Exercise 0.3

Question

\begin{enumerate}[label=(\alph*)]
    \item Show that the composition of homotopy equivalences $X 	o Y$ and $Y 	o Z$ is a homotopy equivalence $X 	o Z$. Deduce that homotopy equivalence is an equivalence relation.
    \item Show that the relation of homotopy among maps $X 	o Y$ is an equivalence relation.
    \item Show that a map homotopic to a homotopy equivalence is homotopy equivalence.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{remark}
  We remark that homotopy equivalences preserve compositions. For, if $f,g \in \operatorname{Hom}(X,Y)$ and $F(x,t)$ is a homotopy $f \simeq g$, and if $e \colon W 	o X$, $h \colon Y 	o Z$, then $h(F(e(x),t))$ is a homotopy $e \circ f \circ h \simeq e \circ g \circ h$ since the composition of continuous maps is continuous and since $h(F(e(x),0)) = e \circ f \circ h$ and $h(F(e(x),1)) = e \circ g \circ h$. This allows us to replace maps by homotopy equivalent ones in $(a)$ and $(c)$.
\end{remark}
\begin{proof}[Proof of $(a)$]
  Let $e\colon X 	o Y$, $f \colon Y 	o X$ such that $e \circ f \simeq \operatorname{id}_X$, $f \circ e \simeq \operatorname{id}_Y$; likewise, let $g \colon Y 	o Z$, $h \colon Z 	o Y$ such that $g \circ h \simeq \operatorname{id}_Y$, $h \circ g \simeq \operatorname{id}_Z$. Then, since composition of maps is associative,
  \begin{alignat*}{5}
    (e \circ g) \circ (h \circ f) &={}& e \circ (g \circ h) \circ f &\simeq{}& e \circ \operatorname{id}_Y \circ\:f &={}& e \circ f &\simeq{}& \operatorname{id}_X\
    (h \circ f) \circ (e \circ g) &={}& h \circ (f \circ e) \circ g &\simeq{}& h \circ \operatorname{id}_Y \circ\:g &={}& h \circ g &\simeq{}& \operatorname{id}_Z
  \end{alignat*}
  and so we have a homotopy equivalence $X \simeq Z$ (note we have to use the fact that homotopy is transitive from $(b)$). Since homotopy equivalence is reflexive (take the identity map) and symmetric (change the roles of $X,Z$), homotopy equivalence is an equivalence relation since we have shown transitivity.
\end{proof}
\begin{proof}[Proof of $(b)$]
  Let $f,g,h \in \operatorname{Hom}(X,Y)$. Homotopy is reflexive since we can take $F(x,t) = f(x)$ for all $t \in [0,1]$, and symmetric since if $F(x,t)$ is a homotopy between $f,g$, then $F(x,1-t)$ is a homotopy between $g,f$ by the fact that $1-t$ is continuous and so the composition $F(x,1-t)$ is continuous.
  \par It remains to show homotopy is transitive. Let $F$ be a homotopy between $f,g$ and $G$ be a homotopy between $g,h$. Then, let
  \begin{equation*}
    H(x,t) = \begin{cases}
      F(x,2t) & 	ext{if}~t \in [0,1/2],\
      G(x,2t-1) & 	ext{if}~t \in [1/2,1].\
    \end{cases}
  \end{equation*}
  This is continuous by the pasting lemma since $H(x,1/2) = F(x,1) = G(x,0) = g(x)$. $H$ is then a homotopy between $f,h$ since $H(x,0) = F(x,0) = f(x)$ while $H(x,1) = G(x,1) = h(x)$. Thus, homotopy is an equivalence relation.
\end{proof}
\begin{proof}[Proof of $(c)$]
  Let $e\colon X 	o Y$, $f\colon Y 	o X$ such that $e \circ f \simeq \operatorname{id}_X$, $f \circ e \simeq \operatorname{id}_Y$, and let $g\colon X 	o Y$ such that $e \simeq g$. Then, $g \circ f \simeq e \circ f \simeq \operatorname{id}_X$, $f \circ g \simeq f \circ e \simeq \operatorname{id}_Y$, and so $g$ is also a homotopy equivalence.
\end{proof}

Exercise 0.3

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Exercise 0.3

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