,-1) (b) $\Gamma_{n,4}$ for $n\geq 1$;

[shift = (-5,0)]
in 1,2,3
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1,2
[edge] (x1) – (x1) ;

in 1,2,3

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[] at (0,-1) (a) $\Gamma_{0,4}$;

For $n\geq 0$,

\begin{equation*}
\ind(\Gamma_{n,4}) \simeq \begin{cases}
\mathbb{S}^0 & \text{if $n =0$,} \\
\mathbb{S}^1\vee \mathbb{S}^1  & \text{if $n =1$,} \\
\Sigma \ind(A_{n-1,4}) \vee \Sigma^{2} \ind(\Lambda_{n-2,4})
& \text{if $n \geq 2$.}
\end{cases}
\end{equation*}

Since $\Gamma_{0,4}$ is a path with three vertices (see <Ref>), $\ind(\Gamma_{0,4})\simeq \mathbb{S}^0$. Now, consider $n\geq 1$. Since $N(g_{11}) \subseteq N(g_{22})$ in the graph $\Gamma_{n,4}$, we have $\ind(\Gamma_{n,4}) \simeq \ind(\Gamma_{n,4} \setminus g_{22})$. Let $\Gamma_{n,4}^1 \coloneq\Gamma_{n,4} \setminus g_{22}$. Since $g_{31}$ is a simplicial vertex in $\Gamma_{n,4}^1$ with $N(g_{31})=\{g_{21},g_{32}\}$,  <Ref> implies that $ \ind(\Gamma_{n,4}^1) \simeq \Sigma \ind(\Gamma_{n,4}^1 \setminus \n[g_{21}]) \vee \Sigma \ind(\Gamma_{n,4}^1 \setminus N[g_{32}])$. Observe that the graph $\Gamma_{n,4}^1 \setminus \n[g_{21}]$ is isomorphic to $A_{n-1,4}$ (see <Ref>). Therefore, $\ind(\Gamma_{n,4}^1)\simeq \Sigma \ind(A_{n-1,4}) \vee \Sigma \ind(\Gamma_{n,4}^1 \setminus N[g_{32}])$. In case of $n=1$, the graph $\Gamma_{1,4}^1 \setminus N[g_{32}]$ is isomorphic to the graph $A_{0,4}$ (see <Ref>). Therefore, $\ind(\Gamma_{1,4}^1)\simeq \Sigma \ind(A_{0,4}) \vee \Sigma \ind(A_{0,4})$. This implies that $\ind(\Gamma_{1,4})\simeq \mathbb{S}^1\vee \mathbb{S}^1$.

Now, consider $n\geq 2$. Since $g_{11}$ is a simplicial vertex and $g_{12}$ is the only neighbor of $g_{11}$ in the graph $\Gamma_{n,4}^1 \setminus N[g_{32}]$, we get  $ \ind(\Gamma_{n,4}^1 \setminus N[g_{32}])\simeq \Sigma\ind(\Gamma_{n,4}^1 \setminus N[g_{32},g_{12}])$. Observe that  the graph $\Gamma_{n,4}^1 \setminus N[g_{32},g_{12}]$ is isomorphic to $\Lambda_{n-2,4}$ (<Ref>) and therefore  $\ind(\Gamma_{n,4}^1 \setminus N[g_{32}]) \simeq \Sigma\ind(\Lambda_{n-2,4})$. Thus, $\ind(\Gamma_{n,4}) \simeq \Sigma \ind(A_{n-1,4}) \vee \Sigma^{2} \ind(\Lambda_{n-2,4})$.

§.§ Graph $\Lambda_{n,4}$

For $n\geq 0$, we define the graph $\Lambda_{n,4}$ as follows:

\begin{equation*}
\begin{split}
V(\Lambda_{n,4}) = & V(\Gamma_{n,4})\cup\{g'_1,g'_2,g'_3\},\\
E(\Lambda_{n,4}) = & E(\Gamma_{n,4})\cup \{g'_1g'_2, g'_1g_{11}, g'_1g_{21}, g'_2g'_3, g'_2g_{21}, g'_2g_{11}, g'_3g_{31} \}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

in 1,2,...,8

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);

in 1,2,3,4,5,...,8
[edge] (1) – (2);
[edge] (2) – (3);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

[evaluate==int(1+)] in 1,3,6

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4,7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

(31) – node[below=6pt] $1$(33);
(33) – node[below=6pt] $2$(35);
(36) – node[below=6pt] $n$(38);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;
[label=[label distance=-3pt]0:$g_{3\,{2n+1}}$] at (38) ;

[vertex] (e1) at (-2,6)  ;
[vertex] (e2) at (-2,4)  ;
[vertex] (e3) at (-2,2)  ;

[edge] (e1) – (e2) ;
[edge] (e2) – (e3) ;
[edge] (e1) – (11) ;
[edge] (e2) – (21) ;
[edge] (e3) – (31) ;
[edge] (e1) – (21) ;
[edge] (e2) – (11) ;

[label=[label distance=-3pt]180:$g'_{1}$] at (e1) ;
[label=[label distance=-3pt]180:$g'_{2}$] at (e2) ;
[label=[label distance=-3pt]180:$g'_{3}$] at (e3) ;
[] at (6,-1) (b) $\Lambda_{n,4}$ for $n\geq 1$;

[shift = (-5,0)]
in 1,2,3
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1,2
[edge] (x1) – (x1) ;

[vertex] (e1) at (-2,6)  ;
[vertex] (e2) at (-2,4)  ;
[vertex] (e3) at (-2,2)  ;

[edge] (e1) – (e2) ;
[edge] (e2) – (e3) ;
[edge] (e1) – (x11) ;
[edge] (e2) – (x21) ;
[edge] (e3) – (x31) ;
[edge] (e1) – (x21) ;
[edge] (e2) – (x11) ;

in 1,2,3

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[label=[label distance=-3pt]180:$g'_{1}$] at (e1) ;
[label=[label distance=-3pt]180:$g'_{2}$] at (e2) ;
[label=[label distance=-3pt]180:$g'_{3}$] at (e3) ;
[] at (-1,-1) (a) $\Lambda_{0,4}$;

[shift = (21,0)]
in 3,...,7

[vertex] (1) at (2*,3*2) ;

in 2,...,7

[vertex] (2) at (2*,2*2) ;

in 1,2,...,7

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 3,5,6

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);

[edge] (31) – (32);
[edge] (32) – (33);
[edge] (22) – (23);
in 3,4,5,6,7
[edge] (1) – (2);
[edge] (2) – (3);

[edge] (22) – (32);
[evaluate==int(1+)] in 6

[edge] (1) – (2);

[evaluate==int(-1)] in 3,7

[edge] (1) – (2);

[evaluate==int(1+)] in 3,5

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4,6

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[vertex] (e'3) at (-0.33,2)  ;
[label=[label distance=-
5pt]-90:$g'_{3}$] at (e'3) ;

[edge] (e'3) – (31) ;

[dotted, thick,shorten >=4pt, shorten <=4pt] (8,5) – (10,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (8,3) – (10,3);

in 3,...,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (17) ;
[label=[label distance=-3pt]90:$\cdots$] at (15) ;

[label=[label distance=-3pt]270:$g_{3\,{2n+1}}$] at (37) ;
[label=[label distance=-3pt]90:$g_{22}$] at (22) ;
[label=[label distance=-3pt]270:$g_{32}$] at (32) ;
[label=[label distance=-3pt]270:$g_{31}$] at (31) ;
[] at (8,-1) (c) $\Lambda_{n,4}^1$;

For $n\geq 0$,

\begin{equation*}
\ind(\Lambda_{n,4}) \simeq \begin{cases}
\mathbb{S}^1 & \text{if $n =0$,} \\
\Sigma \ind(B_{n-1,4}) \vee \Sigma^{2}\ind(\Gamma_{n-1,4})
& \text{if $n \geq 1$.}
\end{cases}
\end{equation*}

Since $N(g'_3)\subseteq N(g_{21})$ in the graph $\Lambda_{n,4}$ (see <Ref>), $\ind(\Lambda_{n,4}) \simeq \ind(\Lambda_{n,4} \setminus g_{21})$. If  $n=0$, then  $N(g_{31})\subseteq N(g'_2)$ in $\Lambda_{0,4} \setminus g_{21}$. Hence $\ind(\Lambda_{0,4}) \simeq \ind(\Lambda_{0,4} \setminus \{g_{21},g'_2\})$. Now, the graph $\Lambda_{0,4} \setminus \{g_{21},g'_2\}$ is the disjoint union of two edges $g'_{1}g_{11}$ and $g'_{3}g_{31}$. Therefore, $\ind(\Lambda_{0,4}) \simeq \mathbb{S}^1$.

Now, let  $n\geq 1$. Since $g'_{1}$ is a simplicial vertex in $\Lambda_{n,4}\setminus g_{21}$, with $N(g'_1)=\{ g'_{2},g_{11}\}$, using <Ref>, we get

\begin{equation*}
\begin{split}
\ind(\Lambda_{n,4} \setminus g_{21}) \simeq \Sigma \ind((\Lambda_{n,4} \setminus g_{21})\setminus N[g'_{2}]) \vee \Sigma \ind((\Lambda_{n,4} \setminus g_{21}) \setminus N[g_{11}]).
\end{split}
\end{equation*}

However, the graph $(\Lambda_{n,4} \setminus g_{21})\setminus N[g'_{2}]$ is isomorphic to $B_{n-1,4}$ (see <Ref>). Therefore,  $\ind(\Lambda_{n,4} \setminus g_{21})\simeq \Sigma \ind(B_{n-1,4}) \vee \Sigma \ind((\Lambda_{n,4}\setminus g_{21}) \setminus  N[g_{11}]).$  Let $\Lambda_{n,4}^1\coloneq (\Lambda_{n,4}\setminus g_{21}) \setminus N[g_{11}]$.
Further, since  $g'_{3}$ is a simplicial vertex in the graph $\Lambda_{n,4}^1$ with $N(g'_3)=\{g_{31}\}$,
$\ind(\Lambda_{n,4}^1)\simeq \Sigma\ind(\Lambda_{n,4}^1\setminus \n[g_{31}])$. Observe that the graph $\Lambda_{n,4}^1\setminus \n[g_{31}]$ is isomorphic to $\Gamma_{n-1,4}$. Therefore, $\ind(\Lambda_{n,4}^1) \simeq \Sigma\ind(\Gamma_{n-1,4})$. This completes the proof.

§.§ Graph $A_{n,4}$

For $n\geq 0$, we define the graph $A_{n,4}$ as follows:

\begin{equation*}
\begin{split}
V(A_{n,4}) = & V(\Gamma_{n,4})\cup\{a_1\},\\
E(A_{n,4}) = & E(\Gamma_{n,4})\cup \{a_1g_{11}, a_1g_{21}\}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

in 1,2,...,8

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);

in 1,2,3,4,5,...,8
[edge] (1) – (2);
[edge] (2) – (3);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

[evaluate==int(1+)] in 1,3,6

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4,7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

(31) – node[below=6pt] $1$(33);
(33) – node[below=6pt] $2$(35);
(36) – node[below=6pt] $n$(38);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;
[label=[label distance=-3pt]0:$g_{3\,{2n+1}}$] at (38) ;
[] at (8,-1) (b) $A_{n,4}$ for $n\geq 1$;
[label=[label distance=-3pt]180:$g_{21}$] at (21) ;
[label=[label distance=-3pt]180:$g_{31}$] at (31) ;

[vertex] (a1) at (-2,6)  ;

[edge] (a1) – (11) ;
[edge] (a1) – (21) ;

[label=[label distance=-3pt]180:$a_{1}$] at (a1) ;

[shift = (-5,0)]
in 1,2,3
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1,2
[edge] (x1) – (x1) ;

[vertex] (a1) at (-2,6)  ;

[edge] (a1) – (x11) ;
[edge] (a1) – (x21) ;

[label=[label distance=-3pt]180:$a_{1}$] at (a1) ;
in 1,2,3

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[] at (-1,-1) (a) $A_{0,4}$;

For $n \geq 0$,

\begin{equation*}
\ind(A_{n,4}) \simeq \begin{cases}
\mathbb{S}^0 & \text{if $n =0$,} \\
\mathbb{S}^1\vee \mathbb{S}^2 & \text{if $n =1$,} \\
\Sigma \ind(A_{n-1,4}) \vee \Sigma^{2}\ind(B_{n-2,4})\vee \Sigma^{3}\ind (C_{n-2,4})
& \text{if $n \geq 2$.}
\end{cases}
\end{equation*}

in 3,4,...,5,6

[vertex] (1) at (2*,3*2) ;

in 2,...,6

[vertex] (2) at (2*,2*2) ;

in 1,2,...,6

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 3,5

[edge] (1) – (1);

[evaluate=as using int(+1)] in 2,3,5

[edge] (2) – (2);

[evaluate=as using int(+1)] in 1,2,3,5

[edge] (3) – (3);

in 3,4,5,6
[edge] (1) – (2);
[edge] (2) – (3);

[edge] (22) – (32);
[edge] (15) – (26);

[evaluate==int(-1)] in 3,6

[edge] (1) – (2);

[evaluate==int(1+)] in 3

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (8,5) – (10,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (8,3) – (10,3);

[label=[label distance=-3pt]90:$g_{22}$] at (22) ;
in 3,4

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (16) ;

[label=[label distance=-3pt]270:$g_{3\,{2n+1}}$] at (36) ;

[label=[label distance=-3pt]270:$g_{32}$] at (32) ;
[label=[label distance=-3pt]180:$g_{31}$] at (31) ;

in 3,...,5,6,7

[vertex] (1) at (2*,3*2) ;

in 5,...,7

[vertex] (2) at (2*,2*2) ;

[vertex] (23) at (6,2*2) ;
in 4,...,7

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 3,...,4,6

[edge] (1) – (1);

[evaluate=as using int(+1)] in 6

[edge] (2) – (2);

[evaluate=as using int(+1)] in 4,6

[edge] (3) – (3);

in 5,...,7
[edge] (1) – (2);
[edge] (2) – (3);

[edge] (13) – (23);
[evaluate==int(1+)] in 4,6

[edge] (1) – (2);

[edge] (17) – (26);
[edge] (23) – (34);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

[label=[label distance=-3pt]180:$g_{23}$] at (23) ;
in 3,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (17) ;

[label=[label distance=-3pt]270:$g_{3\,{2n+1}}$] at (37) ;

[label=[label distance=-3pt]270:$g_{34}$] at (34) ;
[label=[label distance=-3pt]180:$g_{25}$] at (25) ;

in 3,...,5,6,7

[vertex] (1) at (2*,3*2) ;

in 5,...,7

[vertex] (2) at (2*,2*2) ;

[vertex] (23) at (6,2*2) ;
in 4,...,7

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 3,...,4,6

[edge] (1) – (1);

[evaluate=as using int(+1)] in 6

[edge] (2) – (2);

[evaluate=as using int(+1)] in 4,6

[edge] (3) – (3);

in 5,...,7
[edge] (1) – (2);
[edge] (2) – (3);

[edge] (13) – (23);
[evaluate==int(1+)] in 4,6

[edge] (1) – (2);

[edge] (17) – (26);
[edge] (23) – (34);
[edge] (34) – (15);
[edge] (34) – (25);
(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

[label=[label distance=-3pt]180:$g_{23}$] at (23) ;
in 3,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (17) ;

[label=[label distance=-3pt]270:$g_{3\,{2n+1}}$] at (37) ;

[label=[label distance=-3pt]270:$g_{34}$] at (34) ;
[label=[label distance=-3pt]180:$g_{25}$] at (25) ;

Since $\n(g_{31})\subseteq \n(a_1)\cap \n(g_{11})$ in the graph $A_{0,4}$ (see <Ref>), using <Ref>, we get  $\ind(A_{0,4})\simeq  \ind(A_{0,4}\setminus \{a_1,g_{11}\})\simeq \mathbb{S}^0$. So, we assume that $n\geq 1$.  Since $a_1$ is a simplicial vertex in $A_{n,4}$ with $\n(a_1)=\{g_{11},g_{21}\}$, from <Ref>, we get

\begin{align} \label{eq:An41}
\ind(A_{n,4}) & \simeq \Sigma\ind(A_{n,4}\setminus N[g_{11}])\vee \Sigma \ind(A_{n,4}\setminus N[g_{21}]) \nonumber \\
& \simeq  \Sigma \ind(A_{n,4}\setminus N[g_{11}]) \vee  \Sigma \ind (A_{n-1,4}),
\end{align}

where the second homotopy equivalence from the observation that $A_{n,4}\setminus N[g_{21}]$ is isomorphic to  $A_{n-1,4}$. Let $A_{n,4}^1\coloneq A_{n,4}\setminus N[g_{11}]$ (see <Ref>). Since $g_{31}$ is a simplicial vertex in the graph $A_{n,4}^1$ with $N(g_{31})=\{g_{22},g_{32}\}$, we have

\begin{equation}\label{eq:An42}
\ind(A_{n,4}^1)\simeq  \Sigma\ind(A_{n,4}^1\setminus N[g_{22}])\vee \Sigma \ind(A_{n,4}^1\setminus N[g_{32}]).
\end{equation}

In particular, for $n=1$, the graph $A_{1,4}^1\setminus N[g_{22}]$ consists of a single vertex $g_{33}$, and $A_{1,4}^1\setminus N[g_{32}]$ is a graph consisting of exactly one edge $g_{13}g_{23}$. Therefore, $\ind(A_{1,4}^1)\simeq \mathbb{S}^1 $, and it follows that $\ind(A_{1,4})\simeq\mathbb{S}^1\vee \mathbb{S}^2$.

Now, consider $n\geq 2$. Observe that the graph $A_{n,4}^1\setminus N[g_{22}]$ is isomorphic to $B_{n-2,4}$ (see <Ref>). Therefore, $\ind(A_{n,4}^1)\simeq \Sigma \ind (B_{n-2,4}) \vee \Sigma \ind(A_{n,4}^1\setminus N[g_{32}])$.

We now compute $\ind(A_{n,4}^1\setminus N[g_{32}])$. Since $N(g_{13})\subseteq N(g_{24})$ in the graph $A_{n,4}^1\setminus N[g_{32}]$ (see <Ref>), $\ind(A_{n,4}^1\setminus N[g_{32}])\simeq \ind((A_{n,4}^1\setminus N[g_{32}])\setminus g_{24}) $. Let $A^{2}_{n,4}=(A_{n,4}^1\setminus N[g_{32}])\setminus g_{24}$. Then $\ind(A_{n,4}^1\setminus N[g_{32}])\simeq \ind(A^{2}_{n,4})$. Observe that $[g_{34},g_{25};g_{13}]$ and $[g_{34},g_{15};g_{13}]$ are edge-invariant triplets in the graphs $A^{2}_{n,4}$ (see <Ref>) and $A^{2}_{n,4} + \{g_{34}g_{25}\}$, respectively. Therefore, $\ind(A^{2}_{n,4}) \simeq \ind(A^{2}_{n,4} +\{g_{34}g_{25}, g_{34}g_{15} \})$. Let the resultant graph that we obtain be $A^{3}_{n,4}$ (see <Ref>). Then $\ind(A^{2}_{n,4})\simeq\ind(A^{3}_{n,4})$.
Since $[g_{23},g_{34};g_{14}]$ is an edge-invariant triplet in the graph $A^{3}_{n,4}$ (see <Ref>), we find that $\ind(A^{3}_{n,4})\simeq \ind(A^{3}_{n,4}-\{g_{23}g_{34}\})$. Furthermore, $N(g_{23})\subseteq N(g_{14})$ in the graph $A^{3}_{n,4}-\{g_{23}g_{34}\}$ implies that  $\ind(A^3_{n,4})\simeq \ind((A^3_{n,4}-\{g_{23}g_{34}\})\setminus g_{14})$. Since the graph $(A^3_{n,4}-\{g_{23}g_{34}\})\setminus g_{14}$ consists of two connected components: one isomorphic to $K_2$ and the other component is isomorphic to the graph $C_{n-2,4}$ (see <Ref>), we obtain the following

\begin{equation}\label{eq: An43}
\ind(A^{1}_{n,4}\setminus N[g_{32}])\simeq \ind(A^2_{n,4})\simeq \ind(A^3_{n,4})\simeq \Sigma\ind(C_{n-2,4}).
\end{equation}

Now, the result follows from (<ref>), (<ref>) and (<ref>).

§.§ Graph $B_{n,4}$

For $n\geq 0$, we define the graph $B_{n,4}$ as follows:

\begin{equation*}
\begin{split}
V(B_{n,4}) = & V(\Gamma_{n,4})\cup\{b_1,b_2,b_3,b_4\},\\
E(B_{n,4}) = & E(\Gamma_{n,4})\cup \{b_{1}b_3, b_{1}b_4,b_{2}b_{3}, b_{3}b_{4}, b_2g_{11}, b_2g_{21}, b_3g_{11}, b_3g_{21}, b_{4}g_{31} \}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

in 1,2,...,8

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);

in 1,2,3,4,5,...,8
[edge] (1) – (2);
[edge] (2) – (3);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

[evaluate==int(1+)] in 1,3,6

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4,7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

(31) – node[below=6pt] $1$(33);
(33) – node[below=6pt] $2$(35);
(36) – node[below=6pt] $n$(38);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;
[label=[label distance=-3pt]0:$g_{3\,{2n+1}}$] at (38) ;

[vertex] (b1) at (-2,6)  ;
[vertex] (b2) at (-2,4)  ;
[vertex] (b3) at (-2,2)  ;
[vertex] (b4) at (-4,2)  ;

[edge] (b1) – (b2) ;
[edge] (b2) – (b3) ;
[edge] (b1) – (11) ;
[edge] (b2) – (21) ;
[edge] (b3) – (31) ;
[edge] (b1) – (21) ;
[edge] (b2) – (11) ;
[edge] (b4) – (b2) ;
[edge] (b4) – (b3) ;

[label=[label distance=-3pt]180:$b_{2}$] at (b1) ;
[label=[label distance=-3pt]180:$b_{3}$] at (b2) ;
[label=[label distance=-3pt]270:$b_{4}$] at (b3) ;
[label=[label distance=-3pt]180:$b_{1}$] at (b4) ;
[] at (6,-1) (a) $B_{n,4}$ for $n\geq 1$;

[shift = (-7,0)]
in 1,2,3
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1,2
[edge] (x1) – (x1) ;

[vertex] (b1) at (-2,6)  ;
[vertex] (b2) at (-2,4)  ;
[vertex] (b3) at (-2,2)  ;
[vertex] (b4) at (-4,2)  ;

[edge] (b1) – (b2) ;
[edge] (b2) – (b3) ;
[edge] (b1) – (x11) ;
[edge] (b2) – (x21) ;
[edge] (b3) – (x31) ;
[edge] (b1) – (x21) ;
[edge] (b2) – (x11) ;
[edge] (b4) – (b2) ;
[edge] (b4) – (b3) ;

in 1,2,3

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[label=[label distance=-3pt]180:$b_{2}$] at (b1) ;
[label=[label distance=-3pt]180:$b_{3}$] at (b2) ;
[label=[label distance=-3pt]270:$b_{4}$] at (b3) ;
[label=[label distance=-3pt]180:$b_{1}$] at (b4) ;
[] at (-1,-1) (a) $B_{0,4}$;

For $n \geq 0$,

\begin{equation*}
\ind(B_{n,4}) \simeq \begin{cases}
\mathbb{S}^1\vee\mathbb{S}^1 & \text{if  $n =0$,} \\
\vee_{3}\mathbb{S}^2 & \text{if $n =1$,} \\
\Sigma \ind(B_{n-1,4}) \vee \Sigma^{2}\ind(A_{n-1,4})\vee \Sigma^{3}\ind(D_{n-2,4})
& \text{if $n \geq 2$.}
\end{cases}
\end{equation*}

Since $b_{1}$ is a simplicial vertex in $B_{n,4}$ with $\n(b_1)=\{b_3,b_4\}$, we get $\ind(B_{n,4})\simeq \Sigma\ind(B_{n,4}\setminus \n[b_3])\vee \Sigma\ind(B_{n,4}\setminus \n[b_4])$. In the case of $n=0$, the graph $B_{0,4}\setminus \n[b_3]$ consists of an isolated vertex, $g_{31}$, and $B_{0,4}\setminus \n[b_4]$ is isomorphic to the complete graph $K_3$. Therefore,  $\ind(B_{0,4})\simeq \mathbb{S}^1\vee \mathbb{S}^1$.

Let $n = 1$. In the case of $B_{1,4}$, the graph $B_{1,4}\setminus \n[b_3]$ is isomorphic to $B_{0,4}$. Hence, $\ind(B_{1,4}\setminus \n[b_3])\simeq \mathbb{S}^1\vee \mathbb{S}^1$. Let $B'_{1,4}=B_{1,4}\setminus \n[b_4]$. Then, $b_2$ is a simplicial vertex in the graph $B'_{1,4}$ with $\n(b_2)=\{g_{11},g_{21}\}$, implying that $\ind(B'_{1,4})\simeq \Sigma\ind(B'_{1,4} \setminus \n[g_{11}])\vee \Sigma\ind(B'_{1,4}\setminus \n[g_{21}])$. Since $\n(g_{32})\subseteq \n(g_{23})$ in the graph $B'_{1,4}\setminus \n[g_{11}]$,  we get $\ind(B'_{1,4}\setminus \n[g_{11}])\simeq \ind((B'_{1,4}\setminus \n[g_{11}])\setminus g_{23})$. However, the graph $(B'_{1,4}\setminus \n[g_{11}])\setminus g_{23}$ is a path on $4$ vertices, implying that $\ind(B'_{1,4}\setminus n[g_{11}])$ is contractible. On the other hand, the graph $B'_{1,4}\setminus \n[g_{21}]$ is isomorphic to $A_{0,4}$. Therefore, $\ind(B'_{1,4})\simeq \mathbb{S}^1$, and it follows that $\ind(B_{1,4})\simeq \vee_{3} \mathbb{S}^2$.
It is evident that the graphs $B_{n,4}$ and $A_{n,4}$ exhibit similarities. Additionally, the graph $B_{n,4}$ can be derived from $A_{n,4}$ by introducing four vertices, namely $b_1$, $b_2$, $b_3$, and $b_4$, with $b_2$, $b_3$, and $b_4$ forming a vertical line. Consequently, for $n\geq 2$, the proof proceeds analogously to the approach outlined in <Ref>.

§.§ Graph $C_{n,4}$

For $n\geq 0$, we define the graph $C_{n,4}$ as follows:

\begin{equation*}
\begin{split}
V(C_{n,4}) = & V(\Gamma_{n,4})\cup\{c_1\},\\
E(B_{n,4}) = & E(\Gamma_{n,4})\cup \{c_{1}g_{11}, c_{1}g_{21}, c_{1}g_{31}\}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

in 1,2,...,8

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);

in 1,2,3,4,5,...,8
[edge] (1) – (2);
[edge] (2) – (3);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

[evaluate==int(1+)] in 1,3,6

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4,7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

(31) – node[below=6pt] $1$(33);
(33) – node[below=6pt] $2$(35);
(36) – node[below=6pt] $n$(38);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;
[label=[label distance=-3pt]0:$g_{3\,{2n+1}}$] at (38) ;

[vertex] (c1) at (-2,4)  ;

[edge] (c1) – (11) ;
[edge] (c1) – (21) ;
[edge] (c1) – (31) ;

[label=[label distance=-3pt]180:$c_{1}$] at (c1) ;
[] at (6,-1) (b) $C_{n,4}$ for $n\geq 1$;

[shift = (-5,0)]
in 1,2,3
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1,2
[edge] (x1) – (x1) ;

in 1,2,3

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[vertex] (c1) at (-2,4)  ;

[edge] (c1) – (x11) ;
[edge] (c1) – (x21) ;
[edge] (c1) – (x31) ;

[label=[label distance=-3pt]180:$c_{1}$] at (c1) ;
[] at (0,-1) (a) $C_{0,4}$;

For $n\geq 0$,

\begin{equation*}
\ind(C_{n,4}) \simeq \begin{cases}
\mathbb{S}^0\vee\mathbb{S}^0 & \text{for $n =0$,} \\
\mathbb{S}^1 & \text{for $n =1$,} \\
\Sigma \ind(A_{n-1,4}) \vee \Sigma^{2}\ind(B_{n-2,4})\vee \Sigma^{2}\ind(D_{n-2,4})
& \text{for $n \geq 2$.}
\end{cases}
\end{equation*}

Since $\n(g_{11})\subseteq \n(g_{31})$ in the graph $C_{0,4}$,  we get $\ind(C_{0,4})\simeq \ind(C_{0,4}\setminus g_{31})$. However, the graph  $C_{0,4}\setminus g_{31}$ is isomorphic to a cycle with three vertices, and therefore  $\ind(C_{0,4})\simeq \mathbb{S}^0\vee \mathbb{S}^0$.

For $n\geq 1$, the inclusion map $\ind(C_{n,4} \setminus \n[g_{21}]) \hookrightarrow \ind(C_{n,4} \setminus g_{21})$ is null homotopic, because $\ind(C_{n,4} \setminus \n[g_{21}])\ast\{c_1\}\subseteq \ind(C_{n,4} \setminus g_{21})$. Since the graph $C_{n,4}\setminus N[g_{21}]$ is isomorphic to $A_{n-1,4}$, from <Ref>, we get

\begin{equation*}\label{Dn}
\begin{split}
\ind(C_{n,4})\simeq  \ind(C_{n,4}\setminus g_{21})\vee \Sigma \ind(C_{n,4}\setminus N[g_{21}]) & \simeq  \ind(C_{n,4}\setminus g_{21})\vee \Sigma \ind(A_{n-1,4}).
\end{split}
\end{equation*}

in 1,2,...,5

[vertex] (1) at (2*,3*2) ;

in 2,...,5

[vertex] (2) at (2*,2*2) ;

in 1,2,...,5

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 1,2,...,4

[edge] (1) – (1);

[evaluate=as using int(+1)] in 2,3,4

[edge] (2) – (2);

[evaluate=as using int(+1)] in 1,2,...,4

[edge] (3) – (3);

in 2,3,4,5

[edge] (1) – (2);

in 2,3,4,5

[edge] (2) – (3);

[evaluate==int(1+)] in 2,4

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5

[edge] (1) – (2);

[evaluate==int(1+)] in 3

[edge] (2) – (3);

[evaluate==int(-1)] in 4

[edge] (2) – (3);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-8pt]60:$g_{31}$] at (31) ;
[label=[label distance=-8pt]-30:$g_{22}$] at (22) ;
[label=[label distance=-8pt]60:$g_{32}$] at (32) ;
[label=[label distance=-8pt]60:$g_{23}$] at (23) ;

[vertex] (c1) at (-2,4)  ;

[edge] (c1) – (11) ;
[edge] (c1) – (31) ;

[label=[label distance=-3pt]180:$c_{1}$] at (c1) ;

in 2,...,5

[vertex] (1) at (2*,3*2) ;

in 2,...,5

[vertex] (2) at (2*,2*2) ;

in 2,...,5

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 2,...,4

[edge] (1) – (1);

[evaluate=as using int(+1)] in 2,3,4

[edge] (2) – (2);

[evaluate=as using int(+1)] in 2,...,4

[edge] (3) – (3);

in 2,3,4,5

[edge] (1) – (2);

in 2,3,4,5

[edge] (2) – (3);

[evaluate==int(1+)] in 2,4

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5

[edge] (1) – (2);

[evaluate==int(1+)] in 3

[edge] (2) – (3);

[evaluate==int(-1)] in 4

[edge] (2) – (3);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[edge] (12) to[bend right=45] (32);

[label=[label distance=-8pt]-30:$g_{22}$] at (22) ;
[label=[label distance=-8pt]60:$g_{32}$] at (32) ;
[label=[label distance=-8pt]60:$g_{23}$] at (23) ;

in 3,4,5

[vertex] (1) at (2*,3*2) ;

in 2,4,5

[vertex] (2) at (2*,2*2) ;

in 2,...,5

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 3,4

[edge] (1) – (1);

[evaluate=as using int(+1)] in 4

[edge] (2) – (2);

[evaluate=as using int(+1)] in 2,...,4

[edge] (3) – (3);

in 4,5

[edge] (1) – (2);

in 2,4,5

[edge] (2) – (3);

[evaluate==int(1+)] in 4

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5

[edge] (1) – (2);

[evaluate==int(1+)] in

[edge] (2) – (3);

[evaluate==int(-1)] in 4

[edge] (2) – (3);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

in 3,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-8pt]-30:$g_{22}$] at (22) ;
[label=[label distance=-8pt]60:$g_{32}$] at (32) ;
[label=[label distance=-3pt]90:$g_{33}$] at (33) ;

in 3,4,5

[vertex] (1) at (2*,3*2) ;

in 2,4,5

[vertex] (2) at (2*,2*2) ;

in 2,...,5

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 4

[edge] (1) – (1);

[evaluate=as using int(+1)] in 4

[edge] (2) – (2);

[evaluate=as using int(+1)] in 2,...,4

[edge] (3) – (3);

in 4,5

[edge] (1) – (2);

in 2,4,5

[edge] (2) – (3);

[evaluate==int(1+)] in 4

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5

[edge] (1) – (2);

[evaluate==int(1+)] in

[edge] (2) – (3);

[evaluate==int(-1)] in 4

[edge] (2) – (3);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

in 3,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[edge] (33) – (14);

[label=[label distance=-8pt]-30:$g_{22}$] at (22) ;
[label=[label distance=-8pt]60:$g_{32}$] at (32) ;
[label=[label distance=-3pt]90:$g_{33}$] at (33) ;

Let $C_{n,4}^1 \coloneq C_{n,4}\setminus g_{21}$. Since $[g_{31},g_{22};g_{11}]$ is an edge invariant triplet in the graph $C_{n,4}^1$,  $\ind(C_{n,4}^1)\simeq \ind(C_{n,4}^1-\{g_{31}g_{22}\})$. Let $ C_{n,4}^2\coloneq C_{n,4}^1-\{g_{31}g_{22}\}$. Then $\ind(C_{n,4}^1)\simeq \ind(C_{n,4}^2)$.
Since  $[g_{12},g_{32};c_1]$ is an edge-invariant triplet in the graph $C_{n,4}^2$, $\ind(C_{n, 4}^2) \simeq \ind(C_{n,4}^2+\{g_{12}g_{32}\})$. Furthermore,  $[g_{31},g_{32};g_{11}]$ is an edge-invariant triplet in the graph $C_{n,4}^2+\{g_{12}g_{32}\}$. Therefore, $\ind(C_{n,4}^2)\simeq \ind((C_{n,4}^2+\{g_{12}g_{32}\})-\{g_{31}g_{32}\})$. Observe that $\n(g_{31})\subseteq \n(g_{11})$ in the graph $(C_{n,4}^2+\{g_{12}g_{32}\})-\{g_{31}g_{32}\}$. Then, using  <Ref>, we get that $\ind(C_{n,4}^2)\simeq\ind(((C_{n,4}^2+\{g_{12}g_{32}\})-\{g_{31}g_{32}\})\setminus g_{11})$. The resultant graph we obtain has two connected components: one consists of the edge $g_{31}c_1$, and we denote the other component by $C_{n,4}^3$ (see <Ref>). This implies that $\ind(C_{n,4}^2)\simeq \Sigma \ind(C_{n,4}^3)$.

Since  $\ind(C_{n,4}^3 \setminus \n[g_{12}])*\{g_{22}\} \subseteq \ind(C_{n,4}^3\setminus g_{12})$, the inclusion map  $\ind(C_{n,4}^3 \setminus \n[g_{12}]) \hookrightarrow \ind(C_{n,4}^3 \setminus g_{12})$ is null-homotopic. Therefore, by <Ref> we get

\begin{equation}
\begin{split}\label{eq:D3n}
\ind(C_{n,4}^3)\simeq  \ind(C_{n,4}^3\setminus g_{12})\vee \Sigma \ind(C_{n,4}^3\setminus N[g_{12}]).
\end{split}
\end{equation}

Since $g_{33}$ is an isolated vertex in $C_{1,4}^3\setminus N[g_{12}]$, $\ind(C_{n,4}^3\setminus N[g_{12}])$ is contractible. Observe that $\n(g_{32})\subseteq \n(g_{23})$ in the graph $C_{1,4}^3\setminus g_{12}$. Thus, using <Ref>, we find that $\ind(C_{1,4}^3\setminus g_{12})\simeq \ind(C_{1,4}^3\setminus \{g_{12},g_{23}\})$. Since the graph $C_{1,4}^3\setminus \{g_{12},g_{23}\}$ is a path on $4$- vertices, $\ind(C_{1,4}^3\setminus g_{12})$ is contractible. Therefore, $\ind(C_{1,4}^3)$ is contractible. This implies that $\ind(C_{1,4}^1)$ is contractible. Hence, $\ind(C_{1,4})\simeq \Sigma\ind(A_{0,4})\simeq \mathbb{S}^1$.

Now, consider $n\geq 2$. We observe that the graph $C_{n,4}^3 \setminus \n[g_{12}]$ is isomorphic to the graph $B_{n-2,4}$. Thus,
$$\ind(C_{n,4}^3)\simeq  \ind(C_{n,4}^3\setminus g_{12})\vee \Sigma \ind(B_{n-2,4}).$$

Since $\n(g_{32})\subseteq \n(g_{23})$ in $C_{n,4}^3\setminus g_{12}$, using <Ref> we get $\ind(C_{n,4}^3\setminus g_{12})\simeq \ind(C_{n,4}^3\setminus \{g_{12},g_{23}\})$. Let $C_{n,4}^4\coloneq C_{n,4}^3\setminus \{g_{12},g_{23}\}$. Then $\ind(C_{n,4}^3\setminus g_{12})\simeq \ind(C_{n,4}^4)$. Since, $[g_{33},g_{14};g_{22}]$ and $[g_{13},g_{14}; g_{32}]$ are edge-invariant triplets in graphs $C_{n,4}^4$ and $C_{n,4}^4+\{g_{33}g_{14}\}$, respectively, <Ref>, provides us with $C_{n,4}^4\simeq \ind((C_{n,4}^4+\{g_{33}g_{14}\})-\{g_{13}g_{14}\})$. Let  $C_{n,4}^5 \coloneq (C_{n,4}^4+\{g_{33}g_{14}\})-\{g_{13}g_{14}\}$. Then, $\ind(C_{n,4}^4)\simeq C_{n,4}^5$. Since $\n(g_{13})\subseteq \n(g_{32}$) in the graph $C_{n,4}^5$, <Ref> implies that $\ind(C_{n,4}^5)\simeq \ind(C_{n,4}^5\setminus g_{32})$.
The graph $C_{n,4}^5\setminus g_{32}$ consists of two connected components: one component consists of the edge  $g_{22}g_{32}$, and the other component is isomorphic to $D_{n-2,4}$. Therefore, $\ind(C_{n,4}^4)\simeq \ind(C_{n,4}^5)\simeq \Sigma \ind(D_{n-2,4})$.
Therefore, $\ind(C_{n,4}^3\setminus g_{12})\simeq \Sigma \ind(D_{n-2,4})$. Hence, the result follows.

§.§ Graph $D_{n,4}$

For $n\geq 0$, we define the graph $B_{n,4}$ as follows:

\begin{equation*}
\begin{split}
V(D_{n,4}) = & V(\Gamma_{n,4})\cup\{d_1,d_2,d_3,d_4\},\\
E(D_{n,4}) = & E(\Gamma_{n,4})\cup \{d_{1}d_2, d_{1}d_3, d_1d_4, d_{2}d_{3}, d_{3}d_{4}, d_2g_{11}, d_2g_{21}, d_3g_{11}, d_3g_{21}, d_{4}g_{31} \}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

in 1,2,...,8

[vertex] (3) at (2*,1*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);
[edge] (3) – (3);

in 1,2,3,4,5,...,8
[edge] (1) – (2);
[edge] (2) – (3);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

[evaluate==int(1+)] in 1,3,6

[edge] (2) – (3);

[evaluate==int(-1)] in 2,4,7

[edge] (2) – (3);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);
[dotted, thick,shorten >=4pt, shorten <=4pt] (10,3) – (12,3);

(31) – node[below=6pt] $1$(33);
(33) – node[below=6pt] $2$(35);
(36) – node[below=6pt] $n$(38);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;
[label=[label distance=-3pt]0:$g_{3\,{2n+1}}$] at (38) ;

[label=[label distance=-8pt]45:$g_{21}$] at (21) ;
[label=[label distance=-8pt]-45:$g_{22}$] at (22) ;
[label=[label distance=-8pt]135:$g_{31}$] at (31) ;
[label=[label distance=-8pt]135:$g_{33}$] at (33) ;

[vertex] (d1) at (-2,6)  ;
[vertex] (d2) at (-2,4)  ;
[vertex] (d3) at (-2,2)  ;
[vertex] (d4) at (-4,4)  ;

[edge] (d1) – (d2) ;
[edge] (d2) – (d3) ;
[edge] (d1) – (11) ;
[edge] (d2) – (21) ;
[edge] (d3) – (31) ;
[edge] (d1) – (21) ;
[edge] (d2) – (11) ;
[edge] (d4) – (d2) ;
[edge] (d4) – (d3) ;
[edge] (d4) – (d1) ;

[label=[label distance=-3pt]180:$d_{2}$] at (d1) ;
[label=[label distance=-8pt]-135:$d_{3}$] at (d2) ;
[label=[label distance=-3pt]270:$d_{4}$] at (d3) ;
[label=[label distance=-3pt]180:$d_{1}$] at (d4) ;
[] at (6,-1) (b) $D_{n,4}$ for $n\geq 1$;

[shift = (-7,0)]
in 1,2,3
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1,2
[edge] (x1) – (x1) ;

[vertex] (d1) at (-2,6)  ;
[vertex] (d2) at (-2,4)  ;
[vertex] (d3) at (-2,2)  ;
[vertex] (d4) at (-4,4)  ;

[edge] (d1) – (d2) ;
[edge] (d2) – (d3) ;
[edge] (d1) – (x11) ;
[edge] (d2) – (x21) ;
[edge] (d3) – (x31) ;
[edge] (d1) – (x21) ;
[edge] (d2) – (x11) ;
[edge] (d4) – (d2) ;
[edge] (d4) – (d3) ;
[edge] (d4) – (d1) ;

in 1,2,3

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[label=[label distance=-3pt]180:$d_{2}$] at (d1) ;
[label=[label distance=-8pt]-135:$d_{3}$] at (d2) ;
[label=[label distance=-3pt]270:$d_{4}$] at (d3) ;
[label=[label distance=-3pt]180:$d_{1}$] at (d4) ;
[] at (-3,-1) (a) $D_{0,4}$;

For $n\geq 0$,

\begin{equation*}
\ind(D_{n,4}) \simeq \begin{cases}
\mathbb{S}^1\vee \mathbb{S}^1  & \text{if $n =0$,} \\
\bigvee_5 \mathbb{S}^2 & \text{if $n =1$,} \\
\Sigma \ind(B_{n-1,4}) \vee \Sigma^{2}\ind(A_{n-1,4})\vee \Sigma^{2}\ind(C_{n-1,4})
& \text{if $n \geq 2$.}
\end{cases}
\end{equation*}

Since $g_{11}$ is a simplicial vertex in $D_{0,4}$ with $\n(g_{11})=\{d_2,d_3,g_{21}\}$, $ \ind(D_{0,4})\simeq \Sigma \ind(D_{0,4}\setminus\n[d_2])\vee \Sigma\ind(D_{0,4}\setminus\n[d_3])\vee \Sigma(\ind(D_{0,4}\setminus\n[g_{21}])).$

Observe that both the graphs $D_{0,4}\setminus\n[d_2]$ and  $D_{0,4}\setminus\n[g_{21}]$ are paths with two vertices and the graph $D_{0,4}\setminus\n[d_3]$ consists of an isolated vertex $g_{31}$. Hence $\ind(D_{0,4})\simeq \mathbb{S}^1 \vee \Sp^1.$

Now, we compute $\ind(D_{1,4})$.  The inclusion map
$\ind(D_{1,4}\setminus \n[d_3])\hookrightarrow \ind(D_{1,4}\setminus d_3)$ is null homotopic because $\ind(D_{1,4}\setminus \n[d_3])\ast \{d_1\}$ is a subcomplex of $\ind(D_{1,4}\setminus d_3)$. Thus, from <Ref>, we get
$\ind(D_{1,4})\simeq \Sigma\ind(D_{1,4}\setminus \n[d_3])\vee\ind(D_{1,4}\setminus d_3)$.
Observe that the graph $D_{1,4}\setminus \n[d_3]$ is isomorphic to $B_{0,4}$. Therefore, $\ind(B_{0,4})\simeq \mathbb{S}^1\vee\mathbb{S}^1$ implies that
$$\ind(D_{1,4})\simeq \mathbb{S}^2\vee\mathbb{S}^2\vee\ind(D_{1,4}\setminus d_3).$$

Let $D_{1,4}'=(D_{1,4}\setminus d_3)$. Then $g_{13}$ is a simplicial vertex in the graph $D_{1,4}'$ with $\n(g_{13})=\{g_{12}, g_{22},g_{23}\}$. Using <Ref>, we find that $\ind(D_{1,4}')\simeq \Sigma\ind(D_{1,4}'\setminus \n[g_{12}])\vee\Sigma \ind(D_{1,4}'\setminus \n[g_{22}])\vee \Sigma \ind(D_{1,4}'\setminus \n[g_{23}])$.
Since $\n(g_{33})\subseteq \n(g_{21})\cap \n(g_{31})$  in the graph $D_{1,4}'\setminus \n[g_{12}]$,  $\ind(D_{1,4}'\setminus \n[g_{12}])\simeq \ind((D_{1,4}'\setminus \n[g_{12}])\setminus\{g_{21},g_{31}\})$. Observe that the graph $(D_{1,4}'\setminus \n[g_{12}])\setminus\{g_{21},g_{31}\}$ is a disjoint union of two paths with $2$ and $3$ vertices. Hence $\ind(D_{1,4}'\setminus \n[g_{12}])\simeq \mathbb{S}^1$.

Since the graph $D_{1,4}'\setminus \n[g_{22}]$ contains an isolated vertex $g_{33}$,  $\ind(D_{1,4}'\setminus \n[g_{22}])$ is contractible.

Let $D_{1,4}''\coloneq D_{1,4}'\setminus \n[g_{23}]$. Since $g_{32}$ is a simplicial vertex in  $D_{1,4}''$ with $\n(g_{32})=\{g_{21},g_{31}\}$, we have  $\ind(D_{1,4}'')\simeq \Sigma\ind(D_{1,4}''\setminus\n[g_{21}])\vee \Sigma \ind(D_{1,4}''\setminus\n[g_{31}])$. However, the graph $D_{1,4}''\setminus\n[g_{21}]$ is an edge $d_{1}d_4$ and the graph $D_{1,4}''\setminus\n[g_{31}]$
is a path on three vertices $d_1, d_2$ and $g_{11}$ and hence, using <Ref>, we get that $\ind(D_{1,4}'')\simeq \mathbb{S}^1\vee \mathbb{S}^1$. Consequently, $\ind(D_{1,4})\simeq \vee_5\mathbb{S}^2$.
It is evident that the graphs $C_{n,4}$ and $D_{n,4}$ exhibit similarities. Additionally, the graph $D_{n,4}$ can be derived from $C_{n,4}$ by introducing four vertices, namely $d_1$, $d_2$, $d_3$, and $d_4$, with $d_2$, $d_3$, and $d_4$ forming an extra vertical line. Consequently, for $n\geq 2$, the proof proceeds analogously to the approach outlined in <Ref>.

For any graph $G_{n,4}\in\{\Gamma_{n,4},\Lambda_{n,4},A_{n,4},B_{n,4},C_{n,4},D_{n,4}\}$, $n\geq 0$, the complex $\ind(G_{n,4})$ is homotopy equivalent to a wedge of spheres.

The proof follows from Gamman4Fn4, and the induction hypothesis.

§.§ Homotopy type of $\M(P_n \times P_4)$

In this section, we conclude the homotopy type of $\M(P_n \times P_4)$ along with the minimum and maximum dimensions of the independence complexes of the graphs in $\{\Gamma_{n, 4}, \Lambda_{n, 4}, A_{n, 4},$ $ B_{n, 4}, C_{n, 4}, D_{n, 4}\}$.

For $n\geq 1$,  $\M(P_n\times P_4)$ is homotopy equivalent to a wedge of spheres.

For $n\geq 0$, we have relations $M(P_{2n+2}\times P_4)=\ind(\Gamma_{n,4})\ast \ind(\Gamma_{n,4})$ and $M(P_{2n+3}\times P_4)=\ind(\Lambda_{n,4})\ast \ind(\Lambda_{n,4})$. Now, the result follows from <Ref>.

For $n = 7q+r$ with $q \geq 0$ and $0 \leq r \leq 6$, we have

* $\dimin{\Gamma_{n,4}} = n$ and $\displaystyle \dimax{\Gamma_{n,4}} = \begin{cases}
n + 3q& \text{if }r = 0,1, \\
n + 3q+1 & \text{if }r = 2,3,4, \\
n + 3q + 2 & \text{if }r = 5,6.
\end{cases}$

* $\dimin{\Lambda_{n,4}} = n+1$, $\dimax{\Lambda_{0,4}}=1$, and for $n\geq 1$,
$\displaystyle \dimax{\Lambda_{n,4}} = \begin{cases}
n + 3q & \text{if }r = 0, \\
n+3q+1 & \text{if }r = 1,2, \\
n + 3q+2 & \text{if }r = 3,4, \\
n + 3q + 3 & \text{if }r = 5,6. \\
\end{cases}$

* $\dimin{A_{n,4}} = n$ and $\displaystyle \dimax{A_{n,4}} = \begin{cases}
n + 3q & \text{if }r = 0, \\
n + 3q+1 & \text{if }r = 1,2,3, \\
n + 3q + 2 & \text{if }r = 4,5, \\
n + 3q + 3 & \text{if }r = 6. \\
\end{cases}$

* $\dimin{B_{n,4}} = n+1$ and $\displaystyle \dimax{B_{n,4}} = \begin{cases}
n + 3q+1 & \text{if }r = 0,1, \\
n + 3q+2 & \text{if }r = 2,3, \\
n + 3q + 3 & \text{if }r = 4,5,6. \\
\end{cases}$

* $\dimin{C_{n,4}} = n$ and $\displaystyle \dimax{C_{n,4}} = \begin{cases}
n + 3q & \text{if }r = 0,1, \\
n + 3q+1 & \text{if }r = 2,3, \\
n + 3q + 2 & \text{if }r = 4,5, \\
n + 3q + 3 & \text{if }r = 6.
\end{cases}$

* $\dimin{D_{n,4}} = n+1$ and $\displaystyle \dimax{D_{n,4}} = \begin{cases}
n + 3q+1 & \text{if }r = 0,1, \\
n + 3q+2 & \text{if }r = 2,3,4, \\
n + 3q + 3 & \text{if }r = 5,6.
\end{cases}$

We prove the statement using induction on $n$. The base case for $n=0$ follows from Gamman4Fn4. Assume the statement is true for $k < n$. For $\Gamma_{n,4}$, we have

\begin{equation}{\label{eq: max-gamman4-eq1}}
\Gamma_{n,4} = \Sigma \ind(A_{n-1,4}) \vee \Sigma^{2} \ind(\Lambda_{n-2,4}),
\end{equation}

which implies that

\begin{equation*}
\dimin{\Gamma_{n,4}} = \min \set{1 + \dimin{A_{n-1,4}}, 2 + \dimin{\Lambda_{n-2,4}}},
\end{equation*}

\begin{equation}{\label{eq: max-gamman4-eq2}}
\dimax{\Gamma_{n,4}} = \max \set{1 + \dimax{A_{n-1,4}}, 2 + \dimax{\Lambda_{n-2,4}}}.
\end{equation}

In particular, using induction hypothesis, we have $\dimin{\Gamma_{n,4}} = \min\set{1+(n-1), 2+(n-1)} = n$. Further, for $n=7q+r$ with $0 \leq r \leq 6$, (<ref>) and (<ref>) yields the following.

hline1-Z = 0.5pt, vline1-Z = 0.5pt,colspec=cccc,c
$n$                 $1+\dimax{A_{n-1,4}}$                   $2+\dimax{\Lambda_{n-2,4}}$                   $\dimax{\Gamma_{n,4}}$

$7q$                $1+(n-1+3(q-1)+3)$                      $2+(n-2+3(q-1)+3)$                      $n+3q$

$7q+1$              $1+(n-1+3q)$                            $2+(n-2+3(q-1)+3)$                      $n+3q$

$7q+2$              $1+(n-1+3q+1)$                          $2+(n-2+3q)$                          $n+3q+1$

$7q+3$              $1+(n-1+3q+1)$                          $2+(n-2+3q+1)$                      $n+3q+1$

$7q+4$              $1+(n-1+3q+1)$                          $2+(n-2+3q+1)$                      $n+3q+1$

$7q+5$              $1+(n-1+3q+2)$                          $2+(n-2+3q+2)$                      $n+3q+2$

$7q+6$              $1+(n-1+3q+2)$                          $2+(n-2+3q+2)$                      $n+3q+2$

By similar argument, we can conclude the maximum dimensions of $\Lambda_{n,4},A_{n,4},B_{n,4},C_{n,4}$ and $D_{n,4}$.

The maximum and minimum dimension of spheres in the homotopy type of $\M(P_n \times P_4)$ as given in <Ref> can be  concluded from <Ref>.

§ MATCHING COMPLEXES OF  PNXP3

In this section, we determine the homotopy type of the matching complex of  $P_n\times P_3$.
The graph $P_n\times P_3$ is disconnected and has two connected components. Notably, if $n$ is even, then these components are isomorphic. However, for $n$ odd, the components are not isomorphic.
For $n \geq 0$, we denote the (isomorphic) connected components of the line graphs of $P_{2n+2}\times P_3$ by $\Gamma_{n,3}$ (see <Ref>). Furthermore, for $n \geq 0$,  we denote the (non-isomorphic) connected components of the line graphs of  $P_{2n+3}\times P_3$ by $\Lambda_{n,3}$ and $\tilde{\Lambda}_{n,3}$ (see <Ref>), respectively. Therefore, we have the following:

\begin{equation*}
M(P_{2n+2}\times P_3) \simeq \ind(\Gamma_{n,3}) * \ind(\Gamma_{n,3}),
\end{equation*}

\begin{equation*}
M(P_{2n+3}\times P_3) \simeq \ind(\Lambda_{n,3}) * \ind(\Tilde{\Lambda}_{n,3}).
\end{equation*}

During the computation of the homotopy type of the independence complexes of graphs $\Gamma_{n,3}, \Lambda_{n,3}$, and $\Tilde{\Lambda}_{n,3}$ for $n\geq 0$, we encounter an intermediary graph denoted as $A_{n,3}$ (see <Ref>). Using the independence complex of $A_{n,3}$, we illustrate that the independence complexes of $\Gamma_{n,3}, \Lambda_{n,3}$, and $\Tilde{\Lambda}_{n,3}$ are the wedges of spheres.

§.§ Graph $\Gamma_{n,3}$

For $n\geq 0$, we define the graph $\Gamma_{n,3}$ as follows:

\begin{equation*}
\begin{split}
V(\Gamma_{n,3}) = & \set{g_{ij} : i=1,2 \text{ and } j = {1,2,\dots,2n{+}1}},\\
E(\Gamma_{n,3}) = & \ \set{g_{1j}g_{2j} : j=1,2,\dots,2n{+}1}
\cup \set{g_{ij}g_{i\,j+1} : i=1,2 \text{ and } j=1,2,\dots,2n} \\
& \cup \set{g_{ij}g_{i+1\,j+1}, g_{i\,j+1}g_{i+1\,j} :  i=1,2, \; j=1,2,\dots,2n  \text{ and } i+j\equiv 1 \mod{2}}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);

in 1,2,3,4,5,...,8
[edge] (1) – (2);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);

(21) – node[below=6pt] $1$(23);
(23) – node[below=6pt] $2$(25);
(26) – node[below=6pt] $n$(28);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]180:$g_{21}$] at (21) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;
[] at (7,0) (b) $\Gamma_{n,3}$ for $n\geq 1$;

[shift = (-5,0)]
in 1,2
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1
[edge] (x1) – (x1) ;

in 1,2

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[] at (0.8,0) (a) $\Gamma_{0,3}$;

For $n\geq 0$,
$\ind(\Gamma_{n,3}) \simeq \Sp^n$.

Since $N(g_{11}) \subseteq N(g_{22})$ and $N(g_{21}) \subseteq N(g_{12})$,  $ \ind(\Gamma_{n,3}) \simeq \ind(\Gamma_{n,3} \setminus \{g_{12},g_{22}\}).$
It is straightforward to see that $\Gamma_{n,3} \setminus \{g_{12},g_{22}\} $ has two connected components; one component is an edge $g_{11}g_{21}$, and the other is isomorphic to $\Gamma_{n-1,3}$. Therefore, $ \ind(\Gamma_{n,3}) \simeq \Sigma\ind(\Gamma_{n-1,3})$. Consequently, we get $ \ind(\Gamma_{n,3}) \simeq \Sigma^n\ind(\Gamma_{0,3}).$
Since $\Gamma_{0,3}$ is isomorphic to $K_2$, we have $\ind(\Gamma_{0,3}) \simeq \Sp^0$, and result follows.

§.§ Graph $A_{n,3}$

For $n\geq 0$, we define the graph $A_{n,3}$ as follows:

\begin{equation*}
\begin{split}
V(A_{n,3}) = & V(\Gamma_{n,3})\cup\{a_1\},\\
E(A_{n,3}) = & E(\Gamma_{n,3})\cup \{a_1g_{11}, a_1g_{21}\}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);

in 1,2,3,4,5,...,8
[edge] (1) – (2);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);

(21) – node[below=6pt] $1$(23);
(23) – node[below=6pt] $2$(25);
(26) – node[below=6pt] $n$(28);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;

[shift = (-2,0)]
[vertex] (a1) at (2,4) ;

[edge] (a1) – (11);
[edge] (a1) – (21);

[label=[label distance=-3pt]180:$a_{1}$] at (a1) ;
[] at (8,1) (b) $A_{n,3}$ for $n\geq 1$;

[shift = (-5,0)]

[vertex] (a1) at (-2,4) ;

[edge] (a1) – (x11);
[edge] (a1) – (x21);

in 1,2
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1
[edge] (x1) – (x1) ;

in 1,2

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[label=[label distance=-7pt]10:$g_{21}$] at (21) ;
[label=[label distance=-3pt]180:$a_{1}$] at (a1) ;
[] at (0,1) (a) $A_{0,3}$;

For $n \geq 0$, $ \ind(A_{n,3}) \simeq \bigvee_{2}\Sigma \ind(A_{n-1,3}) \simeq \bigvee_{2^{n+1}} \Sp^n.$

Since $a_1$ is a simplicial vertex in the graph $A_{n,3}$ (see <Ref>) with $N(a_1)=\{g_{11},g_{21}\}$,  $ \ind(A_{n,3})\simeq \Sigma \ind(A_{n,3}\setminus \n[g_{11}])\vee \Sigma\ind(A_{n,3}\setminus \n[g_{21}]).$
Since both the graphs $A_{n,3}\setminus \n[g_{11}]$ and $A_{n,3}\setminus \n[g_{21}]$ are isomorphic to $A_{n-1,3}$, $\ind(A_{n,3})\simeq \bigvee_{2}\Sigma \ind(A_{n-1,3})$. Observe that the graph $A_{0,3}$ is isomorphic to $K_3$, and hence $\ind(A_{0,3})\simeq \Sp^0\vee\Sp^0$. Therefore, by  induction, we get $\ind(A_{n,3})\simeq \bigvee_{2^n}(\Sp^n)\bigvee_{2^n}(\Sp^n)$. This completes the proof.

§.§ Graphs $\Lambda_{n,3}$ and $\Tilde{\Lambda}_{n,3}$

For $n\geq 0$, we define the graphs $\Lambda_{n,3}$ (see <Ref>) and $\Tilde{\Lambda}_{n,3}$ (see <Ref>) as follows:

\begin{equation*}
\begin{split}
V(\Lambda_{n,3}) = & V(\Gamma_{n,3})\cup\{g'_{1},g'_{2}\},\\
E(\Lambda_{n,3}) = & E(\Gamma_{n,3})\cup \{g'_{1}g_{11}, g'_{2}g_{21}, g'_{1}g'_{2},g'_{1}g_{21},g'_{2}g_{11}\},\quad
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
V(\Tilde{\Lambda}_{n,3}) = & V(\Gamma_{n,3})\cup\{g_{12n+2},g_{22n+2}\},\\
E(\Tilde{\Lambda}_{n,3}) = & E(\Gamma_{n,3})\cup \{g_{12n+1}g_{12n+2}, g_{22n+1}g_{22n+2}, g_{12n+2}g_{22n+2}\}.
\end{split}
\end{equation*}

in 1,2,...,8

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8

[vertex] (2) at (2*,2*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7

[edge] (1) – (1);
[edge] (2) – (2);

in 1,2,3,4,5,...,8
[edge] (1) – (2);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);

(21) – node[below=6pt] $1$(23);
(23) – node[below=6pt] $2$(25);
(26) – node[below=6pt] $n$(28);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+1}}$] at (18) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+1}}$] at (28) ;

[vertex] (e1) at (-2,6)  ;
[vertex] (e2) at (-2,4)  ;

[edge] (e1) – (e2) ;
[edge] (e1) – (11) ;
[edge] (e2) – (21) ;
[edge] (e1) – (21) ;
[edge] (e2) – (11) ;

[label=[label distance=-3pt]180:$g'_{1}$] at (e1) ;
[label=[label distance=-3pt]180:$g'_{2}$] at (e2) ;
[] at (8,1) (b) $\Lambda_{n,3}$ for $n\geq 1$;

[shift = (-5,0)]

[vertex] (e1) at (-2,6)  ;
[vertex] (e2) at (-2,4)  ;

in 1,2
[vertex] (x1) at (0,2*(4-) ;

[evaluate = as using int(+1)] in 1
[edge] (x1) – (x1) ;

in 1,2

[label=[label distance=-3pt]0:$g_{\y 1}$] at (x1) ;

[edge] (e1) – (e2) ;
[edge] (e1) – (x11) ;
[edge] (e2) – (x21) ;
[edge] (e1) – (x21) ;
[edge] (e2) – (x11) ;

[label=[label distance=-3pt]180:$g'_{1}$] at (e1) ;
[label=[label distance=-3pt]180:$g'_{2}$] at (e2) ;

[label=[label distance=-8pt]45:$g_{21}$] at (21) ;
[] at (0,1) (a) $\Lambda_{0,3}$;

in 1,2,...,8,9

[vertex] (1) at (2*,3*2) ;

in 1,2,...,8,9

[vertex] (2) at (2*,2*2) ;

[evaluate=as using int(+1)] in 1,2,...,4,6,7,8

[edge] (1) – (1);
[edge] (2) – (2);

in 1,2,3,4,5,...,8,9
[edge] (1) – (2);

[evaluate==int(1+)] in 2,4,7

[edge] (1) – (2);

[evaluate==int(-1)] in 3,5,8

[edge] (1) – (2);

(c15) at (15);
(c16) at (16);

[dotted, thick,shorten >=4pt, shorten <=4pt] (10,5) – (12,5);

(21) – node[below=6pt] $1$(23);
(23) – node[below=6pt] $2$(25);
(26) – node[below=6pt] $n$(28);

[label=[label distance=-3pt]90:$g_{11}$] at (11) ;
in 2,...,5

[label=[label distance=-3pt]90:$g_{1\x}$] at (1) ;

[label=[label distance=-3pt]180:$g_{21}$] at (21) ;

[label=[label distance=-3pt]90:$g_{1\,{2n+2}}$] at (19) ;
[label=[label distance=-3pt]90:$\cdots$] at (16) ;

[label=[label distance=-3pt]0:$g_{2\,{2n+2}}$] at (29) ;
[] at (8,1) (b) $\Tilde{\Lambda}_{n,3}$ for $n\geq 1$;

[shift = (-5,0)]
in 1,2
[vertex] (1) at (0,2*(4-) ;
in 1,2
[vertex] (2) at (2,2*(4-) ;

[evaluate = as using int(+1)] in 1
[edge] (1) – (1) ;
[edge] (2) – (2);
[edge] (1) – (1);
[edge] (2) – (2);

in 1,2

[label=[label distance=-3pt]180:$g_{\y 1}$] at (1) ;

in 1,2

[label=[label distance=-3pt]0:$g_{\y 2}$] at (2) ;

[] at (0,1) (a) $\Tilde{\Lambda}_{0,3}$;

For $n\geq0$, $ \ind(\Lambda_{n,3}) \simeq \bigvee_{2^{n+2}-1} \Sp^n.$

The proof is by induction on $n$. Clearly,  $\ind(\Lambda_{0,3}) \simeq \bigvee_{3}\Sp^0$. Assume that $n \geq 1$ and the <Ref> holds for all $0 \leq k < n$. Since $g'_{1}$ is a simplicial vertex and $N(g'_1)=\{g'_2, g_{11},g_{21}\}$ in the graph $\Lambda_{n,3}$ (see <Ref>), we have

\begin{equation*}
\begin{split}
\ind(\Lambda_{n,3}) & \simeq \Sigma\ind(\Lambda_{n,3} \setminus N[g'_2]) \vee \Sigma\ind(\Lambda_{n,3} \setminus N[g_{11}]) \vee \Sigma\ind(\Lambda_{n,3} \setminus N[g_{21}]).
\end{split}
\end{equation*}

Observe that the graph $\Lambda_{n,3} \setminus N[g'_2]$ is isomorphic to $\Lambda_{n-1,3}$, whereas both the graphs $\Lambda_{n,3} \setminus N[g_{11}]$ and $\Lambda_{n,3} \setminus N[g_{21}]$ are isomorphic to $A_{n-1,3}$. Therefore, we get
$$\ind(\Lambda_{n,3})\simeq \Sigma\ind(\Lambda_{n-1,3}) \vee \Sigma\ind(A_{n-1,3}) \vee \Sigma\ind(A_{n-1,3}).$$
Now, from the induction hypothesis and <Ref>, we conclude the following:

\begin{equation*}
\begin{split}
\ind(\Lambda_{n,3}) & \simeq \bigvee_{2^{n+1}-1}(\Sigma \Sp^{n-1})\bigvee_{2^{n}} (\Sigma \Sp^{n-1})\bigvee_{2^{n}} (\Sigma \Sp^{n-1} )
\simeq \bigvee_{2^{n+2}-1}\Sp^{n}.
\end{split}
\end{equation*}

For $n \geq 0 $, $\ind(\Tilde{\Lambda}_{n,3}) \simeq \Sp^{n}$.

It is evident that the graphs $\Gamma_{n,3}$ and $\Tilde{\Lambda}_{n,3}$ exhibit similarities (see <Ref> and <Ref>). Moreover, it is clear from the edge set of $\Tilde{\Lambda}_{n,3}$ that the graph $\Tilde{\Lambda}_{n,3}$ can be obtained by the addition of three edges at rightmost end to the graph $\Gamma_{n,3}$. Therefore, the proof for $\ind(\Tilde{\Lambda}_{n,3})$ is verbatim to the proof of <Ref>.

For $n\geq 0$,  $ M(P_{2n+2} \times P_3) \simeq \Sp^{2n+1}$ and $ M(P_{2n+3} \times P_3) \simeq \bigvee_{2^{n+2} - 1} \Sp^{2n+1}.$

Since the line graph of the categorical product $P_{2n+2} \times P_3$ is the disjoint union of two copies of $\Gamma_{n,3}$, using <Ref> and <Ref>, we get

$$ M(P_{2n+2} \times P_3)\simeq \ind(\Gamma_{n,3})\ast \ind(\Gamma_{n,3}) \simeq \Sp^{2n+1}.$$

Similarly, the line graph of the categorical product $P_{2n+3} \times P_3$ is the disjoint union of $\Lambda_{n,3}$ and $\Tilde{\Lambda}_{n,3}$. Therefore, using <Ref>, <Ref>, and <Ref> we get

$$ M(P_{2n+3} \times P_3) \simeq \ind(\Lambda_{n,3})\ast \ind(\Tilde{\Lambda}_{n,3})\simeq \bigvee_{2^{n+2} - 1} \Sp^{2n+1}.$$

§ FUTURE DIRECTIONS

We proved that for $n \geq 3$ and $3 \leq m \leq 5$, the matching complex of $P_n \times P_m$ is homotopy equivalent to a wedge of spheres. For $m =3$, we have given a closed-form formula for the number and dimension of spheres appearing in the wedge. Further, for $m \in \{4, 5\}$,  we have given the minimum and maximum dimension of spheres appearing in the wedge in $\mathsf{M}(P_n \times P_m)$. This exploration naturally leads to the following question:

Can we determine the dimensions of the spheres appearing in the wedge representing the homotopy types of the matching complexes of categorical products $P_n\times P_4$ and $P_n\times P_5$?

Based on the main result of this article and our computer-based computations, we propose the following:

For $n\geq 3$ and  $m\geq 6$, the matching complex of categorical product $P_n\times P_m$ is homotopy equivalent to a wedge of spheres.

For odd $m, n$, one component of $P_n \times P_m$ is an induced subgraph of $P_n \Box P_m$  (see <Ref>). Since the homotopy type of the matching complexes of $P_n \Box P_m$ is not known for $m \geq 4$, it will be interesting to see if there is any relation between the topology of $\M(P_n \times P_n)$ and $\M(P_n \Box P_m)$.

Is there any correlation between the matching complexes of the cartesian product of path graphs and those of the categorical product of path graphs?

§ ACKNOWLEDGMENT

The authors express gratitude to NCM and CMI for organizing the workshop titled “Cohen Macaulay simplicial complexes in graph theory", which provided an opportunity for the authors to convene and initiate the current work.
The first author is financially supported by CSIR(India). The second author is supported by the Prime Minister's Research Fellowship (PMRF/1401215). The third author is supported by HTRA-fellowship by IIT-Madras. The fourth author is supported by the seed grant project IITM/SG/SMS/95 by IIT Mandi, India.

[1]
Michal Adamaszek.
Splittings of independence complexes and the powers of cycles.
J. Combin. Theory Ser. A, 119(5):1031–1047, 2012.

[2]
Benjamin Braun and Wesley K. Hough.
Matching and independence complexes related to small grids.
Electron. J. Combin., 24(4):Paper No. 4.18, 20, 2017.

[3]
Margaret Bayer, Marija Jelić Milutinović, and Julianne Vega.
General polygonal line tilings and their matching complexes.
Discrete Math., 346(7):Paper No. 113428, 12, 2023.

[4]
A. Björner, L. Lovász, S. T. Vrećica, and R. T. Živaljević.
Chessboard complexes and matching complexes.
J. London Math. Soc. (2), 49(1):25–39, 1994.

[5]
S. Bouc.
Homologie de certains ensembles de $2$-sous-groupes des groupes
J. Algebra, 150(1):158–186, 1992.

[6]
Alexander Engström.
Independence complexes of claw-free graphs.
European J. Combin., 29(1):234–241, 2008.

[7]
Peter Freedman Garst.
Cohen-Macaulay complexes and group actions.
ProQuest LLC, Ann Arbor, MI, 1979.
Thesis (Ph.D.)–The University of Wisconsin - Madison.

[8]
Shuchita Goyal, Samir Shukla, and Anurag Singh.
Matching complexes of $3 \times n$ grid graphs.
Electron. J. Combin., 28(4):Paper No. 4.16, 26, 2021.

[9]
Allen Hatcher.
Algebraic topology.
Cambridge University Press, Cambridge, 2002.

[10]
Jakob Jonsson.
Matching complexes on grids.
unpublished manuscript available at
https://people.kth.se/ jakobj/doc/thesis/grid.pdf, 2005.

[11]
Jakob Jonsson.
Simplicial complexes of graphs, volume 1928 of Lecture
Notes in Mathematics.
Springer-Verlag, Berlin, 2008.

[12]
Dmitry N. Kozlov.
Complexes of directed trees.
J. Combin. Theory Ser. A, 88(1):112–122, 1999.

[13]
Dmitry Kozlov.
Combinatorial algebraic topology, volume 21 of Algorithms
and Computation in Mathematics.
Springer, Berlin, 2008.

[14]
Takahiro Matsushita.
Matching complexes of small grids.
Electron. J. Combin., 26(3):Paper No. 3.1, 8, 2019.

[15]
Takahiro Matsushita.
Matching complexes of polygonal line tilings.
Hokkaido Math. J., 51(3):339–359, 2022.

[16]
Marija Jelić Milutinović, Helen Jenne, Alex McDonough, and Julianne
Matching complexes of trees and applications of the matching tree
Ann. Comb., 26(4):1041–1075, 2022.

[17]
Mario Marietti and Damiano Testa.
A uniform approach to complexes arising from forests.
Electron. J. Combin., 15(1):Research Paper 101, 18, 2008.

[18]
Rade T. Živaljević and Siniša T. Vrećica.
The colored Tverberg's problem and complexes of injective
J. Combin. Theory Ser. A, 61(2):309–318, 1992.

[19]
Michelle L Wachs.
Topology of matching, chessboard, and general bounded degree graph
Algebra Universalis, 49(4):345–385, 2003.

[20]
Douglas B. West.
Introduction to Graph Theory.
Prentice Hall, Upper Saddle River, N.J., second edition, 2001.