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I wonder if my approach is completely wrong. If so, may I request for some hints for heading to the right direction? Thank you!

Show that a retract of a contractible space is contractible.

The previous discussion Proof that retract of contractible space is contractible. used the definition that

The identity map on X is null-homotopic, i.e. if it is homotopic to some constant map.

But as to what I found, a space being contractible is simply defined as

A space having the homotopy type of a point is called contractible.

So following this definition, I can only carried out the proof in such a way:

We retract this space $X$ to $A \subset X$. Since $A$ is a subset of $X$, $A$ is also contractible.

So I am wondering, did I miss the definition somewhere? Or I should just use the other definition?

Thank you...

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1LiterTears
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  • In general, a subspace of a contractible space is not necessarily contractible (take $S^1\subseteq \mathbb{R}^2$). – Aldo Guzmán Sáenz Sep 01 '13 at 04:00
  • Yes @AldoGuzmánSáenz, but $\mathbb{R}^2$ can not retract to $S^1$, hence the retract is contractable. Should I follow this to prove..? – 1LiterTears Sep 01 '13 at 04:59
  • That is why it is important that $A$ is a retract of $X$, not just a subset of $X$. – Tunococ Sep 01 '13 at 05:04
  • @Jellyfish, yes, you are right, of course. I commented it because you wrote "Since $A$ is a subset of $X$, $A$ is also contractible" (emphasis mine), and I thought that maybe you thought that any subset of a contractible space was contractible, which of course is not true. I see now that this was not the case. – Aldo Guzmán Sáenz Sep 01 '13 at 05:19
  • Oh yes, I see what you mean @AldoGuzmánSáenz. I see why my proof won't work - so that means I have to use the other definition..? Thank you! – 1LiterTears Sep 01 '13 at 05:34
  • What is the definition of two spaces having the same homotopy type? The usual one is having two maps such that the compositions are homotopic to the identities of the spaces. In this case, I'd say that the proof in your linked question works. – Aldo Guzmán Sáenz Sep 01 '13 at 05:51

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A hint:

You have maps $A \xrightarrow{i} X \xrightarrow{r} A$ such that $ri = id_A$. You are also given the existence of a map $H : X \times I \to X$ such that, for each $x$, $H(x,0) = x$ and $H(x,1) = *$ for some fixed point $* \in X$. You can show that you might as well choose $* \in A$ (think about why -- must $X$ be path connected?)

You need to show the existence of a map $\widetilde{H} : A \times I \to A$ such that $\widetilde{H}(a,0) = a$ and $\widetilde{H}(a,1) = *$ for every $a \in A$. Try to build $\widetilde{H}$ using the maps $H,i$ and $r$.

Thomas Belulovich
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