MHB Solution: Is a retraction a quotient map?

[Chris L T521]

Here's this week's problem! Sorry about the delay!

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Problem: If $A\subset X$, a retraction of $X$ onto $A$ is a continuous map $r:X\rightarrow A$ such that $r(a)=a$ for each $a\in A$. Show that a retraction is a quotient map.

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[Chris L T521]

I had an emergency to tend to the last couple days, so sorry for the delay in posting the solution!

This week's problem was correctly answered by Deveno. You can find his solution below.

[sp]I will assume that the subspace (or relative) topology on $A$ is what is intended in this problem.

Note that the subspace topology on $A$ automatically makes the function:

$i:A \to X$ given by $i(a) = a$, for all $a \in A$ continuous. Hence:

$r \circ i: A \to A$ is continuous, being the composition of 2 continuous maps. In fact: $(r\circ i)(a) = r(i(a)) = r(a) = a$, so $r \circ i = 1_A$.

Now $r$ is clearly a surjection, since any $a \in A$ is in its own pre-image.

Now suppose $U \subseteq A$ is such that $r^{-1}(U)$ is open in $X$.

Since $i$ is continuous, $i^{-1}(r^{-1}(U))$ is open in $A$.

But $i^{-1}(r^{-1}(U)) = (r \circ i)^{-1}(U) = (1_A)^{-1}(U) = U$.

Hence $U$ is open in $A$, that is, $r$ is a quotient map.[/sp]