Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $a\in A$.

share|improve this question

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer
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 $A$?

share|improve this answer
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 2:52
@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
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 5:58
@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

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.