$\operatorname{\mathcal{S}}_{0}(X_{T})=T$ can be regarded as a constant. Next,
write $\operatorname{\mathbb{B}}_{K}=\\{\beta 1\
\mathrel{\mathop{\ordinarycolon}}\ |\beta|<K\\}$ and take any word $\alpha$ of
length at most $K\geq 2$ and outside of $\operatorname{\mathbb{B}}_{K}$. If
$|\alpha|<K$, then $\operatorname{\mathcal{S}}_{\alpha}(X_{T})$ can be
expressed as a linear combination of
$(\operatorname{\mathcal{S}}_{\gamma}(X_{T}))_{\gamma\in\operatorname{\mathbb{B}}_{K-1}}$
by induction. We can therefore assume that $|\alpha|=K$ and
$\alpha\notin\operatorname{\mathbb{B}}_{K}$, that is $\alpha=\beta 0$ with
$|\beta|=K-1$. Invoking the induction hypothesis, there exist coefficients
$(c_{\gamma})$ such that
$\operatorname{\mathcal{S}}_{\beta}(X_{T})=\sum_{\gamma
1\in\operatorname{\mathbb{B}}_{K-1}}c_{\gamma}\operatorname{\mathcal{S}}_{\gamma
1}(X_{T})$. We thus obtain using Fubini’s theorem and the shuffle product [42]
that

$\displaystyle\operatorname{\mathcal{S}}_{\alpha}(X_{T})$
$\displaystyle=\int_{0}^{T}\operatorname{\mathcal{S}}_{\beta}(X_{t})dt$
$\displaystyle=\sum_{\gamma
1\in\operatorname{\mathbb{B}}_{K-1}}c_{\gamma}\int_{0}^{T}\int_{0}^{t}\operatorname{\mathcal{S}}_{\gamma}(X_{s})dx_{s}dt$
$\displaystyle=\sum_{\gamma
1\in\operatorname{\mathbb{B}}_{K-1}}c_{\gamma}\left(T\operatorname{\mathcal{S}}_{\gamma
1}(X_{T})-\int_{0}^{T}\operatorname{\mathcal{S}}_{1}(X_{s})\operatorname{\mathcal{S}}_{\gamma}(X_{s})dx_{s}\right)$
$\displaystyle=\sum_{\gamma
1\in\operatorname{\mathbb{B}}_{K-1}}c_{\gamma}\left(T\operatorname{\mathcal{S}}_{\gamma
1}(X_{T})-\sum_{j=1}^{K-1}\operatorname{\mathcal{S}}_{\gamma_{1}\cdots\gamma_{j}1\gamma_{j+1}\cdots\gamma_{K-2}1}(X_{T})\right).$

The result follows since each signature term in the last expression belongs to
$\\{\operatorname{\mathcal{S}}_{\beta
1}|_{\Omega_{T}^{\Pi}}\mathrel{\mathop{\ordinarycolon}}|\beta|<K\\}$. ∎

###  A.5 Theorem 3.10

###### Proof.

We carry out an induction on $K\geq 1$. We start with $K=1$ (hence
$f\in\operatorname{\mathbb{C}}^{1,2}$) and define the functional
$\tilde{f}(Y_{r})=f(X_{s}\oplus Y_{r})$, $r\leq u$. As $Y_{u}\in\Omega^{\Pi}$,
we can apply the functional Stratonovich formula (Theorem 3.2) to $\tilde{f}$,
which gives

$\displaystyle f(X_{s}\oplus Y_{u})$ $\displaystyle=\tilde{f}(Y_{u})$
$\displaystyle=\tilde{f}(Y_{0})+\int_{0}^{u}\Delta_{t}\tilde{f}(X_{r})dr+\int_{0}^{t}\Delta_{x}\tilde{f}(Y_{r})\circ
dy_{r}$ $\displaystyle=f(X_{s})+\int_{0}^{u}\Delta_{t}f(X_{s}\oplus
Y_{r})dr+\int_{0}^{u}\Delta_{x}f(X_{s}\oplus Y_{r})\circ dy_{r}$
$\displaystyle=\Delta_{\emptyset}f(X_{s})\operatorname{\mathcal{S}}_{\emptyset}(Y_{u})+\underbrace{\int_{\triangle_{1,u}}\Delta_{t}f(X_{s}\oplus
Y_{t_{1}})\circ dy^{0}+\int_{\triangle_{1,u}}\Delta_{x}f(X_{s}\oplus
Y_{t_{1}})\circ dy^{1}}_{=\,r_{1}(X_{s},Y_{u})}.$

Now take $K\geq 2$ and suppose the result true for all $k\leq K$. As
$f\in\operatorname{\mathbb{C}}^{K,K+1}$ by assumption, then trivially
$f\in\operatorname{\mathbb{C}}^{K-1,K}$ and the induction hypothesis yields

$\displaystyle f(X_{s}\oplus Y_{u})$
$\displaystyle=\sum_{|\alpha|<K-1}\Delta_{\alpha}f(X_{s})\operatorname{\mathcal{S}}_{\alpha}(Y_{u})+r_{K-1}(X_{s},Y_{u}).$

As $\Delta_{\alpha}f$ is at least $\operatorname{\mathbb{C}}^{1,2}$ for all
$|\alpha|=K-1$, we can apply the pathwise Stratonovich formula to the
integrands constituting the remainder functional. That is,

$\displaystyle r_{K-1}(X_{s},Y_{u})$
$\displaystyle=\sum_{|\alpha|=K-1}\int_{\triangle_{K-1,u}}\Delta_{\alpha}f(X_{s}\oplus
Y_{t_{1}})\circ\,dy^{\alpha}$
$\displaystyle=\sum_{|\alpha|=K-1}\int_{\triangle_{K,u}}\left[\Delta_{\alpha}f(X_{s})+\int_{0}^{t_{1}}\Delta_{0\alpha}f(X_{s}\oplus
Y_{t_{0}})\,dt_{0}+\int_{0}^{t_{1}}\Delta_{1\alpha}f(X_{s}\oplus
Y_{t_{0}})\circ\,dy_{t_{0}}\right]\circ dy^{\alpha}$
$\displaystyle=\sum_{|\alpha|=K-1}\Delta_{\alpha}f(X_{s})\operatorname{\mathcal{S}}_{\alpha}(Y_{u})$
$\displaystyle+\underbrace{\sum_{\begin{subarray}{c}|\alpha|=K\\\
\alpha_{1}=\,0\end{subarray}}\int_{\triangle_{K,u}}\Delta_{\alpha}f(X_{s}\oplus
Y_{t_{0}})\circ\,dy^{\alpha}+\sum_{\begin{subarray}{c}|\alpha|=K\\\
\alpha_{1}=\,1\end{subarray}}\int_{\triangle_{K,u}}\Delta_{\alpha}f(X_{s}\oplus
Y_{t_{0}})\circ\,dy^{\alpha}}_{=\,r_{K}(X_{s},Y_{u})}.$

Bundling the terms together completes the proof. ∎

###  A.6 Corollary 3.11

###### Proof.

Equation $\eqref{eq:CorFTE1}$ simply follows from Theorem 3.10 by setting
$u=t-s$ and $Y_{u}=X|_{[s,t]}\in\Omega^{\Pi}$. Although $\eqref{eq:CorFTE2}$
is a consequence of Theorem 3.10 as well, we outline its proof as we deem it
instructive. Indeed, some care is needed when expanding backward. For
simplicity, write $W_{u}=X|_{[s,t]}$ so that
$\overleftarrow{W}=X|_{[t,s]}=\mathrel{\mathop{\ordinarycolon}}Y_{u}$. In
particular, $X_{r}=X_{s}\oplus W_{r-s}$ for all $r\in[s,t]$. We now rearrange
and iterate the pathwise Stratonovich formula as follows:

$\displaystyle f(X_{s})$
$\displaystyle=f(X_{t})-\sum_{|\alpha|=1}\int_{s}^{t}\Delta_{\alpha}f(X_{t_{1}})\circ
dx^{\alpha}$
$\displaystyle=f(X_{t})-\sum_{|\alpha|=1}\int_{0}^{u}\Delta_{\alpha}f(X_{s}\oplus
W_{\\!t_{1}})\circ dw_{t_{1}}^{\alpha}$
$\displaystyle=f(X_{t})-\sum_{|\alpha|=1}\Delta_{\alpha}f(X_{t})\int_{\overleftarrow{\triangle}_{1,u}}\circ
dw^{\alpha}+\sum_{|\alpha|=2}\int_{\overleftarrow{\triangle}_{2,u}}\Delta_{\alpha}f(X_{s}\oplus
W_{t_{1}})\circ dw^{\alpha},$

with the time-reversed simplexes
$\overleftarrow{\triangle}_{k,t}=\\{(t_{1},\ldots,t_{k})\in[0,t]^{k}\,|\,t_{1}\geq\ldots\geq
t_{k}\\}$. Proceeding until $K$ yields

$\displaystyle f(X_{s})=\sum_{|\alpha|<K}\Delta_{\alpha}f(X_{t})\
(-1)^{|\alpha|}\ \int_{\overleftarrow{\triangle}_{|\alpha|,u}}\circ\
dw^{\alpha}+(-1)^{K}\sum_{|\alpha|=K}\int_{\overleftarrow{\triangle}_{K,u}}\Delta_{\alpha}f(X_{s}\oplus
W_{t_{1}})\circ dw^{\alpha}.$ (A.6)

We now verify that

$\int_{\overleftarrow{\triangle}_{|\alpha|,u}}\varphi(X_{s}\oplus
W_{t_{1}})\circ
dw^{\alpha}=(-1)^{|\alpha|}\int_{\triangle_{|\alpha|,u}}\varphi(X_{t}\oplus
Y_{t_{1}})\circ dy^{\alpha},$ (A.7)

for every word $\alpha$ and functional
$\varphi\mathrel{\mathop{\ordinarycolon}}\Lambda\to\operatorname{\mathbb{R}}$
such that the above integrals are well-defined. Write $k=|\alpha|$ and
consider the affine involution
$T_{u}\mathrel{\mathop{\ordinarycolon}}[0,u]^{k}\to[0,u]^{k}$ given by
$T_{u}(t_{1},\ldots,t_{k})=(u-t_{1},\ldots,u-t_{k})$. In particular,
$T_{u}(\overleftarrow{\triangle}_{k,u})=\triangle_{k,u}$,
$w^{\alpha}=y^{\alpha}\circ T_{u}$ and $|\text{det}(\nabla T_{u})|\equiv 1$.
Noticing also that $X_{s}\oplus W_{t_{1}}=X_{t}\oplus Y_{u-t_{1}}$, changing
variables yields

$\displaystyle\int_{\overleftarrow{\triangle}_{k,u}}\varphi(X_{s}\oplus
W_{t_{1}})\circ dw^{\alpha}$
$\displaystyle=\int_{0}^{u}\int_{t_{k}}^{u}\cdots\int_{t_{2}}^{u}\varphi(X_{t}\oplus
Y_{u-t_{1}})\circ d\left(y^{\alpha}\circ T_{u}\right)$
$\displaystyle=\int_{u}^{0}\int_{t_{k}}^{0}\cdots\int_{t_{2}}^{0}\varphi(X_{t}\oplus
Y_{t_{1}})\circ dy_{t_{1}}^{\alpha_{1}}\cdots\circ dy_{t_{k}}^{\alpha_{k}}$
$\displaystyle=(-1)^{k}\int_{\triangle_{k,u}}\varphi(X_{t}\oplus
Y_{t_{1}})\circ dy^{\alpha}.$

This proves the claim. In light of $\eqref{eq:intermediate}$, we choose
$\varphi\equiv 1$ and $\varphi=\Delta_{\alpha}f$ in $\eqref{eq:claim}$, giving
respectively the signature elements
$\operatorname{\mathcal{S}}_{\alpha}(Y_{u})$ and terms in the remainder. ∎

###  A.7 Proposition 3.14

###### Proof.

Fix $\alpha\in\operatorname{\mathbb{A}}$ such that $|\alpha|=K$. Write
$\alpha^{k}=\alpha_{1}\cdots\alpha_{k}$, $k\leq K$, and
$r_{\alpha,k}(X_{t})=\int_{\triangle_{k,t}}\Delta_{\alpha}f(X_{t_{1}})\circ
dx^{\alpha^{k}}$. In particular, $r_{\alpha}=r_{\alpha,K}.$ We prove by
induction on $k$ that

$|r_{\alpha,k}(X_{s})|\leq\frac{2^{|\alpha^{k}|_{01}}}{|\alpha^{k}|_{0}!}c_{\alpha}\
t^{|\alpha^{k}|_{0}}\lVert
X_{t}\rVert_{\infty}^{|\alpha^{k}|_{1}}\quad\forall\ s\leq t,$ (A.8)

so that $\eqref{eq:r_alpha}$ corresponds to $k=K$. Without loss of generality,
we can assume that $s=t$.

  * •

$k=1$: First note that $|\alpha^{k}|_{01}=0$. If $\alpha_{1}=0$, then using
the norm $\lVert\cdot\rVert_{X_{t}}$ defined in $\eqref{eq:funcNorm}$, we
obtain

$|r_{\alpha,1}(X_{t})|\leq\int_{0}^{t}|\Delta_{\alpha}f(X_{s})|ds\leq\lVert\Delta_{\alpha}f\rVert_{X_{t}}\
t\leq c_{\alpha}\ t.$

If $\alpha_{1}=1$, using the expression $\eqref{eq:pathStrat}$ of the
Stratonovich integral and the notation
$\int_{(0,x_{t})}\mathrel{\mathop{\ordinarycolon}}=\int_{x_{t}\wedge
0}^{x_{t}\vee 0}$, this gives

$\displaystyle|r_{\alpha,1}(X_{t})|$
$\displaystyle\leq\int_{(0,x_{t})}|\Delta_{\alpha}f(X_{t}^{(\varepsilon)})|d\varepsilon+\int_{0}^{t}\int_{(0,x_{s})}|\Delta_{t\alpha}f(X_{s}^{(\varepsilon)})|d\varepsilon
ds$ $\displaystyle\leq\lVert\Delta_{\alpha}f\rVert_{X_{t}}\
|x_{t}|+\lVert\Delta_{t\alpha}f\rVert_{X_{t}}\lVert X_{t}\rVert_{\infty}\ \ t\
.$ $\displaystyle\leq c_{\alpha}\lVert X_{t}\rVert_{\infty}.$

Hence $\eqref{eq:inductionRemainder}$ is satisfied for $k=1$.

  * •

$k\geq 2$: There are several cases.

    * –

$\alpha_{k}=0$: As $|\alpha^{k-1}|_{0}=|\alpha^{k}|_{0}-1$ and
$|\alpha^{k-1}|_{01}=|\alpha^{k}|_{01}$, the induction hypothesis gives

$\displaystyle|r_{\alpha,k}(X_{t})|$
$\displaystyle\leq\int_{0}^{t}|r_{\alpha,k-1}(X_{s})|ds$
$\displaystyle\leq\frac{2^{|\alpha^{k}|_{01}}}{(|\alpha^{k}|_{0}-1)!}c_{\alpha}\int_{0}^{t}s^{|\alpha^{k}|_{0}-1}\lVert
X_{s}\rVert_{\infty}^{|\alpha^{k}|_{1}}ds$
$\displaystyle\leq\frac{2^{|\alpha^{k}|_{01}}}{|\alpha^{k}|_{0}!}c_{\alpha}\
t^{{|\alpha^{k}|_{0}}}\lVert X_{t}\rVert_{\infty}^{|\alpha^{k}|_{1}}.$

    * –

$\alpha_{k}=1,\ \alpha_{k-1}=1$: Since $|\alpha^{k}-1|_{0}=|\alpha^{k}|_{0}$
and $|\alpha^{k-1}|_{01}=|\alpha^{k}|_{01}$, we directly obtain

$\displaystyle|r_{\alpha,k}(X_{t})|$
$\displaystyle\leq\int_{(0,x_{t})}|r_{\alpha,k-1}(X^{(\varepsilon)}_{s})|d\varepsilon$
$\displaystyle\leq\frac{2^{|\alpha^{k}|_{01}}}{|\alpha^{k}|_{0}!}c_{\alpha}\
\int_{(0,x_{t})}t^{|\alpha^{k}|_{0}}\lVert
X_{t}^{(\varepsilon)}\rVert_{\infty}^{|\alpha^{k}|_{1}-1}d\varepsilon$
$\displaystyle\leq\frac{2^{|\alpha^{k}|_{01}}}{|\alpha^{k}|_{0}!}c_{\alpha}\ \
t^{{|\alpha^{k}|_{0}}}\lVert X_{t}\rVert_{\infty}^{|\alpha^{k}|_{1}}.$

In the last inequality, we used the fact that $\lVert
X_{t}^{(\varepsilon)}\rVert_{\infty}\leq\lVert X_{t}\rVert_{\infty}$ for all
$\varepsilon\in[0,x_{t}]$.

    * –

$\alpha_{k}=1,\ \alpha_{k-1}=0$: This is the interesting case. First, observe
that $|\alpha^{k-2}|_{0}=|\alpha^{k}|_{0}-1$ and
$|\alpha^{k-2}|_{0}=|\alpha^{k}|_{01}-1$. Thus,

$\displaystyle|r_{\alpha,k}(X_{t})|$
$\displaystyle\leq\int_{(0,x_{t})}|r_{\alpha,k-1}(X^{(\varepsilon)}_{t})|d\varepsilon+\int_{0}^{t}\int_{(0,x_{s})}|r_{\alpha,k-2}(X_{s}^{(\varepsilon)})|d\varepsilon
ds$ $\displaystyle\leq
2\int_{(0,x_{t})}\int_{0}^{t}|r_{\alpha,\alpha^{k}-2}(X_{s}^{(\varepsilon)})|dsd\varepsilon$
$\displaystyle\leq\frac{2^{|\alpha^{k}|_{01}}}{(|\alpha^{k}|_{0}-1)!}c_{\alpha}\int_{0}^{t}s^{|\alpha^{k}|_{0}-1}ds\
\lVert X_{t}\rVert_{\infty}^{|\alpha^{k}|_{1}},$

which concludes the proof of $\eqref{eq:inductionRemainder}$.

Finally, $\eqref{eq:remainder}$ follows by observing that
$\frac{2^{|\alpha|_{01}}}{|\alpha|_{0}!}\leq\frac{2^{|\alpha|_{0}}}{|\alpha|_{0}!}\leq
2$. ∎

###  A.8 Proposition 3.20

###### Proof.

First, $(iii)\Longrightarrow(i)$ is immediate as Itô iterated integrals only
depends on the final value of the path; see $\eqref{eq:Hermite}$. It remains
to show that $(i)\Longrightarrow(ii)$ and $(ii)\Longrightarrow(iii)$.

  1. 1.

$(i)\Longrightarrow(ii)$

The case $k=0$ is trivial so we proceed with $k=1$. Since
$f(X_{T})=g(X_{T})=h(x_{T})$ for some
$h\in\operatorname{\mathcal{C}}^{\infty}(\operatorname{\mathbb{R}})$ and $X$
is Brownian motion, then

$\Delta_{x}f(X_{t})=\frac{d}{dx}\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[h(y_{T})\
|\
x_{t}]=\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[h^{\prime}(x+y_{T-t})]\big{|}_{x=x_{t}}=\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[\Delta_{x}f(Y_{T})\,|\,X_{t}].$
(A.9)

For $k=2$, notice that
$\tilde{g}\mathrel{\mathop{\ordinarycolon}}=\Delta_{x}f\big{|}_{\Lambda_{T}}$
is itself a smooth path-independent payoff. We can thus apply a similar
argument to $\tilde{g}$ and
$\tilde{f}(X_{t})=\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[\tilde{g}(Y_{T})\,|\,X_{t}]$,
which is exactly $\Delta_{x}f(X_{t})$ thanks to $\eqref{eq:martingale}$. Hence
$\Delta_{xx}f(X_{t})=\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[\Delta_{xx}f(Y_{T})\,|\,X_{t}]$
and the same holds for higher derivatives.

  2. 2.

$(ii)\Longrightarrow(iii)$

Recall from Proposition 2.8 that
$g(X_{T})=\sum_{k=0}^{\infty}J_{k}\phi_{k}(X_{T})$ with

$\phi_{0}=\operatorname{\mathcal{M}}_{T}g(X_{0}),\quad\phi_{k}(t_{1},...,t_{k})=\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[\operatorname{\mathcal{M}}_{t_{2}...t_{k}T}g(Y_{t_{1}})].$

If we show that $\phi_{k}\equiv\Delta_{\mathds{1}_{k}}f(X_{0})$ for each
$k\geq 0$, then
$J_{k}\phi_{k}(X_{T})=\Delta_{\mathds{1}_{k}}f(X_{0})J_{k}(X_{T})$ as desired.
We prove by induction the slightly stronger claim,

$\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[(\operatorname{\mathcal{M}}_{t_{2}...t_{k}T}g)(Y_{t_{1}})\,|\,X_{t_{0}}]\equiv\Delta_{\mathds{1}_{k}}f(X_{t_{0}}),\quad(t_{0},t_{1},\ldots,t_{k})\in\triangle_{k+1,T}.$

If $k=0$, then obviously
$\operatorname{\mathcal{M}}_{T}g(X_{0})=f(X_{0})=\Delta_{\emptyset}f(X_{0})$.
For $k\geq 1$, we have

$\displaystyle\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[(\operatorname{\mathcal{M}}_{t_{2}...t_{k}T}g)(Y_{t_{1}})\,|\,X_{t_{0}}]$
$\displaystyle=\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[\Delta_{x}\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[(\operatorname{\mathcal{M}}_{t_{3}...t_{k}T}g)(Y_{t_{2}}^{\prime})\,|\,Y_{t_{1}}]\,|\,X_{t_{0}}]$
$\displaystyle=\operatorname{\mathbb{E}}^{\operatorname{\mathbb{Q}}}[\Delta_{\mathds{1}_{k}}f(Y_{t_{1}})\,|\,X_{t_{0}}]\qquad$
(induction) $\displaystyle=\Delta_{\mathds{1}_{k}}f(X_{t_{0}}),$
(martingality)

Taking $t_{0}=0$ yields the result.

∎

###  A.9 Lemma 3.22

###### Proof.

Fix $k\geq 0$. The Hermite polynomials can be written as

$H_{k}(x)=\sum_{l=0}^{\lfloor k/2\rfloor}c_{k,l}\,x^{k-2l},\quad
c_{k,l}=\frac{k!}{l!(k-2l)!}(-2)^{-l},$ (A.10)

with coefficients $(c_{k,l})_{l=0}^{k}$ retrieved for instance from the
relation $H_{k+1}(x)=xH_{k}(x)-kH_{k-1}(x),\,H_{1}(x)=x,\,H_{0}\equiv 1$; see
[46, Section 5.4]. Thanks to $\eqref{eq:exHermite}$, we have

$J_{k}(X_{t})=\frac{t^{k/2}}{k!}\sum_{l=0}^{\lfloor
k/2\rfloor}c_{k,l}\,\left(\frac{x_{t}}{\sqrt{t}}\right)^{k-2l}=\sum_{2l_{0}+l_{1}=k}(-2)^{-l_{0}}\,\frac{t^{l_{0}}}{l_{0}!}\frac{x_{t}^{l_{1}}}{l_{1}!}.$
(A.11)

Thus, $J_{k}$ is a polynomial in $t$ and $x_{t}$. To conclude, we invoke the
following identity,

$\sum_{|\alpha|_{0}=l_{0},\,|\alpha|_{1}=l_{1}}\operatorname{\mathcal{S}}_{\alpha}(X_{t})=\operatorname{\mathcal{S}}_{\bm{0}_{l_{0}}}(X_{t})\,\operatorname{\mathcal{S}}_{\mathds{1}_{l_{1}}}(X_{t})=\frac{t^{l_{0}}}{l_{0}!}\frac{x_{t}^{l_{1}}}{l_{1}!},$

which is either seen as a consequence of [29, Proposition 5.2.10] or a
particular case of the shuffle product [42]. Since
$\bigcup\limits_{2l_{0}+l_{1}=k}\\{\alpha\,|\,\,|\alpha|_{0}=l_{0},\,|\alpha|_{1}=l_{1}\\}=\operatorname{\mathbb{A}}_{k}$
with $\operatorname{\mathbb{A}}_{k}$ in the statement, the result follows. ∎

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