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\newcommand{\authors}{Alo\"is Rosset}
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\noindent \textbf{Goal :} This document is an assignment aimed to prove the homotopy invariance of relative singular homology for pairs of topological spaces.



\begin{defn} 
	Given pairs of topological spaces $(X, A)$ and $(Y, B)$, two continuous maps $f, g : (X, A) \to (Y, B)$ are said to be \emph{homotopic} if there exists a continuous map $H : (X \times I, A \times I) \to (Y, B)$ such that $H(x, 0) = f (x)$ and $H(x, 1) = g(x), \forall x \in X$. Then the map H is called a \emph{homotopy} from f to g, denoted H : f $\simeq$ g.
\end{defn}



\begin{defn} \thlabel{def chain homotopic}
	Given two chain complexes $C,D$, two chain maps $f,g : C \to D$ are said to be \emph{chain homotopic} if there exists maps $S_n : C_n \to D_n, \forall n \in \mathbb{N}$ s.t 
	\[f_n - g_n = d_{n+1}^D \circ S_n + S_{n+1} \circ d_n^C\]
	\[
	\xymatrix{
		\ldots \phantom{} \ar[r]
			& C_{n+1} \ar[r]^{d} \ar@<-.5ex>[d]_{f} \ar@<.5ex>[d]^{g}
			& C_n \ar[r]^{d} \ar[dl]^S \ar@<-.5ex>[d]_{f} \ar@<.5ex>[d]^{g}
			& C_{n-1} \ar[r]^{d} \ar[dl]^S \ar@<-.5ex>[d]_{f} \ar@<.5ex>[d]^{g}
			& \ldots \\
		\ldots \phantom{} \ar[r]
			& D_{n+1} \ar[r]_{d}
			& D_n \ar[r]_{d}
			& D_{n-1} \ar[r]_{d}
			& \ldots
		} 	
	\]
\end{defn}



\begin{nota} Given a continuous map between topological spaces $f : X \to Y$, we denote by
\begin{itemize}
 \item $f_\# = f_n : C_n(X) \to C_n(Y)$  the induced map defined by $f_\# (\sigma) = f \circ \sigma$, for any singular simplex $\sigma : \Delta^n \to X$
 \item $f_\ast = H_n(f) : H_n(X) \to H_n(Y)$ the induced map defined by $f_\ast(\ov{\alpha})$ where $\ov{\alpha} = \alpha + \ima d_n^X$ denotes the coset $\forall \alpha \in \ker d_n^X$
\end{itemize}
\end{nota}



\begin{prop} \thlabel{homotopic chain maps proposition}
	Given two chain complexes $C,D$ and two homotopic chain maps $f,g : C \to D$, then $H_n(f) = H_n(g), \forall n \in \mathbb{N}$
\end{prop}

\begin{prf}
	With the same notations as in \thref{def chain homotopic},For $\alpha \in \ker d_n^C$ :
	\[
		f_\#(\alpha) - g_\#(\alpha) = d_{n+1}^D \circ S_n (\alpha) + P_{n-1} \circ d_n^C(\alpha) = d_{n+1}^D \circ S_n (\alpha) \in \ima d_{n+1}^D
	\]
	Therefore $\ov{f_\#(\alpha)} = \ov{g_\#(\alpha)}$ as equivalence classes in $H_n(D) = \faktor{\ker d_n^D}{\ima d_{n+1}^D}$. And as $\alpha$ is arbitrary, $H_n(f)=H_n(g)$.
\end{prf}



\begin{thm} \thlabel{homotopy invariance}
	Let $f,g : (X,A) \to (Y,B)$ be two homotopic maps between pairs of spaces. Then $f_\#,g_\# : C(X,A) \to C(Y,B)$ are chain homotopic. And in particular $H_n(f) = H_n(g), \forall n \in \mathbb{N}$
\end{thm}

\begin{prf}
	As $f,g$ are homotopic, there exists a homotopy $H : (X \times I, A \times I) \to (Y, B)$ from $f$ to $g$. For a n-simplex $\sigma : \Delta^n \to X$, we can consider
	\[
		\Delta^n \times I \xrightarrow{\sigma \times \id_I} X \times I \xrightarrow{H} Y
	\] 
	Observe that one can subdivide $\Delta^n \times I$ into $n+1$ $(n+1)$-simplices .
	Indeed if we denote $\Delta^n \times \{0\} = [v_0,\ldots,v_n]$ and $\Delta^n \times \{1\} = [w_0,\ldots,w_n]$, then one let the reader convince oneself that $R_i = [v_0,\ldots,v_i,w_i,\ldots,w_n]$, for $i = 0,\ldots,n$, are $(n+1)$-simplices, with
	\[ 
		\Delta^n \times I = \displaystyle \bigcup_{i=0,\ldots,n} R_i
	\]
	And as $\forall i=0,\ldots,n$, $R_i \cong \Delta^{n+1}$, let's say $\phi_i : \Delta^{n+1} \to R_i$ is the isomorphism.\\
	Let's define for $n \in \mathbb{N}$ :
	\[
		S_n : C_n(X) \to C_{n+1}(Y) : \sigma \mapsto \sum_{i=0}^n (-1)^n H \circ (\sigma \times \id_I) \circ \phi_i
	\]
	Observe that for $\sigma : \Delta^n \to A$, a singular simplex :
	\[ 
		H \circ (\sigma \times \id_I) \circ \phi_i : \Delta^{n+1} \to B \in C_{n+1}(B), \forall i=0,\ldots n
	\]
	Hence it induces a relative map $\tilde{S_n} : C_n(X,A) \to C_{n+1}(Y,B)$.
	From there we can get the relation $f_\# - g_\# = d_{n+1}^Y \circ S_n + S_{n-1} \circ d_n^X$ by computing both sides and obtaining the same sums. And by passing to quotients, the relation being valid for the representatives, it remains so with the $\tilde{S_n}$'s too. This implies that $f_\#,g_\# : C_\ast(X,A) \to C_\ast(Y,B)$ are in fact  chain homotopic. Therefore by \thref{homotopic chain maps proposition}, we get that $\forall n \in \mathbb{N}$, $H_n(f)=H_n(g)$.
\end{prf}



\begin{cor}
	Let $f:(X,A) \to (Y,B)$ be a homotopy equivalence. Then $H_n(f)$ is an isomorphism $\forall n \in \mathbb{N}$
\end{cor}

\begin{proof}
	Let $g : (Y,B) \to (X,A)$ be the homotopy inverse of $f$, i.e $g \circ f \simeq \id_{(X,A)}$, and $f \circ g \simeq \id_{(Y,B)}$. We get the following diagrams :
	\[
	\xymatrix{
		X \ar@/^/[r]^f & Y \ar@/^/[l]^g & & H_n(X,A) \ar@/^1pc/[r]^{H_n(f)} & H_n(Y,B) \ar@/^1pc/[l]^{H_n(g)}
	}
	\]
	Therefore, by \thref{homotopy invariance} and the fact that $H_n$'s are functors :
	\[
		H_n(g) \circ H_n(f) = H_n(g \circ f) = H_n(\id_{(X,A)}) = \id_{H_n(X,A)}
	\]
	Similarly $H_n(f) \circ H_n(g) = \id_{H_n(Y,B)}$ and they are therefore isomorphisms.
\end{proof}
\end{document}

