# Every retraction is a quotient map?

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 $a\in A$.

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## 3 Answers

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.

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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 $Y$. – omar 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 0:02

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 $A$?

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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 3:07
@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 6:01
@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 0:43
@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 0:57
@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 1:03

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.

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