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I have to proof that every retraction is a quotient map.. I have no idea where to start or what to use! A retraction $r:X \rightarrow A$ is a continuous map s.t. $r(a)=a$ for every $a\in A$.

user57012
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3 Answers3

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You could use the following theorem; Let $f: X \rightarrow Y$ be a continous map and suppose there is a continuous map $g : Y \rightarrow X$ such that $f \circ g$ is the identity. Then $f$ is a quotient map.

Now let $r:X \rightarrow A$ be your retraction, you could now easiy find a map $s$ such that $r \circ s$ is the identity map on A.

onimoni
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    proof of the theorom used: First of all note that $f$ is surjective because $g$ is a right inverse for $f$. Let $V \subset Y$ and suppose $f^{-1}(V)$ is open in $X$. Because $g$ is continuous, $g^{-1}(f^{-1}(V))$ is open in $Y$. But $g^{-1}(f^{-1}(V)) = (f \circ g)^{-1}(V) = U$, thus $V$ is open in $Y$. – onimoni Jan 13 '13 at 16:30
  • Another nice result that follows from $f\circ g$ being the identity on Y, is that $g$ is an embedding, i.e. $g(Y)\cong Y$. So by replacing $f$ with the "equivalent" $g\circ f$ we get a retraction in the topological sense whenever $f$ is retraction in the categorical sense. – Stefan Hamcke Jan 14 '13 at 00:02
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HINT: By definition the map $r$ is a quotient map if and only if the following is true:

$\qquad\qquad\qquad$ a set $U\subseteq A$ is open in $A$ if and only if $r^{-1}[U]$ is open in $X$.

Suppose that $U\subseteq A$ is open in $A$; then $r^{-1}[U]$ is certainly open in $X$, simply because $r$ is continuous. Now suppose that $U\subseteq A$ is not open in $A$, and let $S=r^{-1}[U]$; you want to prove that $S$ is not open in $X$. If $S$ were open in $X$, then $S\cap A$ would be open in $A$, by definition of the subspace topology on $A$. What is $S\cap A$? Is it open in $A$?

Brian M. Scott
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  • This, however, is only true if we assume A has the subspace topology, is it not? I assume that if it doesn't, a continuous retraction is not necessarily a quotient map. Or am I missing something? – Ryker Feb 13 '13 at 02:52
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    @Ryker: If you’re talking about a retraction of $X$ to $A$, $A$ simply does have the subspace topology. – Brian M. Scott Feb 13 '13 at 03:07
  • Ah, I see. But from the problem as stated above, this would not be clear, would it? I mean, "retraction" is just a tag, after all. Oh, and the reason why I brought this up is that I wanted to prove the same thing the OP did, and was given the same starting conditions. So since there was no additional info on what a retraction is, it wasn't clear to me A does indeed have the subspace topology. – Ryker Feb 13 '13 at 05:58
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    @Ryker: Yes, it’s absolutely clear from the problem statement. A retraction $r:X\to A$ is a continuous map from $X$ to its subspace $A$ with the property that $f(a)=a$ for each $a\in A$. – Brian M. Scott Feb 13 '13 at 06:01
  • why do we need to assume that U is closed? If r is continuous and U is closed (not open in A), then the pre-image $r^{-1}(U)$ is not open in X by definition of continuous functions pg 104 Munkres. which is what we wanted. – sophie-germain Oct 21 '14 at 00:35
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    @Anthony: At no point do I assume that $S$ is closed. I don’t even assume that $S$ is not open, which is a very different thing. – Brian M. Scott Oct 21 '14 at 00:43
  • I made a typo, and updated it, I meant your comment "Now suppose U $\subseteq A$ is not open in A" ... I claim that since r is continuous then the pre image of U (not open in Y) is not open in X. please forgive confusion. – sophie-germain Oct 21 '14 at 00:52
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    @Anthony: I’m proving an equivalence. First I showed that if $U$ is open in $A$, then $S$ is open in $X$. Now I have to prove the opposite implication: if $S$ is open in $X$, then $U$ is open in $A$. As it happens I could have proved this directly (and essentially did), but when I wrote that up for some reason I chose to write it as a proof of the (logically equivalent) contrapositive, i.e., that if $U$ is not open in $A$, then $S$ is not open in $X$. – Brian M. Scott Oct 21 '14 at 00:57
  • There is a theorem in Munkres pg 104 thm 18.1 that says "for every closed set B of Y, the set $f^{-1}(B)$ is closed in X , such that $f : X \to Y$ is continuous ... this seems that this theorem will do exactly what we want. Do you agree? – sophie-germain Oct 21 '14 at 01:00
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    @Anthony: No. Why do you think so? (In fact $f$ is continuous if and only if $f^{-1}[B]$ is closed in $X$ for each closed $B$ in $Y$. This is of course also true if you substitute open for closed in both places, and that’s what I used to prove the first direction of the equivalence.) – Brian M. Scott Oct 21 '14 at 01:03
  • Let us continue this discussion in chat. – sophie-germain Oct 21 '14 at 01:38
  • @Anthony: Okay, I’m there. – Brian M. Scott Oct 21 '14 at 01:40
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If you don't want to play around with open sets, you could use the universal property of quotient maps: a continuous map $q:X\to A$ is a quotient map if and only if, whenever $f:X\to Z$ is a continuous map such that $q(x)=q(y)\Rightarrow f(y)=f(y)$, there is a unique continuous map $f':A\to Z$ such that $f = f'\circ q$. I suggest that in the case where $q$ is a retraction onto $A\subset X$, given such an $f$, there is a clear choice of $f'$ to try.

bradhd
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