I am wondering if my proof is correct? Thank you very much.

Show that a retract of a contractible space is contractible.

Given $X$ contracts to $x \in X$, we know there is a family of maps $f_t: X \to X, t \in I$, such that $f_0 = \mathbb{I}$ (the identity map), $f_1(X) = x$, and f_t|x= \mathbb{I} for all t.

Consider a retract on $X$ to $A$, we know there is a map $r: X \to A$, such that $r(X) = A$, $r|_A = \mathbb{I}|_A$.

And now we set out to show that $A$ contracts to any $a \in A$, that is, there exists $\hat{f}$ such that $\hat{f}_t: A \to A, t \in I$, such that $\hat{f}_0 = \mathbb{I}$, $\hat{f}_1(A) = a$ , and f_t|a= \mathbb{I} for all t .

But since $X$ retracts to $A$, that means $r$ brings any point $x \in X$ to some $a^\prime \in A$ homotopically. Therefore, we have a map from $X$ to $a^\prime$, which is the $\hat{f}$ we want when restricts to $A$. That is, $$\hat{f}_t = r \circ f_t,$$ because it satisfies all the criterion we want: $\hat{f}_0|_A = r \circ f_0|_A = r \circ \mathbb{I}|_A = \mathbb{I}|_A$, $\hat{f}_1(A) = r \circ f_1(A) = r \circ x = a^\prime$ which satisfy the condition that \hat{f}_1(A) = a for any a \in A, and f_t|a^\prime= \mathbb{I} for all t.

  • 1
    $\begingroup$ You got the wrong definitions: $A$ being a retract of $X$ means only that there is a map $r:X\to A$ restricting to the identity on $A$. Also, you assumed a strong deformation retract to $x$, but contractible means that the identity on $X$ is null-homotopic, that is homotopic to the constant map $X\to x$, the homotopy need not fix $x$ during the time interval. $\endgroup$ – Stefan Hamcke Sep 1 '13 at 21:06
  • $\begingroup$ Dear @StefanH. - that is very helpful, thank you. I fixed it, does it look ok now? Thanks. $\endgroup$ – 1LiterTears Sep 1 '13 at 21:31
  • $\begingroup$ I think you should add more details and explain what you did to explicitly verify that $\hat f_t=r\circ f_t$ is the homotopy we want (which it is indeed). $\endgroup$ – Stefan Hamcke Sep 1 '13 at 22:02
  • $\begingroup$ I see, thank you very much @StefanH. I added at the last, and I hope it is better now. $\endgroup$ – 1LiterTears Sep 1 '13 at 22:12
  • $\begingroup$ You have to show that $\hat f_0(a)=a$ for each $a\in A$ and not such $\hat f_0(A)=A.$ $\endgroup$ – Stefan Hamcke Sep 1 '13 at 22:17

There are still some inconsistencies in your text:

Consider a retract on $X$ to $A$, we know there is a map $r:X→A,$ such that $r(X)=A,$ $r(A)=A.$

This still has to be corrected. It should better say: "$r:X\to A$ such that $r|_A=\Bbb I|_A$"

$\hat f_1(A)=a$ for any $a∈A$

That can be deleted. You already know that $\hat f_1(A)=\{a'\}$ and that is all you want.

  • $\begingroup$ I see, thank you very much Stefan! $\endgroup$ – 1LiterTears Sep 2 '13 at 2:48
  • $\begingroup$ Also the formula "$r(X)=A$" is not wrong but unnecessary, because $r|_A=\Bbb I|_A$ tells you that each $a$ in $A$ is image of itself under $r$, so the range of $r$ contains $A$ and, conversely, is contained in $A$ since $r:X\to A$. $\endgroup$ – Stefan Hamcke Sep 2 '13 at 13:53

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