Homotopy and Homotopy Equivalence

The projection $p_X:X\times I\to X$ is continuous by [L1]. For every open $V\subseteq Y$, $(f\circ p_X{)}^{-1}[V]=p_X^{-1}[f^{-1}[V]]$ is open, so $K_f(x,t):=f(x)$ is continuous by [L2].

The map $r:I\to I$, $r(t)=1-t$, is continuous: for $t_0\in I$ and an open neighbourhood $V=O\cap I$ of $r(t_0)$, with $O$ open in $\mathbb{R}$, [L3] gives $\epsilon >0$ with $(r(t_0)-\epsilon ,r(t_0)+\epsilon )\subseteq O$; then $U=(t_0-\epsilon ,t_0+\epsilon )\cap I$ is an open neighbourhood of $t_0$ and $r[U]\subseteq V$, since $\mathrm{\mid }r(t)-r(t_0)\mathrm{\mid }=\mathrm{\mid }t-t_0\mathrm{\mid }$.

The homotopy $K_f$ has $K_f(x,0)=K_f(x,1)=f(x)$ and $K_f(a,t)=f(a)$ for $a\in A$, so it is a homotopy from $f$ to itself rel $A$.

The map $R:X\times I\to X\times I$, $R(x,t)=(x,r(t))$, is continuous because its components are continuous by step 1.2 and [L1]. For every open $V\subseteq Y$, $(H\circ R{)}^{-1}[V]=R^{-1}[H^{-1}[V]]$ is open, so $\bar{H}:=H\circ R$ is continuous by [L2].

One has $\bar{H}(x,0)=H(x,1)=g(x)$ and $\bar{H}(x,1)=H(x,0)=f(x)$; for $a\in A$, $\bar{H}(a,t)=H(a,1-t)=f(a)=g(a)$. Hence $\bar{H}$ is a homotopy from $g$ to $f$ rel $A$.

The sets $I_0=[0,\frac{1}{2}]$ and $I_1=[\frac{1}{2},1]$ are closed in $I$: their complements are respectively $I\cap (\frac{1}{2},\frac{3}{2})$ and $I\cap (-\frac{1}{2},\frac{1}{2})$, traces of open intervals of $\mathbb{R}$. Hence $D_0=X\times I_0$ and $D_1=X\times I_1$ are closed in $X\times I$, because they are the preimages of $I_0,I_1$ under the continuous time projection.

The maps $a_0:I_0\to I$, $a_0(t)=2t$, and $a_1:I_1\to I$, $a_1(t)=2t-1$, are continuous. Indeed, at any $t_0$ and for any ambient open interval of radius $\epsilon$ about $a_j(t_0)$, the relative interval of radius $\epsilon \mathrm{/}2$ about $t_0$ maps into it, since $\mathrm{\mid }{a}_j(t)-a_j(t_0)\mathrm{\mid }=2\mathrm{\mid }t-t_0\mathrm{\mid }$; [L5] turns these intervals into the required subspace neighbourhoods.

Define $r_j:D_j\to X\times I$ by $r_j(x,t)=(x,a_j(t))$. The first component is the restricted product projection and the second is $a_j$ after the time projection, so $r_j$ is continuous by [L2], step 1.2 and [L3].

The maps $K_0:=F\circ r_0:D_0\to Y$ and $K_1:=G\circ r_1:D_1\to Y$ are continuous: for every closed $C\subseteq Y$, $K_j^{-1}[C]=r_j^{-1}[F^{-1}[C]]$ or $r_j^{-1}[G^{-1}[C]]$, which is closed by [L1]. On $D_0\cap D_1=X\times \{\frac{1}{2}\}$ they agree, since $K_0(x,\frac{1}{2})=F(x,1)=g(x)=G(x,0)=K_1(x,\frac{1}{2})$.

Thus the displayed clauses define one function $K:X\times I\to Y$. If $C\subseteq Y$ is closed, then $K^{-1}[C]$ is the union of $K_0^{-1}[C]$, regarded as a closed subset of $X\times I$ through the closed subspace $D_0$, and $K_1^{-1}[C]$, regarded likewise through $D_1$. Each is closed by [L2] and steps 1.1 and 3.1, so their union is closed by [L4]. Hence $K$ is continuous by [L1].

At $t=0$ the first clause gives $K(x,0)=F(x,0)=f(x)$, and at $t=1$ the second gives $K(x,1)=G(x,1)=h(x)$. If $a\in A$, both clauses give the common value $f(a)=g(a)=h(a)$ for every $t$. Therefore $K$ is a homotopy from $f$ to $h$ rel $A$.

By [L1] and [L2], ${\simeq }_A$ is reflexive, symmetric and transitive, so it is an equivalence relation by [L3]. Taking $A=\mathrm{\varnothing }$ gives ordinary homotopy.

Paths from $y_0$ to $y_1$ are continuous maps $I\to Y$ with one prescribed restriction to the subspace $\{0,1\}$, and their path homotopies are exactly homotopies rel $\{0,1\}$. Hence step 1.1 applies to them.

Define $U:W\times I\to X\times I$ by $U(w,t)=(u(w),t)$. Its components are $u$ after the first projection and the second projection, so $U$ is continuous by [L1].

The composite $v\circ H:X\times I\to Z$ is continuous by the same preimage calculation, and its endpoints are $v\circ f$ and $v\circ g$; for $a\in A$ it has the fixed value $v(f(a))=v(g(a))$. Thus it is a homotopy from $v\circ f$ to $v\circ g$ rel $A$.

The composite $H\circ U$ is continuous: for every open $V\subseteq Y$, $(H\circ U{)}^{-1}[V]=U^{-1}[H^{-1}[V]]$ is open by [L2]. Its endpoints are $(f\circ u)(w)$ and $(g\circ u)(w)$; if $b\in B$, then $u(b)\in A$, so $(H\circ U)(b,t)=f(u(b))=g(u(b))$. Thus it is a homotopy from $f\circ u$ to $g\circ u$ rel $B$.

Steps 2.1 and 1.2 prove the two claims.

If $X$ is contractible, apply [A1] to the continuous map ${\mathrm{id}}_X:X\to X$ to conclude that ${\mathrm{id}}_X$ is nullhomotopic.

Conversely suppose ${\mathrm{id}}_X\simeq c_{x_0}$ for some $x_0\in X$, and let $f:X\to Y$ be any continuous map. Postcomposition by $f$ gives $f=f\circ {\mathrm{id}}_X\simeq f\circ c_{x_0}=c_{f(x_0)}$ by [L1]. Thus $f$ is nullhomotopic.

Since $Y$ and $f$ in step 1.2 were arbitrary, every continuous map out of $X$ is nullhomotopic, so $X$ is contractible by [A1]. Together with step 1.1 this proves the equivalence.

The identity ${\mathrm{id}}_X:X\to X$ is a homotopy equivalence with itself as homotopy inverse, since both composites equal, and hence are homotopic to, ${\mathrm{id}}_X$.

If $f:X\to Y$ has homotopy inverse $g:Y\to X$, the same two homotopies show that $g$ has homotopy inverse $f$. Thus the relation is symmetric.

For transitivity, the composite $p\circ f:X\to Z$ has candidate homotopy inverse $g\circ q:Z\to X$. By [A1], $q\circ p\simeq {\mathrm{id}}_Y$ and $g\circ f\simeq {\mathrm{id}}_X$.

Similarly, $f\circ g\simeq {\mathrm{id}}_Y$ gives $p\circ (f\circ g)\circ q\simeq p\circ q$, and $p\circ q\simeq {\mathrm{id}}_Z$ then gives $(p\circ f)\circ (g\circ q)\simeq {\mathrm{id}}_Z$.

Applying [L1] to $q\circ p\simeq {\mathrm{id}}_Y$ gives $g\circ (q\circ p)\circ f\simeq g\circ f$, and [L2] with $g\circ f\simeq {\mathrm{id}}_X$ gives $(g\circ q)\circ (p\circ f)\simeq {\mathrm{id}}_X$.

Steps 2.1 and 1.4 make $g\circ q$ a homotopy inverse of $p\circ f$, so the relation is transitive. With steps 1.1 and 1.2, [L3] makes it an equivalence relation.

Postcomposing $f\simeq f_0$ by $g$ gives $g\circ f\simeq g\circ f_0$ by [L1], and [A1] with transitivity gives $g\circ f\simeq {\mathrm{id}}_X$.

Precomposing $f\simeq f_0$ by $g$ gives $f\circ g\simeq f_0\circ g$ by [L1], and [A1] with transitivity gives $f\circ g\simeq {\mathrm{id}}_Y$.

Steps 1.1 and 1.2 show that $g$ is a homotopy inverse of $f$, so $f$ is a homotopy equivalence.

The equality $r\circ i={\mathrm{id}}_A$ is in particular a homotopy $r\circ i\simeq {\mathrm{id}}_A$, while $i\circ r{\simeq }_A{\mathrm{id}}_X$ is in particular an ordinary homotopy.

Hence $r$ satisfies both homotopy-inverse conditions for $i$, so $i$ is a homotopy equivalence with homotopy inverse $r$.

By [L1], ${\mathrm{id}}_X\simeq c_{x_0}$ for some $x_0\in X$. Precompose this homotopy by $i$ and postcompose by $r$. By [L2], $r\circ {\mathrm{id}}_X\circ i\simeq r\circ c_{x_0}\circ i$.

The left side of step 1.1 is $r\circ i={\mathrm{id}}_A$ by [A1], and the right side is the constant map on $A$ with value $r(x_0)$. Thus ${\mathrm{id}}_A$ is nullhomotopic.

Since $A$ is nonempty, step 2.1 and [L1] applied to $A$ show that $A$ is contractible.

If a path $\gamma :I\to X$ joins $x$ to $x^{\mathrm{\prime }}$, then $f\circ \gamma$ is continuous because $(f\circ \gamma {)}^{-1}[V]={\gamma }^{-1}[f^{-1}[V]]$ for every open $V\subseteq Y$; it joins $f(x)$ to $f(x^{\mathrm{\prime }})$. Hence points in one path component of $X$ have images in one path component of $Y$, so $f_{\ast }$ is well defined. The same argument defines $g_{\ast }$.

If continuous maps $u,v:X\to Y$ are homotopic, then $u(x)$ and $v(x)$ lie in the same path component for every $x\in X$: precompose the homotopy by the continuous map from a one-point space selecting $x$, using [L3]; the resulting homotopy of two maps from a point is exactly a path from $u(x)$ to $v(x)$.

Apply step 1.2 to $g\circ f\simeq {\mathrm{id}}_X$. For every $x\in X$, $g(f(x))$ and $x$ lie in the same path component, so $(g_{\ast }\circ f_{\ast })(P_X(x))=P_X(x)$. Thus $g_{\ast }\circ f_{\ast }$ is the identity on ${\pi }_0(X)$.

Applying step 1.2 to $f\circ g\simeq {\mathrm{id}}_Y$ similarly gives $f_{\ast }\circ g_{\ast }={\mathrm{id}}_{{\pi }_0(Y)}$.

Therefore $f_{\ast }$ and $g_{\ast }$ are mutually inverse functions, so $f_{\ast }$ is a bijection.

Addition and multiplication ${\mathbb{R}}^2\to \mathbb{R}$ are continuous. Indeed the coordinate projections are continuous by [L1], and by [L2] may be read as continuous scalar functions on the Euclidean metric space ${\mathbb{R}}^2$. The identity map $z\mapsto z$ and the constant map $z\mapsto (1,1)$ are continuous, so [L3] makes their inner product $z_0+z_1$ continuous. The maps $z\mapsto (z_0,0)$ and $z\mapsto (z_1,0)$ are continuous by the componentwise part of [L3], and their inner product $z_0{z}_1$ is continuous by its algebra part.

Consequently, if $a,b:Z\to \mathbb{R}$ are continuous on an arbitrary topological space $Z$, then $a+b$ and $ab$ are continuous: the pair $(a,b):Z\to {\mathbb{R}}^2$ is continuous by [L1] and [L2], and composing it with the two maps of step 1.1 is continuous because the preimage of an open set under a composite is an iterated preimage, which is open by [L5]. Constant functions and additive inverses are continuous by the same argument, using a constant component and the continuous scalar multiple supplied by [L3].

Let $\iota :C\hookrightarrow {\mathbb{R}}^n$ be the inclusion and put $F=\iota \circ f$, $G=\iota \circ g$. These ambient maps are continuous by [L4]. For $Z=X\times I$, let $p_X:Z\to X$ and $\tau :Z\to I\subseteq \mathbb{R}$ be the projections. Each scalar coordinate $F_k\circ p_X$ and $G_k\circ p_X$ is continuous: coordinate projections on ${\mathbb{R}}^n$ are continuous by [L1] and [L2], and composites preserve continuity by the preimage calculation of step 2.1. The scalar map $\tau$ is continuous, also as a map into $\mathbb{R}$, by [L1] and [L4].

By step 2.1, for every $k<n$ the function $(x,t)\mapsto (1-t)F_k(x)+t{G}_k(x)$ is continuous on $Z$. Therefore the ambient map $\tilde{H}:Z\to {\mathbb{R}}^n$ with these coordinates is continuous by [L1] and [L2].

Convexity [A1] gives $\tilde{H}(x,t)\in C$ for every $(x,t)\in Z$. Since the composite of $H:Z\to C$ with the inclusion $\iota$ is $\tilde{H}$, [L4] makes $H$ continuous into $C$.

The map $H(x,t)=(1-t)f(x)+tg(x)$ is continuous by [L1].

Substitution gives $H(x,0)=f(x)$ and $H(x,1)=g(x)$ for every $x\in X$.

Thus $H$ satisfies the continuity and endpoint conditions of [A1], so it is a homotopy from $f$ to $g$.

Apply [L1] to ${\mathrm{id}}_C:C\to C$ and the constant map $c_c:C\to C$. The resulting homotopy is $H(x,t)=(1-t)x+tc$.

Thus ${\mathrm{id}}_C$ is nullhomotopic, so $C$ is contractible by [L2].

Let $H:X\times I\to X$ be a homotopy from ${\mathrm{id}}_X$ to $c_{x_0}$. For each $z\in X$, the map $j_z:I\to X\times I$, $j_z(t)=(z,t)$, is continuous by [L2], its components being constant and the identity.

The map ${\gamma }_z:=H\circ j_z:I\to X$ is continuous because $(H\circ j_z{)}^{-1}[V]=j_z^{-1}[H^{-1}[V]]$ is open for every open $V\subseteq X$. It has ${\gamma }_z(0)=z$ and ${\gamma }_z(1)=x_0$, so it is a path from $z$ to $x_0$.

Step 2.1 gives a path from $x$ to $x_0$ and a path from $y$ to $x_0$. Reversing the latter and concatenating it with the former gives a path from $x$ to $y$ by [A1].

Since $x,y\in X$ were arbitrary, $X$ is path-connected.

If $s\in S^{n-1}$ then $\parallel s{\parallel }_2=1$, so $r(s)=s$. Thus the continuous map $r$ of [L1] is a retraction.