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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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  • $\begingroup$ In general, a subspace of a contractible space is not necessarily contractible (take $S^1\subseteq \mathbb{R}^2$). $\endgroup$ – Aldo Guzmán Sáenz Sep 1 '13 at 4:00
  • $\begingroup$ Yes @AldoGuzmánSáenz, but $\mathbb{R}^2$ can not retract to $S^1$, hence the retract is contractable. Should I follow this to prove..? $\endgroup$ – 1LiterTears Sep 1 '13 at 4:59
  • $\begingroup$ That is why it is important that $A$ is a retract of $X$, not just a subset of $X$. $\endgroup$ – Tunococ Sep 1 '13 at 5:04
  • $\begingroup$ @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. $\endgroup$ – Aldo Guzmán Sáenz Sep 1 '13 at 5:19
  • $\begingroup$ 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! $\endgroup$ – 1LiterTears Sep 1 '13 at 5:34
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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 $r$.

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  • $\begingroup$ So that means I have to use the other definition..? Thank you Thomas! $\endgroup$ – 1LiterTears Sep 1 '13 at 5:36

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