Homotopy and Homotopy Equivalence — Examples

For $u,v\in\mathbb R^n$ and $t\in I$, the vector $(1-t)u+tv$ lies in $\mathbb R^n$, so $\mathbb R^n$ is convex.

The map $H$ is continuous by [L1].

Substitution gives $H(x,0)=f(x)$ and $H(x,1)=g(x)$, so [L2] identifies $H$ as a homotopy from $f$ to $g$.

For $x,c\in J$ and $t\in I$, $(1-t)x+tc$ lies between $x$ and $c$, so it lies in $J$ by [A1]. Thus $J$, viewed as a subset of $\mathbb R^1$, is convex.

The whole space $\mathbb R^n$ is convex, since it is closed under vector addition and scalar multiplication.

Apply [L1] to step 1.1 and the point $c$ to obtain the stated contraction of $J$, and apply it to step 1.2 and any chosen point of $\mathbb R^n$ to obtain the Euclidean contraction.

Both $i$ and $q$ are continuous, since the preimage of any open set is either empty or the entire one-point source or target preimage.

No bijection exists from the finite set $P$ to the uncountable set $\mathbb R$, so no homeomorphism exists by [L3].

One has $q\circ i=\operatorname{id}_P$ and $i\circ q=c_0\simeq\operatorname{id}_{\mathbb R}$ by [L1]. Hence $i$ and $q$ are homotopy inverses by [L2].

Steps 2.1 and 1.2 verify the claimed contrast, agreeing with [L4].

The map $r(0)=r(1)=0$ is continuous by [L1] and fixes the point of $A$, so it is a retraction.

If a deformation retraction existed, [A1] would make $\operatorname{id}_X$ homotopic to the constant map at $0$; the full contradiction with the separation $X=\{0\}\cup\{1\}$ is established in [L2].

Therefore $A$ is a retract but not a deformation retract of $X$.

The map $r:Z\to A$, $r(x,t)=(x,0)$, is continuous: its composite with the inclusion $A\hookrightarrow Z$ has the continuous components $(x,t)\mapsto x$ and the constant $0$, so [L1] applies. It fixes every $(x,0)\in A$, hence is a retraction.

On the source $Z$, the second projection $p_I:Z\to I$ and the constant zero map are continuous. Since $I$ is convex, [L2] makes $L:Z\times I\to I$, $L((x,t),s)=(1-s)t$, continuous.

One has $K((x,t),0)=(x,t)$, $K((x,t),1)=(x,0)$, and $K((x,0),s)=(x,0)$ for every $s\in I$.

The first component $((x,t),s)\mapsto x$ is continuous as a composite of product projections, since the preimage of an open set is an iterated preimage and hence open by [L3]. Together with step 1.2, [L1] makes $K((x,t),s)=(x,L((x,t),s))$ continuous into $Z$.

Steps 1.1, 2.1 and 1.3 satisfy [A1], so $(r,K)$ is a deformation retraction of $Z$ onto $A$.

Substituting $t=0$ gives $H(x,0)=x$, and substituting $t=1$ gives $H(x,1)=x/\lVert x\rVert_2$.

If $s\in S^{n-1}$ then $\lVert s\rVert_2=1$, so $H(s,t)=((1-t)+t)s=s$ for all $t\in I$.

Continuity and avoidance of the origin are supplied by [L1]. Thus steps 1.1 and 1.2 explicitly verify the endpoint and fixed-sphere clauses of the deformation retraction [L2].

The map $H$ is continuous by [L1].

One has $H(s,0)=\alpha(s)$ and $H(s,1)=\beta(s)$. At the endpoints, $H(0,t)=(1-t)\alpha(0)+t\beta(0)=\alpha(0)$ and $H(1,t)=(1-t)\alpha(1)+t\beta(1)=\alpha(1)$.

Thus $H$ satisfies all clauses of [A1], so it is a path homotopy from $\alpha$ to $\beta$ relative to the endpoints.