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Definitions:

A subspace $A \subset X$ is called a retract of $X$ if there is a map $r: X \rightarrow A$ such that $r(a) = a$ for all $a \in A$. (Such a map is called a retraction.)

Problem statement:

Show that a retract of a Hausdorff space is closed.

Proof:

Proof. Let $x \notin A$ and $a =r(x) \in A$. Since $X$ is Hausdorff, $x$ and $a$ have disjoint neighborhoods $U$ and $V$, respectively. Then $r^{−1}(V \cap A) \cap U$ is a neighborhood of $x$ disjoint from $A$. (*) Hence, $A$ is closed.

Question:

I do not understand how "$r^{−1}(V \cap A) \cap U$ is a neighborhood of $x$ disjoint from $A$" implies that $A$ is closed. I would be grateful if someone could point me in te right direction.

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    $\begingroup$ Actually, I think it is more natural to prove this by noting that $A$ is the set of fixed points of $r$. $\endgroup$ Jan 14, 2013 at 19:12
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    $\begingroup$ And the set of fixed points of $r$ is necessarily closed in $X$? $\endgroup$
    – onimoni
    Jan 14, 2013 at 19:29
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    $\begingroup$ Yes. The set of fixed points is the inverse image of the diagonal in $X\times X$ under the continuous map $x\mapsto(x,r(x))$, and the diagonal is itself closed. (There are more direct proofs too, none of them really difficult.) $\endgroup$ Jan 15, 2013 at 11:55
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    $\begingroup$ @mathreader: Since the image of $r$ is (contained in) $A$ already, $r^{-1}(V\cap A)=r^{-1}(V)$. $\endgroup$ Aug 10, 2013 at 8:48
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    $\begingroup$ @OliverG I don't think $V \cap A=V$ is implied. Clearly, $LHS \subseteq RHS$. Conversely, $x \in RHS \Rightarrow r(x) \in V$. But the image of $r$ is in $A$, so $r(x) \in V \cap A$ and $x \in r^{-1}(V \cap A)$. $\endgroup$
    – Bryan Shih
    Jul 5, 2018 at 12:55

2 Answers 2

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The proof shows that the complement of $A$ is open.

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    $\begingroup$ Why is it true that $r^{−1}(V \cap A) \cap U$ is a neighborhood of $x$ disjoint from $A$, though? How do we know we can find a neighborhood U disjoint from A? $\endgroup$
    – Ryker
    Mar 6, 2013 at 4:07
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    $\begingroup$ @Ryker It is true because if $x$ belongs to that set, by definition $x\in U$ and $r(x)\in V$. Therefore $r(x)\ne x$, so $x\notin A$. $\endgroup$ Mar 6, 2013 at 7:07
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    $\begingroup$ Thanks! I actually managed to figure it out on my own in the meantime, but I figured I'd leave the question, so that if you answered someone else might find it useful, as well. $\endgroup$
    – Ryker
    Mar 6, 2013 at 7:37
  • $\begingroup$ You should use another variable. Since a person may confuse the $x$ you used for the arguement, and $x$ used in the main arguement of the question @HaraldHanche-Olsen $\endgroup$ Apr 20, 2023 at 17:04
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    $\begingroup$ @TrystwithFreedom You have a point there, but I can't edit such an old comment. In any case, the reader coming here will be able to see your notice and take corrective action. $\endgroup$ Apr 21, 2023 at 7:03
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Why $r^{-1}(V\cap A)\cap U$ is disjoint from $A$?

$$ \begin{array}{rcl} y\in r^{-1}(V\cap A)\cap U & \Leftrightarrow & y\in\{ z\in X: r(z)\in V\cap A \}\cap U \\ & \Leftrightarrow & y\in\{ z\in U: r(z)\in V\cap A \}. \end{array} $$ So $y\notin A$. If the opposite was true $\;$($y\in A$)$\;$ it should be $y\in U \Rightarrow r(y)\in U$, $\;$but $r(y)\in V\cap A\subset V$ and $U\cap V=\varnothing$.

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