Let $\mu_{K}\to W_{D}$ be the inclusion of the meridian $\mu_{K}$ of $K$. Then
we have $H_{*}(W_{D},*\times\mu_{K};\mathbb{Z})=H_{*}(W_{D},D\times
S^{1};\mathbb{Z})=H_{*}(X,D\times D^{2})=0$ by excision and the hypothesis
that $X$ be a homology $4$-ball. By the homology long exact sequence for the
pair $(W_{D},\mu_{K})$, the meridional map $\mu_{K}\to W_{D}$ induces a
homology equivalence, so $W_{D}$ is a homology circle. ∎

######  Proposition 15.4.

The exterior $W_{D}$ of a slice disc $D$ is homotopy equivalent to a finite
$3$-dimensional CW complex. In particular the homology groups

$H_{*}(W_{D};\mathbb{Z}[t^{\pm 1}]),\,\,H_{*}(W_{D},N_{K};\mathbb{Z}[t^{\pm
1}])\text{ and }H_{*}(N_{K};\mathbb{Z}[t^{\pm 1}])$

are all finitely generated.

###### Proof.

Note that $W_{D}$ is a compact 4-manifold with nonempty boundary. It follows
from Theorem 4.5 that $W_{D}$ is homotopy equivalent to a 3-dimensional CW
complex. The statements regarding the homology groups follow from Proposition
A.9. ∎

######  Proposition 15.5.

  1. (1)

For any knot $K$ in a homology 3-sphere the modules
$H_{*}(N_{K};\mathbb{Z}[t^{\pm 1}])$ are $\mathbb{Z}[t^{\pm 1}]$-torsion.

  2. (2)

If $D$ is a slice disc, then all the modules $H_{*}(W_{D};\mathbb{Z}[t^{\pm
1}])$ are $\mathbb{Z}[t^{\pm 1}]$-torsion.

###### Proof.

We start out with the proof of the second statement. Let
$P\subset\mathbb{Z}[t^{\pm 1}]$ be the multiplicative subset of Laurent
polynomials that augment to $\pm 1$, that is $p(1)=\pm 1$ if and only if $p\in
P$. We shall prove the slightly stronger statement, that
$H_{k}(W_{D};\mathbb{Z}[t^{\pm 1}])$ is $P$-torsion for $k>0$. Since
$H_{0}(W_{D};\mathbb{Z}[t^{\pm 1}])\cong\mathbb{Z}[t^{\pm 1}]/(t-1)$ is
$\mathbb{Z}[t^{\pm 1}]$-torsion, the result will follow. We write
$\pi=\pi_{1}(W_{D})$. Let $Q:=P^{-1}\mathbb{Z}[t^{\pm 1}]$ be the result of
inverting the polynomials in $P$. By Proposition A.5 there exists a chain
complex $C_{*}$ of finite length consisting of finitely generated free left
$\mathbb{Z}[\pi]$-modules such that for any ring $R$ and any
$(R,\mathbb{Z}[\pi])$-bimodule $A$ we have

$H_{k}(W_{D},\mu_{K};A)\,\,\cong\,\,H_{k}(A\otimes_{\mathbb{Z}[\pi]}C_{*}).$

By Proposition 15.3 we know that
$H_{k}(\mathbb{Z}\otimes_{\mathbb{Z}[\pi]}C_{*})=H_{k}(W_{D},\mu_{K};\mathbb{Z})=0$.
Since $C_{*}$ is a chain complex of finite length consisting of finitely
generated free left $\mathbb{Z}[\pi]$-modules we obtain from chain homotopy
lifting [COT03, Proposition 2.10], see also [NP17, Lemma 3.1], that
$H_{k}(Q\otimes_{\mathbb{Z}[\pi]}C_{*})=0$. A straightforward calculation
shows that $H_{*}(S^{1},\operatorname{pt};Q)=0$. It follows that
$H_{*}(W_{D},\operatorname{pt};Q)=0$, so that $H_{k}(W_{D};\mathbb{Z}[t^{\pm
1}])$ is $P$-torsion for $k>0$.

The first statement is very well known. One of the many proofs would be to use
the above argument and the fact that $S^{1}\to Y\setminus\nu K$ is a homology
equivalence to show that the modules $H_{*}(Y\setminus\nu K;\mathbb{Z}[t^{\pm
1}])$ are torsion. A basic Mayer-Vietoris argument then shows that the modules
$H_{*}(N_{K};\mathbb{Z}[t^{\pm 1}])$ are also torsion. ∎

We want to recall the definition of the Alexander polynomial of a knot. To do
so we need the notion of the order of a module.

######  Definition 15.6.

Let $H$ be a finitely generated free abelian group and let $M$ be a finitely
generated $\mathbb{Z}[H]$-module. By [Lan02a, Corollary IV.9.5] the ring
$\mathbb{Z}[H]$ is Noetherian which implies that $M$ admits a free resolution

$\mathbb{Z}[H]^{r}\,\,\xrightarrow{\cdot
A}\,\,\mathbb{Z}[H]^{s}\,\,\to\,\,M\,\,\to\,\,0.$

Without loss of generality we can assume that $r>s$. Since $\mathbb{Z}[H]$ is
unique factorisation domain, see [Lan02a, Lemma IV.2.3], the order
$\operatorname{ord}(M)$ is defined as the greatest common divisor of the
$s\times s$-minors of $A$. By [Tur01, Lemma 4.4] the order is well-defined,
i.e. independent of the choice of the free resolution, up to multiplication by
a unit in $\mathbb{Z}[H]$.

In the proof of the Fox-Milnor theorem we will need the following lemma,
collecting basic facts about orders of finitely generated
$\mathbb{Z}[H]$-modules.

######  Lemma 15.7.

Let $H$ be a finitely generated free abelian group.

  1. (1)

If $0\to A\to B\to C\to 0$ is a short exact sequence of finitely generated
$\mathbb{Z}[H]$-modules, then
$\operatorname{ord}(B)=\operatorname{ord}(A)\cdot\operatorname{ord}(C)$.

  2. (2)

If $0\to C_{k}\to\dots\to C_{0}\to 0$ is an exact sequence of finitely
generated $\mathbb{Z}[H]$-torsion modules, then the alternating product of the
orders is a unit in $\mathbb{Z}[H]$.

  3. (3)

For any finitely generated $\mathbb{Z}[H]$-module $A$ we have
$\operatorname{ord}(\overline{A})=\overline{\operatorname{ord}(A)}$.

  4. (4)

For any finitely generated torsion $\mathbb{Z}[H]$-module $A$ we have
$\operatorname{ord}(\operatorname{Ext}^{1}_{\mathbb{Z}[H]}(A,\mathbb{Z}[H]))=\operatorname{ord}(A)$.

###### Proof.

Statement (1) is proved for $H\cong\mathbb{Z}$ in [Lev67, Lemma 5]. The
general case follows from [Hil12, Theorem 3.12]. Note that (2) is an immediate
consequence of (1), by separating the long exact sequence into short exact
sequences such at $0\to\operatorname{Im}C_{j}\to
C_{j-1}\to\operatorname{Im}C_{j-1}\to 0$, applying (1), and performing
substitutions using the resulting equations involving orders.

Next (3) follows immediately from the definition. Finally (4) is well-known to
the experts, but we could not find a reference, therefore we sketch the key
ingredients in the proof. We introduce the following notation.

  1. (a)

Given any prime ideal $\mathfrak{p}$ of $\mathbb{Z}[H]$, let
$\mathbb{Z}[H]_{\mathfrak{p}}$ be the localisation at $\mathfrak{p}$, that is
we invert all elements that do not lie in $\mathfrak{p}$. We view
$\mathbb{Z}[H]$ as a subring of $\mathbb{Z}[H]_{\mathfrak{p}}$.

  2. (b)

Given a ring $R$ and $f,g\in R$ we write $f\doteq_{R}g$ if $f$ and $g$ differ
by multiplication by a unit in $R$.

Now we sketch the proof of (4). We will use the following five observations.

  1. (i)

First note that since $\mathbb{Z}[H]$ is a unique factorisation domain, for
any prime element $p\in\mathbb{Z}[H]$ the ideal $(p)$ is a prime ideal.

  2. (ii)

Note that being unique factorisation domain and being Noetherian is preserved
under localisation, see [Pes96, Theorem 7.53] and [Row06, Corollary 8.8’]. In
particular each $\mathbb{Z}[H]_{\mathfrak{p}}$ is a noetherian unique
factorisation domain. This allows us, by the same definitions as above, to
define the order of a finitely generated module over
$\mathbb{Z}[H]_{\mathfrak{p}}$.

  3. (iii)

Localisation is flat [Lan02a, Proposition XVI.3.2]. It follows that for any
finitely generated $\mathbb{Z}[H]$-module $M$ and any prime element
$p\in\mathbb{Z}[H]$ one has
$\operatorname{ord}(M)\doteq_{\mathbb{Z}[H]_{(p)}}\operatorname{ord}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}M)$
and
$\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}\operatorname{Ext}_{\mathbb{Z}[H]}^{1}(M,\mathbb{Z}[H])\cong\operatorname{Ext}_{\mathbb{Z}[H]_{(p)}}^{1}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}M,\mathbb{Z}[H]_{(p)})$
as $\mathbb{Z}[H]_{(p)}$-modules.

  4. (iv)

By [Osb00, Corollary A.14] any commutative ring that has the property that
each prime ideal is principal, is a PID. It follows easily that for each prime
element $p$, the localisation $\mathbb{Z}[H]_{(p)}$ is a PID.

  5. (v)

Let $L$ be a torsion $\mathbb{Z}[H]_{(p)}$-module. Since $\mathbb{Z}[H]_{(p)}$
is a PID every two elements have a greatest common divisor. We can therefore
perform row and column operations to find a resolution for $L$ such that the
presentation matrix is diagonal. From this observation one easily deduces that
$L\cong\operatorname{Ext}_{\mathbb{Z}[H]_{(p)}}^{1}(L,\mathbb{Z}[H]_{(p)})$ as
left $\mathbb{Z}[H]_{(p)}$-modules. Since $L$ is torsion the presentation
matrix is injective and so its transpose presents the $\operatorname{Ext}$
group. To convert the $\operatorname{Ext}$ group to a left module, we use the
trivial involution, which we may do since $\mathbb{Z}[H]_{(p)}$ is a
commutative ring.

  6. (vi)

Suppose that $f$ and $g$ are in $\mathbb{Z}[H]$. If
$f\doteq_{\mathbb{Z}[H]_{(p)}}g$ for all prime elements $p\in\mathbb{Z}[H]$,
then since $\mathbb{Z}[H]$ is a unique factorisation domain we must have
$f\doteq_{\mathbb{Z}[H]}g$.

Now with $L=\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}A$ a finitely generated
$\mathbb{Z}[H]$-torsion module, we have

$\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}A\cong\operatorname{Ext}_{\mathbb{Z}[H]_{(p)}}^{1}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}A,\mathbb{Z}[H]_{(p)})$

for every prime element $p$, by (iv). On the other hand, again for each prime
element $p$, we have

$\operatorname{ord}(A)\doteq_{\mathbb{Z}[H]_{(p)}}\operatorname{ord}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}A)$

by (iii). Combining these two observations yields

$\operatorname{ord}(A)\doteq_{\mathbb{Z}[H]_{(p)}}\operatorname{ord}(\operatorname{Ext}_{\mathbb{Z}[H]_{(p)}}^{1}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}A,\mathbb{Z}[H]_{(p)})).$

By the second part of (iii) we have that

$\operatorname{ord}(\operatorname{Ext}_{\mathbb{Z}[H]_{(p)}}^{1}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}A,\mathbb{Z}[H]_{(p)}))\doteq_{\mathbb{Z}[H]_{(p)}}\operatorname{ord}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}\operatorname{Ext}^{1}_{\mathbb{Z}[H]}(A,\mathbb{Z}[H])).$

By the first part of (iii) again we have

$\operatorname{ord}(\mathbb{Z}[H]_{(p)}\otimes_{\mathbb{Z}[H]}\operatorname{Ext}^{1}_{\mathbb{Z}[H]}(A,\mathbb{Z}[H]))\doteq_{\mathbb{Z}[H]_{(p)}}\operatorname{ord}(\operatorname{Ext}^{1}_{\mathbb{Z}[H]}(A,\mathbb{Z}[H])).$

Thus combining the last three equalities we have

$\operatorname{ord}(A)\doteq_{\mathbb{Z}[H]_{(p)}}\operatorname{ord}(\operatorname{Ext}^{1}_{\mathbb{Z}[H]}(A,\mathbb{Z}[H]))$

for all prime elements $p$. Now (4) follows by applying (vi). ∎

We use the notion of order to define the Alexander polynomial of a knot in a
homology $3$-sphere.

######  Definition 15.8.

The _Alexander polynomial $\Delta_{K}(t)$_ of a knot $K$ is defined as the
order of the Alexander module $H_{1}(N_{K};\mathbb{Z}[t^{\pm 1}])$. Note that
this polynomial is only well-defined up to units in $\mathbb{Z}[t^{\pm}]$.

After these preparations we turn to the actual proof of the Fox-Milnor Theorem
15.2. We need the following elementary lemma.

######  Lemma 15.9.

Let $\pi$ be a group, let $C_{*}$ be a chain complex of left free
$\mathbb{Z}[\pi]$-modules and let $\phi\colon\pi\to\langle t\rangle$ be a
homomorphism. The map

$\begin{array}[]{rcl}\operatorname{Hom}_{\operatorname{right-}\mathbb{Z}[\pi]}(\overline{C_{*}},\mathbb{Z}[t^{\pm
1}])&\to&\overline{\operatorname{Hom}_{\operatorname{left-}\mathbb{Z}[t^{\pm
1}]}(\mathbb{Z}[t^{\pm 1}]\otimes_{\mathbb{Z}[\pi]}C_{*},\mathbb{Z}[t^{\pm
1}])}\\\ f&\mapsto&(p\otimes\sigma\mapsto
p\cdot\overline{f(\sigma)})\end{array}$

is well-defined and is an isomorphism of left $\mathbb{Z}[t^{\pm 1}]$-cochain
complexes.∎

###### First proof of the Fox-Milnor theorem 15.2.

In this proof we abbreviate $\Lambda:=\mathbb{Z}[t^{\pm 1}]$. We start out
with the following three observations.

  1. (a)

We have $H_{0}(W_{D};\Lambda)\cong H_{0}(N_{K};\Lambda)\cong\Lambda/(t-1)$.

  2. (b)

We have $H_{0}(W_{D},N_{K};\Lambda)=0$,

  3. (c)

By Proposition 15.5 and Proposition 15.4 we know that for all $k$

$\operatorname{Ext}^{0}_{\Lambda}(H_{k}(W_{D},N_{K};\Lambda),\Lambda)\,\,=\,\,\operatorname{Hom}_{\Lambda}(H_{k}(W_{D},N_{K};\Lambda),\Lambda)\,\,=\,\,0.$

######  Claim.

For any $i\in\mathbb{N}$ we have

$\begin{array}[]{rcl}H_{i}(N_{K};\Lambda)&\cong&\overline{\operatorname{Ext}^{1}_{\Lambda}(H_{2-i}(N_{K};\Lambda),\Lambda)}\\\
H_{i}(W_{D};\Lambda)&\cong&\overline{\operatorname{Ext}^{1}_{\Lambda}(H_{3-i}(W_{D},N_{K};\Lambda),\Lambda)}.\end{array}$

We prove the second statement of the claim. The proof of the first statement
is almost identical. By the Poincaré duality theorem A.15 we have an
isomorphism $H_{i}(W_{D};\Lambda)\cong H^{4-i}(W_{D},N_{K};\Lambda)$ of
$\Lambda$-modules. By Lemma 15.9, applied to
$C_{*}=C_{*}(W_{D},N_{K};\mathbb{Z}[\pi])$, we know that

$H^{4-i}(W_{D},N_{K};\Lambda)\,\,\cong\,\,\overline{H_{4-i}(\operatorname{Hom}_{\Lambda}(\Lambda\otimes_{\mathbb{Z}[\pi]}C_{*}(W_{D},N_{K};\mathbb{Z}[\pi]),\Lambda))}.$

Finally we apply the universal coefficient spectral sequence [Lev77, Theorem
2.3] to the $\Lambda$-module chain complex $C_{*}(W_{D},N_{K};\Lambda)$. It
follows from the above observations (b) and (c) that the spectral sequence
collapses and that we have an isomorphism

$H_{4-i}\big{(}\operatorname{Hom}_{\Lambda}(\Lambda\otimes_{\mathbb{Z}[\pi]}C_{*}(W_{D},N_{K};\Lambda)\big{)}\,\,\cong\,\,\operatorname{Ext}^{1}_{\Lambda}(H_{3-i}(W_{D},N_{K};\Lambda),\Lambda).$

This concludes the proof of the claim.

Next we consider the long exact sequence of the pair $(W_{D},N_{K})$ of
twisted homology with $\Lambda$-coefficients:

$\dots\to H_{2}(W_{D};\Lambda)\to H_{2}(W_{D},N_{K};\Lambda)\to
H_{1}(N_{K};\Lambda)\to H_{1}(W_{D};\Lambda)\to
H_{1}(W_{D},N_{K};\Lambda)\to\dots$

It follows from Propositions 15.4 that all the above modules are finitely
generated modules. Thus it makes sense to consider their orders. Also note
that in Proposition 15.5 we saw that the modules for $N_{K}$ and $W_{D}$ are
all $\Lambda$-torsion. It follows from the long exact sequence that the
relative homology groups $H_{*}(W_{D},N_{K};\Lambda)$ are also
$\Lambda$-torsion. By Lemma 15.7 (3) the alternating product of the orders
equals $\pm t^{k}$.

By the above claim and Lemma 15.7 (3) and (4) the orders are anti-symmetric
around $H_{1}(N_{K};\Lambda)$. More precisely, we have

$\operatorname{ord}(H_{2}(W_{D},N_{K};\Lambda))\,\,=\,\,\operatorname{ord}\big{(}\overline{\operatorname{Ext}_{\Lambda}^{1}(H_{1}(W_{D};\Lambda),\Lambda)}\big{)}\,\,=\,\,\overline{\operatorname{ord}(H_{1}(W_{D};\Lambda))},$

and the same type of relation holds as we progress further from the middle
term $H_{1}(N_{K};\Lambda)$ in the above long exact sequence. But this implies
that there exist non-zero polynomials $f,g\in\Lambda$ with
$f\cdot\overline{f}=\Delta_{K}(t)\cdot g\cdot\overline{g}$. By considering
irreducible factors, we obtain the desired result. ∎

We conclude with an alternative argument for the Fox-Milnor theorem in the
topological category using Reidemeister torsion. The advantage of the
Reidemeister torsion invariant is that proofs are often easier, and it has in
general a smaller indeterminacy than the order of homology, although this will
not manifest itself in the upcoming proof.

###### Second proof of Theorem 15.2.

We continue with the notation introduced above. As before we have a
homomorphism $\alpha\colon\pi_{1}(W_{D})\to
H_{1}(W_{D};\mathbb{Z})\xrightarrow{\simeq}\mathbb{Z}$, sending an oriented
meridian of $K$ to $1\in\mathbb{Z}$. As usual $\mathbb{Q}(t)$ denotes the
field of fractions of the Laurent polynomial ring $\mathbb{Z}[t,t^{-1}]$. We
take $d=1$, and so obtain a representation
$\phi\colon\pi_{1}(W_{D})\to\operatorname{GL}(1,\mathbb{Q}(t))$, that sends
$g\mapsto(t^{\alpha(g)})$. In the previous proof we had already seen that the
modules $H_{*}(N_{K};\mathbb{Z}[t^{\pm 1}])$, $H_{*}(W_{D};\mathbb{Z}[t^{\pm
1}])$ and $H_{*}(N_{K};\mathbb{Z}[t^{\pm 1}])$ are $\mathbb{Z}[t^{\pm
1}]$-torsion. Since $\mathbb{Q}(t)$ is flat over $\mathbb{Z}[t^{\pm 1}]$ it
follows that the corresponding twisted homology groups with
$\mathbb{Q}(t)$-coefficients are zero.

By the discussion in Section 14.3 we can consider the Reidemeister torsions
$\tau(W_{D},\phi)$, $\tau(N_{K},\phi)$ and $\tau(W_{D},N_{K},\phi)$. By
Theorem 14.14,
$\tau(W_{D},N_{K},\phi)=\overline{\tau(W_{D},\phi)}^{(-1)^{5}}=\overline{\tau(W_{D},\phi)}^{-1}$.
Since the torsion is multiplicative in short exact sequences by Theorem 14.11,
we have that

$\tau(W_{D},\phi)=\tau(N_{K},\phi)\cdot\tau(W_{D},N_{K},\phi)=\tau(M_{K},\phi)\cdot\overline{\tau(W_{D},\phi)}^{-1}.$

By [Tur01, Theorem 14.12] the torsion of the zero surgery of a knot is equal
to $\Delta_{K}(t)/((t-1)(t^{-1}-1))$. It follows that $\Delta_{K}(t)$ is a
norm as claimed. ∎

######  Remark 15.10.

The two proofs presented above avoid the use of the smooth category, and so
are in keeping with the spirit of this article. However, one can give a
further alternative proof by allowing smooth techniques. First one can use
Theorem 8.6 to find a simply connected 4-manifold $W^{\prime}$ such that
$W:=V\\#W^{\prime}$ is smoothable. Then one can triangulate $W$ and apply
Reidemeister torsion machinery without appealing to [KS77, Essay III]. The
disadvantage of this approach is that typically $H_{2}(W^{\prime};\mathbb{Z})$
will be nontrivial, so that $W$ is not acyclic over $\mathbb{Q}(t)$. One can
proceed by choosing a self-dual basis for homology, so that one can still
obtain a torsion invariant that is well-defined up to norms. Apply [CF13,
Theorem 2.4], and argue that since the intersection form of $W$ is
nonsingular, the contribution of $W^{\prime}$ to the torsion is a norm.

##  Appendix A Poincaré Duality with twisted coefficients

Surveying the literature, we felt it would be of benefit to have a more
detailed proof of Poincaré duality with twisted coefficients for manifolds
with boundary, but without a smooth or $\operatorname{PL}$ structure, so we
offer one in this appendix. One can find other proofs of Poincaré duality for
some subsets of these conditions.

###  A.1. Twisted homology and cohomology groups

We start out with the following notation.

######  Notation A.1.

Given a group $\pi$ and a left $\mathbb{Z}[\pi]$-module $A$, write
$\overline{A}$ for the right $\mathbb{Z}[\pi]$-module that has the same
underlying abelian group but for which the right action of $\mathbb{Z}[\pi]$
is defined by $a\cdot g:=g^{-1}\cdot a$ for $a\in A$ and $g\in\pi$. The same
notation is also used with the rôles of left and right reversed and $g\cdot
a:=a\cdot g^{-1}$.

We recall the definition of twisted homology and cohomology groups.

######  Definition A.2.

Let $X$ be a connected topological space that admits a universal cover
$p\colon\widetilde{X}\to X$. Write $\pi=\pi_{1}(X)$. Let $Y$ be a subset of
$X$, let $A$ be a right $\mathbb{Z}[\pi]$-module. Let $\pi$ act on
$\widetilde{X}$ by deck transformations, which is naturally a left action.
Thus, the singular chain complex $C_{*}(\widetilde{X},p^{-1}(Y))$ becomes a
left $\mathbb{Z}[\pi]$-module chain complex. Define the _twisted chain
complex_

$C_{*}(X,Y;A)\,\,:=\,\,\big{(}A\otimes_{\mathbb{Z}[\pi]}C_{*}(\widetilde{X},p^{-1}(Y)),\operatorname{Id}\otimes\partial_{*}\big{)}.$

The corresponding _twisted homology groups_ are $H_{k}(X,Y;A)$. With
$\delta^{k}=\operatorname{Hom}(\partial_{k},\operatorname{Id})$ define the
_twisted cochain complex_ to be

$C^{*}(X,Y;A)\,\,:=\,\,\big{(}\operatorname{Hom}_{\operatorname{\textnormal{right-}}\mathbb{Z}[\pi]}\big{(}\overline{C_{*}(\widetilde{X},p^{-1}(Y)}),A\big{)},\delta^{*}\big{)}.$

The corresponding _twisted cohomology groups_ are $H^{k}(X,Y;A)$.

Note that if $R$ is some ring (not necessarily commutative) and if $A$ is an
$(R,\mathbb{Z}[\pi])$-bimodule, then the above twisted homology and cohomology
groups are naturally left $R$-modules.

Given a CW complex one can similarly define twisted cellular (co-) chain
complexes and twisted cellular (co-) homology groups. The following
proposition implies that twisted singular (co-) homology groups are isomorphic
to twisted cellular (co-) homology groups.

######  Proposition A.3.

Let $(X,Y)$ be a CW complex pair and write $\pi=\pi_{1}(X)$. The singular
chain complexes $C_{*}^{\operatorname{sing}}(X,Y;\mathbb{Z}[\pi])$ and
$C_{*}^{\operatorname{cell}}(X,Y;\mathbb{Z}[\pi])$ are chain equivalent as
chain complexes of left $\mathbb{Z}[\pi]$-modules.

The proof of Proposition A.3 relies on the following very useful lemma.

######  Lemma A.4.

Let $f\colon C_{*}\rightarrow D_{*}$ be a chain map of free
$\mathbb{Z}[\pi]$-module chain complexes $($here chain complexes are
understood to start in degree 0$)$ that induces an isomorphism on homology.
Then $f$ is a chain equivalence.

###### Proof.

Since $f$ induces an isomorphism of homology groups we know that the
$\operatorname{cone}(f)_{\ast}$ is acyclic. By assumption $C_{\ast}$ and
$D_{\ast}$ are free $\mathbb{Z}[\pi]$-modules and so is
$\operatorname{cone}(f)_{\ast}$. But this guarantees the existence of a chain
homotopy $\operatorname{Id}_{\operatorname{cone}{f}_{\ast}}\simeq_{P}0$, since
we can view $\operatorname{cone}(f)_{*}$ as a free resolution of $0$ and any
two such resolutions are chain homotopic. Recall that chain homotopy means

(A.1) $\displaystyle\partial^{\operatorname{cone}{f}_{\ast}}\circ
P+P\circ\partial^{\operatorname{cone}{f}_{\ast}}=\operatorname{Id}_{\operatorname{cone}{f}_{\ast}}$

If we write $P$ as a matrix

$P_{n}=\begin{pmatrix}P^{11}_{n}&P^{12}_{n}\\\
P^{21}_{n}&P^{22}_{n}\end{pmatrix}\colon\begin{array}[]{c}C_{n-1}\\\
D_{n}\end{array}\to\begin{array}[]{c}C_{n}\\\ D_{n+1}\end{array}$

then one easily verifies using Equation (A.1), that $P^{12}_{*}\colon D_{*}\to
C_{*}$ is a chain homotopy inverse of $f_{*}$, where the chain homotopies are
given by $P^{11}_{*}$ and $P^{22}_{*}$. ∎

###### Proof of Proposition A.3.

In [Sch68, p. 303] (see also [Lüc98, Lemma 4.2]) it is shown that there exists
a natural chain homotopy equivalence
$C_{*}^{\operatorname{sing}}(\widetilde{X},p^{-1}(Y))\to
C_{*}^{\operatorname{cell}}(\widetilde{X},p^{-1}(Y))$ of $\mathbb{Z}$-modules,
where $p\colon\widetilde{X}\to X$ denotes the universal cover. Since the chain
homotopy is natural it is in particular $\pi$-equivariant. In other words, it
is a chain homotopy equivalence of $\mathbb{Z}[\pi]$-modules. Now the
proposition follows from Lemma A.4. ∎

######  Proposition A.5.

Let $M$ be a compact $n$-manifold and let $N\subset M$ be a subspace that is a
compact manifold in its own right. Write $\pi=\pi_{1}(M)$. There exists a
chain complex $C_{*}$ of finite length consisting of finitely generated free
left $\mathbb{Z}[\pi]$-modules such that for any ring $R$ and for any
$(R,\mathbb{Z}[\pi])$-bimodule $A$ we have left $R$-module isomorphisms

$H_{k}(M,N;A)\,\,\cong\,\,H_{k}(A\otimes_{\mathbb{Z}[\pi]}C_{*})$

and

$H^{k}(M,N;A)\,\,\cong\,\,H_{k}(\operatorname{Hom}_{\mathbb{Z}[\pi]}(\overline{C_{*}},A)).$

######  Remark A.6.

Note that we do _not_ demand that $N$ be a submanifold of $M$. For example $N$
could be a union of boundary components of $M$, or $N$ could be a submanifold
of the boundary. Evidently $N$ could also be the empty set.

###### Proof.

By Theorem 4.5 the manifolds $M$ and $N$ are homotopy equivalent to finite CW
complexes $X$ and $Y$ respectively. Let $i\colon N\to M$ be the inclusion map.
By the cellular approximation theorem there exists a cellular map $j\colon
X\to Y$ such that the following diagram commutes up to homotopy

$\textstyle{N\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\scriptstyle{\simeq}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j}$$\textstyle{X.}$

Next we replace $M$ and $X$ by the mapping cylinders of $i$ and $j$
respectively, to create cofibrations. Given a map $f\colon U\to V$ between
topological spaces let $\operatorname{cyl}(f)$ be the mapping cylinder. We
view $U$ as a subset of $\operatorname{cyl}(f)$ in the obvious way. With this
notation we have

$H_{k}(M,N;A)\,\cong\,H_{k}(\operatorname{cyl}(i\colon N\to
M),N;A)\,\cong\,H_{k}(\operatorname{cyl}(j\colon Y\to X),Y;A).$

The mapping cylinder $Z:=\operatorname{cyl}(j\colon Y\to X)$ admits the
structure of a finite CW complex such that $Y$ is a subcomplex. Thus we can
compute the twisted homology groups $H_{k}(\operatorname{cyl}(j\colon X\to
Y);A)$ using the relative twisted cellular chain complex, and similarly for
cohomology. Put differently,
$C_{*}=C_{*}^{\operatorname{cell}}(Z,Y;\mathbb{Z}[\pi])$ has the desired
properties. ∎

In order to give a criterion for twisted homology modules to be finitely
generated, we need the notion of a Noetherian ring.

######  Definition A.7.

A ring $R$ is said to be _left Noetherian_ if for any descending chain

$R\,\,\supseteq\,\,I_{1}\,\,\supseteq\,\,I_{2}\,\,\supseteq\,\,I_{3}\,\,\supseteq\,\,\dots$

of left $R$-ideals the inclusions eventually become equality.

######  Example A.8.

The following rings are left Noetherian:

  1. (1)

The ring $\mathbb{Z}$ is Noetherian.

  2. (2)

Any (skew) field is left Noetherian.

  3. (3)

If $A$ is a commutative Noetherian ring, then the multivariable Laurent
polynomial ring $A[t_{1}^{\pm 1},\dots,t_{k}^{\pm 1}]$ is also Noetherian
[Lan02a, Corollary IV.9.5].

The following theorem is often implicitly used.

######  Proposition A.9.

Let $M$ be a compact $n$-manifold, let $N\subset M$ be a subspace that is a
compact manifold in its own right, let $R$ be a ring and let $A$ be an
$(R,\mathbb{Z}[\pi])$-bimodule. If $R$ is Noetherian and if $A$ is a finitely
generated $R$-module, then all the twisted homology modules $H_{*}(M,N;A)$ are
finitely generated left $R$-modules.

In the proof of Proposition A.9 we will need the following lemma; cf. [Lam91,
Proposition 1.21] or [Lan02a, Proposition X.1.4].

######  Lemma A.10.

Let $R$ be a Noetherian ring. If $P$ is a finitely generated left $R$-module,
then any submodule of $P$ is also a finitely generated left $R$-module.

###### Proof of Proposition A.9.

By Proposition A.5, there exists a chain complex $C_{*}$ of finite length
consisting of finitely generated free left $\mathbb{Z}[\pi]$-modules such that

$H_{k}(M,N;A)\,\,\cong\,\,H_{k}(A\otimes_{\mathbb{Z}[\pi]}C_{*}).$

Given $k\in\mathbb{N}_{0}$ we denote the rank of $C_{k}$ as a free left
$\mathbb{Z}[\pi]$-module by $r_{k}$. Then we have
$A\otimes_{\mathbb{Z}[\pi]}C_{k}\cong
A\otimes_{\mathbb{Z}[\pi]}\mathbb{Z}[\pi]^{r_{k}}\cong A^{r_{k}}$. In
particular $H_{k}(M,N;A)$ is isomorphic to a quotient of a submodule of a
finitely generated $R$-module. The desired statement follows from Lemma A.10.
∎

###  A.2. Cup and cap products on twisted (co-) chain complexes

Throughout this section let $X$ be a connected topological space admitting a
universal cover, and write $\pi=\pi_{1}(X)$. We want to introduce the cup
product and the cap product on twisted (co-) chain complexes. Given an
$n$-simplex $\sigma$, define the $p$-simplices $\sigma\lfloor_{p}$ and
$\sigma\rfloor_{p}$ by

$\displaystyle\sigma\rfloor_{p}(t_{0},\ldots,t_{p})$
$\displaystyle:=\sigma(t_{0},\ldots,t_{p},0,\ldots,0),$
$\displaystyle\sigma\lfloor_{p}(t_{0},\ldots,t_{p})$
$\displaystyle:=\sigma(0,\ldots,0,t_{0},\ldots,t_{p}).$

Given two right $\mathbb{Z}[\pi]$-modules $A$ and $B$ we view
$A\otimes_{\mathbb{Z}}B$ as a right $\mathbb{Z}[\pi]$-module via the diagonal
action of $\pi$.

First we introduce the cup product on twisted cohomology. The following lemma
can be verified easily by hand, say along the lines of the proof of [Hat02,
Lemma 3.6].

######  Lemma A.11.

Let $Y$ be a subset of $X$. We consider the map

$\displaystyle{\acup}\colon C^{p}(X;A)\times C^{q}(X;B)$
$\displaystyle\longrightarrow C^{p+q}(X;{A}\otimes_{\mathbb{Z}}B)$
$\displaystyle(\phi,\psi)$
$\displaystyle\longmapsto\big{(}\sigma\mapsto\varphi(\sigma\lfloor_{p})\otimes_{\mathbb{Z}}\psi(\sigma\rfloor_{k-p})\big{)}.$

$($Note that the right-hand side is indeed $\mathbb{Z}[\pi]$-homomorphism,
i.e. it defines an element $C^{p+q}(X;{A}\otimes_{\mathbb{Z}}B)$.$)$
Furthermore the map descends to a well defined map

$\acup\colon H^{p}(X,Y;A)\times
H^{q}(X,Y;B)\,\,\longrightarrow\,\,H^{p+q}(X,Y;A\otimes_{\mathbb{Z}}B).$

We refer to this map as the _cup product_.

Next we introduce the cap product. As with cup product, first we define it on
the chain level.

######  Lemma A.12.

Let $X$ be a topological space and let $S,T\subset X$ be subsets. We write
$C_{k}(X,\\{S,T\\})=C_{k}(X)/(C_{k}(S)+C_{k}(T))$. The map

$\displaystyle\acap\colon C^{p}(X,S;A)\times C_{k}(X,\\{S,T\\};A)$
$\displaystyle\longrightarrow C_{k-p}(X,T;{A}\otimes_{\mathbb{Z}}B)$
$\displaystyle(\psi,b\otimes_{\mathbb{Z}[\pi]}\sigma)$
$\displaystyle\longmapsto(\psi(\sigma\lfloor_{p})\otimes_{\mathbb{Z}}b)\otimes_{\mathbb{Z}[\pi]}\sigma\rfloor_{k-p}.$

is well-defined. We refer to this map as the _cap product_.

###### Proof.

We verify that the given map respects the tensor product. Thus let $\psi\in
C^{p}(X;A)$, $\sigma\in C_{k}(\widetilde{X})$, $\gamma\in\pi$ and $b\in B$. We
calculate that

$\begin{array}[]{rll}\psi\acap
b\otimes_{\mathbb{Z}[\pi]}\gamma\sigma&=\,\,\left(\psi(\gamma\sigma\lfloor_{p})\otimes_{\mathbb{Z}}b\right)\otimes_{\mathbb{Z}[\pi]}\gamma\sigma\rfloor_{k-p}&=\,\,(\gamma\psi(\sigma\lfloor_{p})\otimes_{\mathbb{Z}}b)\cdot\gamma\otimes_{\mathbb{Z}[\pi]}\sigma\rfloor_{k-p}\\\\[2.84544pt]
&=\,\,(\gamma^{-1}\gamma\psi(\sigma\lfloor_{p})\otimes_{\mathbb{Z}}b\gamma)\otimes_{\mathbb{Z}[\pi]}\sigma\rfloor_{k-p}&=\,\,\psi\acap
b\gamma\otimes_{\mathbb{Z}[\pi]}\sigma.\end{array}$

It follows easily from the definitions that the cap product descends to the
given quotient (co-) chain complexes. ∎

######  Lemma A.13.

Let $f\in C^{p}(X;A)$ and let $c\in C_{k}(X;B)$. We have

$\partial(f\acap c)\,\,=\,\,(-1)^{p}\cdot(-\delta(f)\acap c+f\acap\partial
c)\in C_{k-1}(X;A\otimes_{\mathbb{Z}}B).$

###### Proof.

The lemma follows from a calculation using the definition of the cap product
and the boundary maps, see e.g. [Fri19, Lemma 59.1] for details. Note that the
precise signs differ from similar formulas in some textbooks in algebraic
topology since there are many different sign conventions in usage. ∎

######  Corollary A.14.

Let $X$ be a connected topological space, let $S,T\subset X$ be subsets, let
$R$ be a ring and let $A$ be an $(R,\mathbb{Z}[\pi_{1}(M)])$-bimodule. For any
cycle $\sigma\in C_{n}(X,\\{S,T\\};\mathbb{Z})$ the cap product

$\begin{array}[]{rcl}\acap[\sigma]\colon
H^{k}(X,S;A)&\to&H_{n-k}(X,T;A)=H_{n-k}(X,T;A\otimes_{\mathbb{Z}}\mathbb{Z})\\\
{}[\varphi]&\mapsto&[\varphi\acap\sigma]\end{array}$

is well-defined. Furthermore this map only depends on the homology class
$[\sigma]\in H_{n}(X,S\cup T;\mathbb{Z})$.

###  A.3. The Poincaré duality theorem

The following theorem is a generalisation of the familiar Poincaré duality for
untwisted coefficients to the case of twisted coefficients.

######  Theorem A.15 (Twisted Poincaré duality).

Let $M$ an compact, oriented, connected $n$-dimensional manifold. Let $S$ and
$T$ be codimension 0 compact submanifolds of $\partial M$ such that $\partial
S=\partial T=S\cap T$ and $\partial M=S\cup T$. Let $[M]\in H_{n}(M,\partial
M;\mathbb{Z})$ be the fundamental class of $M$. If $R$ is a ring and if $A$ is
an $(R,\mathbb{Z}[\pi_{1}(M)])$-bimodule, then the map

$-\acap[M]\colon H^{k}(M,S;{A})\,\,\rightarrow\,\,H_{n-k}(M,T;A)$

defined by Lemma A.13 is an isomorphism of left $R$-modules.

We also have the following Poincaré Duality statement on the (co-) chain
level.

######  Theorem A.16 (Universal Poincaré duality).

Let $M$ an compact, oriented, connected $n$-dimensional manifold. Let $S$ and
$T$ be codimension 0 compact submanifolds of $\partial M$ such that $\partial
S=\partial T=S\cap T$ and $\partial M=S\cup T$. Let $\sigma\in
C_{n}(M,\\{S,T\\};\mathbb{Z})$ be a representative of the fundamental class of
$M$. If $R$ is a ring and if $A$ is an $(R,\mathbb{Z}[\pi_{1}(M)])$-bimodule,
then the map

$-\acap\sigma\colon C^{k}(M,\partial M;\mathbb{Z}[\pi_{1}(M)])\rightarrow
C_{n-k}(M;\mathbb{Z}[\pi_{1}(M)])$

defined by Lemma A.13 is a chain homotopy equivalence of left $R$-chain
complexes.

Note that Theorem 14.9 can be used to give an alternative proof that the chain
complexes of the theorem are chain homotopy equivalent.

###### Proof.

The Universal Poincaré Duality Theorem A.16 follows immediately from the
Twisted Poincaré Duality Theorem A.15 together with Lemma A.4. ∎

Below we will provide a proof of the Twisted Poincaré Duality Theorem A.15.
But just for fun we would like to show that the Universal Poincaré Duality
Theorem A.16 also implies the Twisted Poincaré Duality Theorem A.15.

###### Proof of Theorem A.15 using Theorem A.16.

Let $M$ be an compact, oriented, connected $n$-dimensional manifold. We pick a
representative $\sigma$ for $[M]$ and we write $\pi=\pi_{1}(M)$. Let $A$ be an
$(R,\mathbb{Z}[\pi])$-bimodule. Given a chain complex $D_{*}$ of right
$\mathbb{Z}[\pi]$-modules we consider the cochain map

$\begin{array}[]{rcl}\Xi\colon
A\otimes_{\mathbb{Z}[\pi]}\operatorname{Hom}_{\operatorname{\textnormal{right-}}\mathbb{Z}[\pi]}(D_{*};\mathbb{Z}[\pi]))&\to&\operatorname{Hom}_{\operatorname{\textnormal{right-}}\mathbb{Z}[\pi]}(D_{*},A)\\\
a\otimes f&\mapsto&(\sigma\mapsto a\cdot f(\sigma)).\end{array}$

Note that $\Xi$ is an isomorphism if each $D_{k}$ is a finitely generated free
$\mathbb{Z}[\pi]$-module. But in general $\Xi$ is not an isomorphism.

Furthermore we consider the following diagram

---
$\textstyle{H^{k}(A\otimes_{\mathbb{Z}[\pi]}C^{*}(M,\partial
M;\mathbb{Z}[\pi]))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{Id}_{A}\otimes(\acap\sigma)}$$\scriptstyle{\Xi_{*}}$$\textstyle{H_{n-k}(A\otimes_{\mathbb{Z}[\pi]}C_{*}(M;\mathbb{Z}[\pi]))=H_{n-k}(M;A)}$$\textstyle{H^{k}(C^{*}(M,\partial
M;A))=H^{k}(M,\partial
M;A).\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\acap[M]}$

One easily verifies that the diagram commutes. The top horizontal map is an
isomorphism by the Universal Poincaré Duality Theorem A.16. It remains to show
that the vertical map is an isomorphism. As we had pointed out above, on the
chain level $\Xi$ is in general not an isomorphism.

As in the proof of Proposition A.5 we can use Theorem 4.5 to find a pair
$(X,Y)$ of finite CW-complexes and a homotopy equivalence
$f\colon(X,Y)\to(M,\partial M)$. By Proposition A.3 there exists a homotopy
equivalence $\Theta\colon C_{*}^{\operatorname{cell}}(X,Y;\mathbb{Z}[\pi])\to
C_{*}(X,Y;\mathbb{Z}[\pi])$ of $\mathbb{Z}[\pi]$-chain complexes. We consider
the following diagram where all tensor products and homomorphism are over
$\mathbb{Z}[\pi]$:

$\textstyle{A\otimes
C^{*}_{\operatorname{cell}}(X,Y;\mathbb{Z}[\pi])\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Xi_{*}}$$\scriptstyle{\Theta^{*}}$$\textstyle{A\otimes
C^{*}(X,Y;\mathbb{Z}[\pi])\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Xi_{*}}$$\scriptstyle{f^{*}}$$\textstyle{A\otimes
C^{*}(M,\partial
M;\mathbb{Z}[\pi])\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Xi_{*}}$$\textstyle{C^{*}_{\operatorname{cell}}(X,Y;A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Theta^{*}}$$\textstyle{C^{*}(X,Y;A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{*}}$$\textstyle{C^{*}(M,\partial
M;A).}$

One easily verifies that the diagram commutes. As pointed out above, the
horizontal maps are chain homotopy equivalences over $\mathbb{Z}[\pi]$. Since
$X$ is a finite CW-complex we see that each
$C_{k}^{\operatorname{cell}}(X,Y;\mathbb{Z}[\pi])$ is a finitely generated
free $\mathbb{Z}[\pi]$-module. Thus we obtain from the above that the left
vertical map is an isomorphism. Therefore we see that the right vertical map
is a chain homotopy equivalence. In particular it induces an isomorphism of
homology groups. ∎

The remainder of this appendix is dedicated to the proof of the Twisted
Poincaré Duality Theorem A.15. Even though the theorem is well-known and often
used, there are not many satisfactory proofs in the literature. The proof
which is closest to ours in spirit is the proof of Sun [Sun17]. For closed
manifolds Kwasik-Sun [KS18] provide a proof by using the work of Kirby-
Siebenmann to reduce the proof to the case of triangulated manifolds.

The proof of the Twisted Poincaré Duality Theorem A.15 is modelled on the
proof of untwisted Poincaré Duality that is given in Bredon’s book [Bre97,
Chapter VI Section 8]. The logic of his proof is unchanged, but some arguments
and definitions have to be adjusted for the twisted setting.

###  A.4. Preparations for the proof of the Twisted Poincaré Duality Theorem
A.15

We fix some notation that we will use for the remainder of this appendix. Let
$M$ be a _connected_ manifold and denote by $\pi:=\pi_{1}(M,x_{0})$ the
fundamental group. Finally let $R$ be a ring and let $A$ be an
$(R,\mathbb{Z}[\pi])$-bimodule.

We write $p\colon\widetilde{M}\rightarrow M$ for the universal cover of $M$.
For a subset $X\subset M$ (not necessarily connected) we consider the
(co)-homology of $X$ with respect to the coefficient system coming from $M$ by
setting

$\displaystyle C_{*}(X;A)$
$\displaystyle:=A\otimes_{\mathbb{Z}[\pi]}C_{*}(p^{-1}(X);\mathbb{Z}),$
$\displaystyle C^{*}(X;A)$
$\displaystyle:=\operatorname{Hom}_{\mathbb{Z}[\pi]}\left(\overline{C_{*}(p^{-1}(X);\mathbb{Z})},A\right),$

with generalisation to pairs $Y\subset X\subset M$ by

$\displaystyle C_{*}(X,Y;A)$
$\displaystyle:=A\otimes_{\mathbb{Z}[\pi]}C_{*}(p^{-1}(X),p^{-1}(Y);\mathbb{Z}),$
$\displaystyle C^{*}(X,Y;A)$
$\displaystyle:=\operatorname{Hom}_{\mathbb{Z}[\pi]}\left(\overline{C_{*}(p^{-1}(X),p^{-1}(Y);\mathbb{Z})},A\right).$

We summarise the basic properties of twisted coefficients in the following
theorem, which should be compared to the untwisted case.

######  Theorem A.17.

Let $M$ be a connected manifold with fundamental group $\pi$, and let $A$ be
an $(R,\mathbb{Z}[\pi])$-bimodule.

  1. (1)

Given $Y\subset X\subset M$ there is a long exact sequence of pairs in
homology

${\cdots}$${H_{k}(Y;A)}$${H_{k}(X;A)}$${H_{k}(X,Y;A)}$${H_{k-1}(Y;A)}$${\cdots}$

and cohomology

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  2. (2)

Suppose we have a chain of subspaces $Z\subset Y\subset X\subset M$ such that
the closure of $Z$ is contained in the interior of $Y$. Then the inclusion
$(X\setminus Z,Y\setminus Z)\to(X,Y)$ induces an isomorphism in homology and
cohomology i.e.

$H_{k}(X\setminus Z,Y\setminus
Z;A)\xrightarrow{\sim}H_{k}(X,Y;A)\quad\mbox{and}\quad H^{k}(X\setminus
Z,Y\setminus Z;A)\xleftarrow{\sim}H^{k}(X,Y;A)$

  3. (3)

If $U_{1}\subset U_{2}\subset M$ and $V_{1}\subset V_{2}\subset M$ are open
subsets in $M$, then there are long exact sequences in homology

${\ldots}$${H_{k}(U_{1}\cap V_{1},U_{2}\cap
V_{2};A)}$${\begin{array}[]{c}H_{k}(U_{1},U_{2};A)\\\ \oplus\\\
H_{k}(V_{1},V_{2};A)\end{array}}$${H_{k}(U_{1}\cup V_{2},U_{2}\cup
V_{2};A)}$${H_{k-1}(U_{1}\cap V_{1},U_{2}\cap V_{2};A)}$${\ldots}$

and cohomology

${\ldots}$${H_{k}(U_{1}\cup V_{1},U_{2}\cup
V_{2};A)}$${\begin{array}[]{c}H_{k}(U_{1},U_{2};A)\\\ \oplus\\\
H_{k}(V_{1},V_{2};A)\end{array}}$${H_{k}(U_{1}\cap V_{2},U_{2}\cap
V_{2};A)}$${H_{k-1}(U_{1}\cup V_{1},U_{2}\cup V_{2};A)}$${\ldots}$

  4. (4)

Suppose the inclusion $Y\to X$ is a homotopy equivalence, then the inclusion
induced isomorphisms

$H_{k}(Y;A)\xrightarrow{\sim}H_{k}(X;A)\quad\mbox{and}\quad
H^{k}(Y;A)\xleftarrow{\sim}H^{k}(X;A).$

  5. (5)

Let $U_{1}\subset U_{2}\subset\ldots$ be a sequence of open sets in $M$ and
let $U=\bigcup_{i\in\mathbb{N}}U_{i}$, then inclusions induce an isomorphism

${\varinjlim}_{i\in\mathbb{N}}C_{*}(U_{i};A)\,\,\xrightarrow{\cong}\,\,C_{*}(U;A).$

The proofs are essentially the same as in the classical case. Therefore we
will only sketch the arguments and focus on what is different. We also warn
the reader that we give the “wrong proof” of statement (4). This is due to the
fact that we developed the theory of twisted coefficients only for inclusions
and hence a homotopy inverse does not fit in our theory. Therefore statement
(4) will be deduced in a slightly round-about way using the following
elementary lemma [Bre97, Chapter III Theorem 3.4 & remark after proof].

######  Lemma A.18 (The Covering Homotopy Theorem).

Given a covering $p\colon\widetilde{X}\rightarrow X$, a homotopy $H\colon
Y\times I\rightarrow X$, and a lift $\widetilde{h}\colon Y\to\widetilde{X}$ of
$H(-,0)$, there exists a unique lift $\widetilde{H}\colon Y\times
I\rightarrow\widetilde{X}$ of $H$ with $\widetilde{h}=\widetilde{H}(-,0)$.

###### Proof of Theorem A.17.

Recall that $p\colon\widetilde{M}\rightarrow M$ denotes the universal cover.

For statement (1) we consider the short exact sequence $0\to
C_{*}(Y;\mathbb{Z}[\pi])\to C_{*}(X;\mathbb{Z}[\pi])\to
C_{*}(X,Y;\mathbb{Z}[\pi])\to 0$ of _free_ $\mathbb{Z}[\pi]$-modules. Since
the modules are free the sequence stays exact after applying the functors
$A\otimes_{\mathbb{Z}[\pi]}-$ and $\operatorname{Hom}_{\mathbb{Z}[\pi]}(-,A)$.

Recall the proof of statement (2) and (3) in the classical case as it is done
for example in Bredon’s book [Bre97, Chapter IV Section 17]. The main
ingredient is to show that the inclusion of chain complexes
$C^{\mathcal{U}}_{*}(X;\mathbb{Z}[\pi])\to C^{*}(X;\mathbb{Z}[\pi])$ induces
an isomorphism on homology [Bre97, Theorem 17.7]. Here $\mathcal{U}$ is an
open cover of $X$ and $C^{\mathcal{U}}_{*}(X;\mathbb{Z}[\pi])$ is the free
abelian group generated by simplices $\sigma$ for which there is a
$U\in\mathcal{U}$ such that $\sigma\colon\Delta^{*}\to p^{-1}(U)$. This is
done by defining the barycentric subdivision $\Upsilon_{*}\colon
C_{*}(\widetilde{X};\mathbb{Z})\rightarrow C_{*}(\widetilde{X};\mathbb{Z})$
and a chain homotopy $T$ between $\Upsilon_{*}$ and the identity [Bre97, Lemma
17.1]. The important thing for us to observe is that both maps are natural
[Bre97, Claim (1) in proof of Lemma 17.1]. Hence for a twisted chain
$\Upsilon(e\otimes_{\mathbb{Z}[\pi]}\sigma):=e\otimes_{\mathbb{Z}[\pi]}\Upsilon(\sigma)$
is well-defined, because

$\displaystyle\Upsilon(e\otimes_{\mathbb{Z}[\pi]}\gamma\sigma)$
$\displaystyle=e\otimes_{\mathbb{Z}[\pi]}\Upsilon(\gamma\sigma)$
$\displaystyle=e\otimes_{\mathbb{Z}[\pi]}\gamma\Upsilon(\sigma)\quad\mbox{(naturality
of $\Upsilon$)}$
$\displaystyle=e\gamma\otimes_{\mathbb{Z}[\pi]}\Upsilon(\sigma)=\Upsilon(e\gamma\otimes_{\mathbb{Z}[\pi]}\sigma)$

The same holds for $T$ and from now on one can follow the classical proofs.
Alternatively, one could invoke Lemma A.4.

Next we prove statement (4). Let $f\colon X\rightarrow Y\subset X$ be a
homotopy inverse of the inclusion and $H\colon X\times I\rightarrow X$ a
homotopy between $\operatorname{Id}_{X}$ and $f$. Since
$p\colon\widetilde{X}\rightarrow X$ is a covering and
$\operatorname{Id}_{{\widetilde{X}}}$ is a lift of $H(p(-),0)$, we get by
Lemma A.4 a lift $\widetilde{H}\colon\widetilde{X}\times
I\rightarrow\widetilde{X}$ of the homotopy $H$. One easily verifies that the
inclusion $\widetilde{Y}\to\widetilde{X}$ induces a homotopy equivalence where
a homotopy inverse is given by $\widetilde{H}(-,1)$. Hence the inclusion
induced map $H_{k}(C_{*}(\widetilde{Y};\mathbb{Z}))\to
H_{k}(C_{*}(\widetilde{X};\mathbb{Z}))$ is an isomorphism for every $k$. Thus
the claim follows from Lemma A.4.

The proof of Statement (5) is almost verbatim the same proof as in the
classical case. ∎

###  A.5. The main technical theorem regarding Poincaré Duality

Given a group $\pi$ we can view $\mathbb{Z}$ as a $\mathbb{Z}[\pi]$-module
with trivial $\pi$-action. We denote this module by
$\mathbb{Z}^{\operatorname{triv}}$. Let $p\colon\widetilde{M}\to M$ be the
covering projection. We have the following useful lemma, concerning the chain
map $C_{*}(X;\mathbb{Z}^{\operatorname{triv}})\to C_{*}(X;\mathbb{Z})$ defined
by $k\otimes_{\mathbb{Z}[\pi]}\widetilde{\sigma}\mapsto k\cdot p(\sigma)$.

######  Lemma A.19.

Given any subset $X\subset M$ the chain map above is an isomorphism between
$C_{*}(X;\mathbb{Z}^{\operatorname{triv}})$ and $C_{*}(X;\mathbb{Z})$, and
induces one between $C^{*}(X;\mathbb{Z})$ and
$C^{*}(X;\mathbb{Z}^{\operatorname{triv}})$, where $C_{*}(X;\mathbb{Z})$ and
$C^{*}(X;\mathbb{Z})$ are the untwisted singular chain complexes.

###### Proof.

The isomorphism is given by lifting a simplex, which is always possible since
a simplex is simply connected. If one has two different choices of lifts, then
they differ by an element in $\pi$. But the action of $\mathbb{Z}[\pi]$ on
$\mathbb{Z}$ is trivial and hence this indeterminacy vanishes. ∎

We will keep the notational difference between $C_{*}(X;\mathbb{Z})$ and
$C_{*}(X;\mathbb{Z}^{\operatorname{triv}})$ to emphasise where our simplices
live.

As above let $R$ be a ring and let $A$ is an $(R,\mathbb{Z}[\pi])$-bimodule.
Let $K\subset M$ be a compact subset of $M$. We define the (twisted) Cech
cohomology groups

$\displaystyle\check{H}^{p}(K;A):=\lim_{\underset{K\subset U\subset
M}{\longrightarrow}}H^{p}(U;A),$

where the direct limit runs over all open sets in $M$ containing $K$. Since
cohomology is contravariant, we define the order on open sets in the reversed
way i.e. $U\leq V$ if $V\subset U$.

Now we assume that $M$ is oriented. Being oriented gives us for any closed
subset $Z\subset M$ a preferred element $\theta_{Z}\in H_{n}(M,M\setminus
Z;\mathbb{Z}^{\operatorname{triv}})\cong H_{n}(M,M\setminus Z;\mathbb{Z})$,
which restricts for all $x\in Z$ to the generator in
$H_{n}(M,M\setminus\left\\{x\right\\};\mathbb{Z}^{\operatorname{triv}})$.

For any open set $U\subset M$ containing $K$ let $\operatorname{ex}_{U}\colon
H_{n}(M,M\setminus K;\mathbb{Z}^{\operatorname{triv}})\to H_{n}(U,U\setminus
K;\mathbb{Z}^{\operatorname{triv}})$ be the inverse of the inclusion given by
the excision isomorphism i.e. if $j\colon U\rightarrow M$ is the inclusion
then $j_{*}\circ\operatorname{ex}_{U}=\operatorname{Id}$. We then obtain a map

$\displaystyle D_{U}\colon H^{p}(U;A)$ $\displaystyle\longrightarrow
H_{n}(M,M\setminus K;A)$ $\displaystyle\phi$ $\displaystyle\longmapsto
j_{*}(\phi\acap\operatorname{ex}_{U}(\theta_{K})).$

Given another open set $V\subset U$ denote by $i\colon V\rightarrow U$ the
inclusion. Then one easily calculates:

$\displaystyle\operatorname{PD}_{V}(i^{*}\phi)=j_{*}i_{*}(i^{*}\phi\acap\operatorname{ex}_{V}(\theta_{K}))=j_{*}(\phi\acap
i_{*}\operatorname{ex}_{V}(\theta_{K}))=j_{*}(\phi\acap\operatorname{ex}_{U}(\theta_{K}))=\operatorname{PD}_{U}(\phi).$

Or, with other words, the following diagram commutes:

${H^{p}(U;A)}$${H_{n-p}(M,M\setminus
K;A)}$${H^{p}(V;A)}$$\scriptstyle{i^{*}}$$\scriptstyle{\operatorname{PD}_{U}}$$\scriptstyle{\operatorname{PD}_{V}}$

By the universal property of the direct limit we obtain the _dualising_ map
$\operatorname{PD}_{K}\colon\check{H}^{p}(K;A)\rightarrow H_{n-p}(M,M\setminus
K;A)$.

In the remainder of this section we will prove the following theorem.

######  Theorem A.20 (Poincaré duality).

The map $\operatorname{PD}_{K}\colon\check{H}^{p}(K;A)\rightarrow
H_{n-p}(M,M\setminus K;A)$ is a left $R$-module isomorphism for all compact
subsets $K\subset M$.

Here, as above, $A$ is an $(R,\mathbb{Z}[\pi])$-bimodule. In the subsequent
section we will see that the Twisted Poincaré Duality Theorem A.15 is a
reasonably straightforward consequence of Theorem A.20.

The proof of Theorem A.20 will be an application of the following lemma.

######  Lemma A.21 (Bootstrap lemma).

Let $P_{M}(K)$ be a statement about compact sets $K$ in $M$. If $P_{M}(\cdot)$
satisfies the following three conditions:

  1. (1)

$P_{M}(K)$ holds true for all compact subsets $K\subset M$ with the property
that for all $x\in K$ the inclusions $\left\\{x\right\\}\to K$ and $M\setminus
K\to M\setminus\left\\{x\right\\}$ are deformation retracts,

  2. (2)

If $P_{M}(K_{1}),P_{M}(K_{2})$ and $P_{M}(K_{1}\cap K_{2})$ is true, then
$P_{M}(K_{1}\cup K_{2})$ is true,

  3. (3)

If $\cdots\subset K_{2}\subset K_{1}$ and $P_{M}(K_{i})$ is true for all
$i\in\mathbb{N}$, then $P_{M}(\bigcap_{i\in\mathbb{N}}K_{i})$ is true.

Then $P_{M}(K)$ is true for all $K\subset M$.

###### Proof.

See [Bre97, Chapter VI Lemma 7.9]. ∎

The idea is to apply the bootstrap lemma to the statements that the conclusion
of Theorem A.20 holds for a given compact set $K$. It turns out that condition
(3) is the easiest to verify. It follows from formal properties about direct
limits. For the verification of condition (1) we have do to one explicit
calculation. This is the content of the next lemma.

######  Lemma A.22.

Let $x\in M$ be a point. The map
$\operatorname{PD}_{\left\\{x\right\\}}\colon\check{H}^{0}(\left\\{x\right\\};A)\to
H_{n}(M,M\setminus\left\\{x\right\\};A)$ is an $R$-module isomorphism.

###### Proof.

Let $p\colon\widetilde{M}\rightarrow M$ be the universal cover. Since $x$ is a
point in a manifold we can calculate the dualising map
$\operatorname{PD}_{\left\\{x\right\\}}$ by taking the limit over open
neighbourhoods $U$ of $x$ with the following two properties:

  1. (1)

$U$ is contractible,

  2. (2)

for any connected component $\overline{U}\subset p^{-1}(U)$ the map
$p|_{\overline{U}}$ is a homeomorphism.

This can be done, since any neighbourhood of $x$ contains a neighbourhood with
these two properties. Let $U$ be such a neighbourhood of $x$ and
$\overline{U}\subset p^{-1}(U)$ a fixed connected component. This choice of
connected component gives us an isomorphism $H^{0}(U;A)\cong A$ as follows.
Let $f\in H^{0}(U;A)$ be arbitrary and $\overline{x}\in\overline{U}$ be a
point in our connected component, then we get an element in $A$ by evaluating
$f([\overline{x}])$. Conversely, given an element $e\in A$ we can construct a
function in $H^{0}(U;A)$ by setting $f([\overline{x}])=e$ for all
$\overline{x}\in\overline{U}$. Note that there is a unique way to extend $f$
equivariantly to $C_{0}(p^{-1}(U);\mathbb{Z})$.

We are now going to construct a representative of the orientation class
$\theta_{K}\in
H_{n}(M,M\setminus\left\\{x\right\\};\mathbb{Z}^{\operatorname{triv}})$ for
which it is very simple to calculate the dualising map. Let $\overline{x}$ be
the preimage of $x$ in $\overline{U}$. Now take a cycle
$\sum_{i=1}^{d}k_{i}\sigma_{i}$ which generates
$H_{n}(\overline{U},\overline{U}\setminus\left\\{\overline{x}\right\\};\mathbb{Z})$.
By excision and Lemma A.19 one easily sees that
$1\otimes_{\mathbb{Z}[\pi]}\sum_{i=1}^{d}k_{i}\sigma_{i}$ is a generator of
$H_{n}(M,M\setminus\left\\{x\right\\};\mathbb{Z}^{\operatorname{triv}})$.

Using the isomorphism $H^{0}(U;A)\cong A$ from above the dualising map becomes
$\operatorname{PD}_{\left\\{x\right\\}}\colon A\to
H_{n}(M,M\setminus\left\\{x\right\\};A),e\mapsto
e\otimes_{\mathbb{Z}[\pi]}\sum_{i=1}^{d}k_{i}\sigma_{i}$. This is clearly an
isomorphism, since on the chain level we have:

$C_{*}(U,U\setminus\left\\{x\right\\};A)=A\otimes_{\mathbb{Z}[\pi]}\bigoplus_{\gamma\in\pi}C_{*}(\gamma\overline{U},\gamma\overline{U}\setminus\left\\{\gamma\overline{x}\right\\};\mathbb{Z})\cong
A\otimes_{\mathbb{Z}}C_{*}(\overline{U},\overline{U}\setminus\left\\{\overline{x}\right\\};\mathbb{Z}).\qed$

In order to verify condition (2) of the bootstrap lemma we will need the
following lemma (compare [Bre97, Lemma 8.2]).

######  Lemma A.23.

If $K$ and $L$ are two compact subsets of $M$, then the diagram

${\cdots}$${\check{H}^{p}(K\\!\cup\\!L;A)}$${\begin{array}[]{c}\check{H}^{p}(K;A)\\\
\oplus\\\ \check{H}^{p}(L;A)\end{array}}$${\check{H}^{p}(K\cap
L;A)}$${\check{H}^{p+1}(K\\!\cup\\!L;A)}$${\cdots}$${\cdots}$${H_{n-p}(M,M\\!\setminus\\!(L\\!\cup\\!K);A)}$${\begin{array}[]{c}H_{n-p}(M,M\\!\setminus\\!K;A)\\\
\oplus\\\
H_{n-p}(M,M\\!\setminus\\!L;A)\end{array}}$${H_{n-p}(M,M\\!\setminus\\!(L\cap
K);A)}$${H_{n-p-1}(M,M\\!\setminus\\!(L\\!\cup\\!K);A)}$${\cdots}$$\scriptstyle{\operatorname{PD}_{K\cup
L}}$$\scriptstyle{\operatorname{PD}_{K}\oplus\operatorname{PD}_{L}}$$\scriptstyle{\operatorname{PD}_{K\cap
L}}$$\scriptstyle{\operatorname{PD}_{K\cup L}}$

has exact rows and it commutes up to a sign depending only on $p$.

###### Proof.

The rows are exact by Mayer-Vietoris and the fact that direct limit is an
exact functor. The commutativity of the squares is clear except for the last
one involving the boundary map. This will be a painful diagram chase. Let
$U\supset K$ and $V\supset L$ be open neighbourhoods containg $K$ resp. $L$.
The sequence in the top row comes from the short exact sequence
($\mathcal{U}=\left\\{U,V\right\\}$):

${0}$${C^{*}_{\mathcal{U}}(U\cup V;A)}$${C^{*}(U;A)\oplus
C^{*}(V;A)}$${C^{*}(U\cap V;A)}$${0.}$

An element $\phi\in\check{H}^{p}(K\cap L;A)$ will already be represented by
same element $f\in C^{p}(U\cap V;A)$ for some $U$ and $V$ as above. We can
extend $f$ to an element $\overline{f}\in C^{p}(M;A)$ by

$\displaystyle\overline{f}(\sigma)=\begin{cases}f(\sigma)&\mbox{if
}\operatorname{Im}\sigma\subset\widetilde{U}\cap\widetilde{V}\\\
0&\mbox{else.}\end{cases}$

Note that $\overline{f}\in C^{p}(M;A)$ since $p^{-1}(U\cap V)$ is an
equivariant subspace and hence $\overline{f}$ is equivariant. If we consider
$\overline{f}$ as an element in $C^{p}(U;A)$ then the cohomology class
$\delta(\phi)$ is represented by the cochain $h\in C^{p+1}(U\cup V;A)$ which
is given by

$\displaystyle h(\sigma)=\begin{cases}\delta(\overline{f})(\sigma)&\mbox{if
}\operatorname{Im}\sigma\subset\widetilde{U}\\\ 0&\mbox{else.}\end{cases}$

Since $\phi$ is a cocycle we have $\delta(f)(\sigma)=0$ for $\sigma\in
C_{*}(U\cap V;A)$. It follows in particular that if $\sigma$ is a simplex
whose image is completely contained in $\widetilde{V}$, then $h(\sigma)=0$. We
can represent our orientation class $\theta\in H_{n}(M,M\setminus(K\cup L))$
by a cycle

$\displaystyle a=b+c+d+e\quad\mbox{with}\quad$ $\displaystyle b\in C_{n}(U\cap
V;\mathbb{Z}^{\operatorname{triv}})\quad c\in C_{n}(U\setminus(U\cap
L);\mathbb{Z}^{\operatorname{triv}})$ $\displaystyle d\in
C_{n}(V\setminus(V\cap K);\mathbb{Z}^{\operatorname{triv}}),\quad e\in
C_{n}(M\setminus(K\cup L);\mathbb{Z}^{\operatorname{triv}}).$

Obviously $e$ does not play a role since we kill it in the end. With these
representatives one computes that $\delta(\phi)(\theta)$ is represented by

$h\acap(b+c+d)=\delta(\overline{f})\cap c+h\acap d+\delta(f)\cap
b=\delta(\overline{f})\cap c.$

The pairing of $h$ with $b$ is zero since $f$ was a cocycle in $C^{*}(U\cap
V;A)$ and the pairing of $h$ with $d$ is zero since $d$ consist of simplices
with image in $\widetilde{V}$.

The lower sequence comes from the short exact sequence:

${0}$${C_{*}(M,M\setminus(K\cup L);A)}$${\begin{array}[]{c}C_{*}(M,M\setminus
K;A)\\\ \oplus\\\ C_{*}(M,M\setminus
L);A)\end{array}}$${C_{*}(M,M\setminus(K\cap L);A)}$${0.}$

Before we compute the other side $\partial(\phi\acap\operatorname{ex}_{U\cap
V}(\theta))$ we want to recall that the cap product is natural on the chain
complex level i.e. the following diagram commutes:

${C^{p}(U;A)\times C_{n}(U,U\setminus
K;\mathbb{Z}^{\operatorname{triv}})}$${C_{*}(U,U\setminus
K;A)}$${C^{p}(M;A)\times C_{n}(M,M\setminus
K;\mathbb{Z}^{\operatorname{triv}})}$${C_{*}(M,M\setminus K;A).}$

Therefore we use the representatives from above. To construct the boundary map
$\partial$, we take as the preimage of $\overline{f}\acap a\in
C_{*}(M,M\setminus(K\cap L);A)$ the element $(\overline{f}\acap a,0)\in
C_{*}(M,M\setminus K;A)\oplus C_{*}(M,M\setminus L;A)$. Then one computes in
$C_{*}(M,M\setminus K;A)$

$\displaystyle\partial(\overline{f}\acap a)$
$\displaystyle=(-1)^{p+1}\cdot\delta(\overline{f})\acap a\pm f\acap\partial
a\quad(\mbox{by Lemma~{}\ref{lem:capboundary}})$
$\displaystyle=(-1)^{p+1}\cdot\delta(\overline{f})\acap a\quad(\mbox{since
$f\acap\partial a\in C_{n-p-1}(M\setminus(K\cup L);A)$})$
$\displaystyle=(-1)^{p+1}\cdot\delta(\overline{f})\acap b+c+d+e$
$\displaystyle=(-1)^{p+1}\cdot\delta(\overline{f})\acap(c+d)\quad\mbox{(same
reason as above)}$ $\displaystyle=(-1)^{p+1}\cdot\delta(\overline{f})\acap
c\quad\mbox{(since $d\in C_{n-p}(V\setminus(K\cap V);A)$)}$

Therefore the element $\partial(\phi\acap\operatorname{ex}_{U\cap V}(\theta))$
is also represented by $(-1)^{p+1}\cdot\delta(\overline{f})\acap c\in
C_{n-p-1}(M,M\setminus(K\cup L);A)$. ∎

###### Proof of Theorem A.20.

Let $P_{M}(K)$ be the statement that the map $\operatorname{PD}_{K}$ is an
isomorphism. Then it is sufficient to verify condition (1),(2) and (3) of the
bootstrap lemma. We start with verifying (1). In the case that
$K=\left\\{x\right\\}$ is just a point we have already seen in Lemma A.22 that
the statement holds true. For a general compact $K$ with the property of (1)
the statement follows from the following commutative diagram:

${\check{H}^{p}(K;A)}$${H_{n-p}(M,M\setminus
K;A)}$${\check{H}^{p}(\left\\{x\right\\};A)}$${H_{n-p}(M,M\setminus\left\\{x\right\\};A),}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$

where the vertical maps are isomorphisms by the homotopy invariance and the
bottom row by the observation above. Hence condition (1) is verified.

Condition (2) follows immediately from the five-lemma and Lemma A.23.

Let $K_{i}$ be a sequence of compact subsets such that $P_{M}(K_{i})$ holds
for all $i\in\mathbb{N}$. We set $K=\bigcap_{i\in\mathbb{N}}K_{i}$. It is an
exercise in pointset topology of manifolds that each $K_{i}$ has a fundamental
system $U_{i,j}$ of open neighbourhoods. Fundamental system means that
$U_{i,j}\subset U_{i,k}$ if $j<k$ and that for each open set $U$ containing
$K_{i}$ there is a $j$ such that $U_{i,j}\subset U$. Another exercise in point
set topology of manifolds shows, that one can construct these sets such that
$U_{1,j}\supset U_{2,j}\supset U_{3,j}\supset\ldots$ for all $j\in\mathbb{N}$.
Then $U_{i,j}$ is a fundamental system of open neighbourhoods of $K$ with the
order $(i,j)\leq(k,l)\Leftrightarrow i\leq k\ \wedge j\leq l$. One has the
natural isomorphism [Bre97, Appendix D5]:

$\varinjlim_{i\in\mathbb{N}}\check{H}^{p}(K_{i};A)=\varinjlim_{i\in\mathbb{N}}\varinjlim_{j\in\mathbb{N}}H^{p}(U_{i,j};A)\xrightarrow{\
\simeq\
}\varinjlim_{i,j\in\mathbb{N}}H^{p}(U_{i,j};A)\cong\check{H}^{p}(K;A).$

And hence the theorem follows from the commutativity of the diagram:

${{\varinjlim}_{i\in\mathbb{N}}\check{H}^{p}(K_{i};A)}$${{\varinjlim}_{i\in\mathbb{N}}H_{n-p}(M,M\setminus
K_{i};A)}$${\check{H}^{p}(K;A)}$${H_{n-p}(M,M\setminus K;A).}$

∎

###  A.6. Proof of the Twisted Poincaré Duality Theorem A.15

For the reader’s convenience we recall the main theorem from the last section.
Here, as above, $R$ is a ring and $A$ is an $(R,\mathbb{Z}[\pi])$-bimodule.

Theorem A.20. _Let $M$ be a compact, oriented, connected $n$-dimensional
manifold. The map $\operatorname{PD}_{K}\colon\check{H}^{p}(K;A)\rightarrow
H_{n-p}(M,M\setminus K;A)$ is an isomorphism of left $R$-modules for all
compact subsets $K\subset M$._

Furthermore, we also recall that we need to prove the following theorem.

Theorem A.15. _Let $M$ an compact, oriented, connected $n$-dimensional
manifold. Let $S$ and $T$ be codimension 0 compact submanifolds of $\partial
M$ such that $\partial S=\partial T=S\cap T$ and $\partial M=S\cup T$. Let
$[M]\in H_{n}(M,\partial M;\mathbb{Z})$ be the fundamental class of $M$. The
map_

$-\acap[M]\colon H^{k}(M,S;{A})\,\,\rightarrow\,\,H_{n-k}(M,T;A)$

_defined by Lemma A.13 is an isomorphism of left $R$-modules._

In the remainder of this appendix we will explain how to deduce Theorem A.15
from Theorem A.20. First note that if $M$ is a closed manifold, then we can
set $K=M$ in Theorem A.20. Evidently we have $\check{H}^{p}(M;A)=H(M;A)$. Thus
we obtain precisely the statement of Theorem A.15 in the closed case.

Next let $M$ be a compact oriented manifold with non-empty boundary. First we
consider the case $R=\varnothing$ and $S=\partial M$. By the Collar
neighbourhood theorem 2.5 there exists a collar $\partial M\times[0,2]\subset
M$ of the boundary such that $\partial M=\partial M\times\left\\{0\right\\}$.
We obtain the following chain of isomorphisms:

$\displaystyle H^{p}(M;A)$ $\displaystyle\cong H^{p}(M\setminus(\partial
M\times[0,1));A)\quad\mbox{(homotopy)}$
$\displaystyle\cong\check{H}^{p}(M\setminus(\partial
M\times[0,1));A)\quad\mbox{(follows from considering the open}$ neighborhoods
$M\setminus(\partial M\times[0,1-\frac{1}{n}])$) $\displaystyle\cong
H_{n-p}(M\setminus\partial M,\partial M\times(0,1);A)\quad\mbox{(duality
$K=M\setminus(\partial M\times[0,1)))$}$ $\displaystyle\cong
H_{n-p}(M,\partial M\times[0,1);A)\quad\mbox{(excision $U=\partial M$)}$
$\displaystyle\cong H_{n-p}(M,\partial M;A),$

It follows from the definition of the dualising map and naturality of cap
product that these isomorphisms are given by capping with a generator $[M]\in
H_{n}(M,\partial M;\mathbb{Z}^{\operatorname{triv}})\cong H_{n}(M,\partial
M;\mathbb{Z})$ as in the classical case.

The proof of the general case of Theorem A.15 relies on the following lemma.

######  Lemma A.24.

Let $M$ a compact, oriented, connected $n$-dimensional manifold. Let $R$ and
$S$ be compact codimension 0 submanifolds of $\partial M$ such that $\partial
R=\partial S=R\cap S$ and $\partial M=R\cup S$. Then the following diagram
commutes up to a sign:

${\ldots}$${H^{p}(M,R;A)}$${H^{p}(M;A)}$${H^{p+1}(R;A)}$${H^{p+1}(M,R;A)}$${\ldots}$${H_{n-p-1}(R,\partial
R;A)}$${\ldots}$${H_{n-p}(M,S;A)}$${H_{n-p}(M,\partial M;A)}$${H_{n-p-1}(R\cup
S,S;A)}$${H_{n-p-1}(M,S;A)}$${\ldots}$$\scriptstyle{\acap{[M]}}$$\scriptstyle{\acap{\left[M\right]}}$$\scriptstyle{\acap{\left[R\right]}}$$\scriptstyle{\acap{[M]}}$$\scriptstyle{\operatorname{excision\,\,isom.}}$

###### Proof.

The commutativity is a more or less direct consequence of Lemma A.13 and the
observation that $\partial_{*}[M]=[\partial M]$. More precisely, the proof in
the untwisted case is given in detail in [Fri19, Proof of Theorem 61.1]. The
proof in the twisted case is basically the same. ∎

The proof of the general case of Theorem A.15 follows from the previously
discussed Poincaré Duality isomorphisms $\acap[M]\colon
H^{p}(M;A)\xrightarrow{\cong}H_{n-p}(M,\partial M;A)$, $\acap[R]\colon
H^{p+1}(R,A)\xrightarrow{\cong}H_{n-p-1}(R,\partial R;A)$ together with Lemma
A.24 and the five lemma.

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