Prove that if $X$ is contractible, then the suspension $SX$ is also contractible

Question

I am trying to prove the following:

If X is contractible, then the suspension SX is also contractible.

I want to prove this by showing the identity on $SX$ is nullhomotopic. Below is my attempt. I was able to construct a homotopy $SX\times I\to SX$ that starts at the identity on $SX$, but it appears to not end up at a constant map.

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Proof. Let $X$ be contractible and consider $SX:=(X\times I)/\sim$ with quotient map $q\colon X\times I\to SX$. Since $X$ is contractible, there is a homotopy $H\colon X\times I\to X$ such that $H(x,0)=x$ and $H(x,1)=x_0$ for every $x\in X$ with $x_0\in X$ fixed. Define $F\colon (X\times I)\times I\to X\times I$ by $F(x,s,t):=(H(x,t),s)$. We have a diagram (I can't seem to draw diagrams, if someone could edit and produce the linked diagram here that would be great!)

https://q.uiver.app/#q=WzAsNCxbMCwwLCIoWFxcdGltZXMgSSlcXHRpbWVzIEkiXSxbMCwxLCJTWFxcdGltZXMgSSJdLFsxLDEsIlNYIl0sWzEsMCwiWFxcdGltZXMgSSJdLFswLDEsInFcXHRpbWVzXFx0ZXh0e0lkfV9JIiwyXSxbMywyLCJxIl0sWzEsMiwiXFx3aWRldGlsZGV7cVxcY2lyYyBGfSIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDMsIkYiXV0=

where the map $\widetilde{q\circ F}$ making the diagram commute exists if $q\times\text{Id}_I$ is a quotient map and $q\circ F$ is continuous that is constant on the fibers of $q\times\text{Id}_I$. The fact that $q\times\text{Id}_I$ is a quotient map follows from $q$ being a quotient map and $I$ being locally compact, and as both $q$ and $F$ are continuous so is the composition $q\circ F$. Now suppose $(q\times\text{Id}_I)(x,s,t)=(q\times\text{Id}_I)(x',s',t')$. Then $(q(x,s),t)=(q(x',s'),t')$, hence $t=t'$ and $q(x,s)=q(x',s')$ so either $s=s'=0$ or $s=s'=1$. If $t=t'$ and $s=s'=0$ then $$ (q\circ F)(x,s,t)=q(F(x,0,t))=q(H(x,t),0)=q(H(x',t'),0)=q(F(x',0,t'))=(q\circ F)(x',s',t'). $$ If $t=t'$ and $s=s'=1$ then $$ (q\circ F)(x,s,t)=q(F(x,1,t))=q(H(x,t),1)=q(H(x',t'),1)=q(F(x',1,t'))=(q\circ F)(x',s',t'). $$ Therefore $q\circ F$ is constant on the fibers of $q\times\text{Id}_I$, so the map $\widetilde{q\circ F}$ making the diagram commute exists. In particular, we have $$ \widetilde{q\circ F}(q(x,s),0)=(\widetilde{q\circ F})\circ(q\times\text{Id}_I)(x,s,0)=(q\circ F)(x,s,0)=q(F(x,s,0))=q(H(x,0),s)=q(x,s) $$ and $$ \widetilde{q\circ F}(q(x,s),1)=(\widetilde{q\circ F})\circ(q\times\text{Id}_I)(x,s,1)=(q\circ F)(x,s,1)=q(F(x,s,1))=q(H(x,1),s)=q(x_0,s)... $$

Answer 1

I always like to reason about this kind of thing diagramatically. In the below I use the following notation:

• $\Gamma:X\times I\to X$ is a given contraction of $X$ (at time zero) onto the point $p\in X$ (at time $1$)
• $\gamma:I\times I\to I$ is the following map: $$(s,t)\mapsto\begin{cases}(2t,s)&0\le t\le1/2\\(1,2(1-t)s)&1/2\le t\le 1\end{cases}$$
• $\sigma:\partial I\times I\to I$ is the following map: $$(s,t)\mapsto\begin{cases}0&s=0\\1&s=1,\quad0\le t\le1/2\\2(1-t)&s=1,\quad1/2\le t\le1\end{cases}$$

Because the second diagram remains a pushout ($(-)\times I$ is cocontinuous because $I$ is LCH) you can check that a unique $\Phi:SX\times I\to SX$ is induced, which you can further check is a contraction of $SX$ (at time zero) to the point $q(X\times\{0\})$ (at time one).

This diagram works in stages: first, the suspension is collapsed onto a single spine, the image of $\{p\}\times I$ in $SX$, and then this spine is contracted down to its basepoint.

Answer 2

Let $x_0 \in X$ and $H$ a homotopy from the identity of $X$ to the constant map $x_0$. You can construct a retract of $\Sigma X$ in two steps : first retracting on $I \times \{x_0\}$ then retracting to $x_0$. The first retraction is simply $H'((x,s),t)=(H(x,t),s)$ and you can check this is well defined on the suspension. $H'((x,s),0))$ is the identity while $H'((x,s),1)) = (x_0,s)$. Then the second retraction consists of retracting the segment to the point. It is given by : $H((x,s),t) = (x_0,(1-t)s)$ You have $H((x,s),1) = (x_0,0) $ constant while $H((x,s),0)= (x_0,s)$ which is the retraction on the interval.