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Let $p:X\rightarrow Y$ be an open map and let $A$ be a subspace of $X$. Then, is it true that $p|_A:A\rightarrow p(A)$ is open?

I think so, but am struggling to show it.

My thoughts: Let $O$ be an open set in $A$. Then, since $A$ has the subspace topology, there exists a $U$ open in $X$ such that $U\cap A=O$. But then, $p(O)=p(U\cap A)$. Yet, I only know that $p(O)=p(U\cap A)\subseteq p(U)\cap p(A)$, which doesn't really help.

If I assume $p$ is also a quotient map, does this help? Or can I do it without that assumption?

  • An open map is already a quotient map (if it's onto,as this is assumed for most quotient maps). So the quotient condition does not add information really. – Henno Brandsma Nov 14 '17 at 09:45
  • https://math.stackexchange.com/q/1513717/322548 I think my problem was solved here. –  Nov 15 '17 at 02:55
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    That’s a different problem from what you stated above. $A$ is also open which does make it true. – Henno Brandsma Nov 15 '17 at 04:46

3 Answers3

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The claim holds when $A = p^{-1}(Y')$ and $Y' \subset Y$.

Let $U' \subset A$ be open in $A$. By subspace topology, there exists $U \subset X$ open in $X$, such that $U' = U \cap A$. Then

$$p[U'] = p[U \cap A] = p[U \cap p^{-1}(Y')] = p[U] \cap Y'.$$

Therefore $p|(A \to Y')$ is open.

Further, we have $p(A) = p[p^{-1}(Y')] \subset Y'$. Then

$$p[U'] = p[U'] \cap p[A] = p[U] \cap Y' \cap p[A] = p[U] \cap p[A].$$

Therefore $p|(A \to p[A])$ is open.

kaba
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  • How do you justify $p[U \cap p^{-1}(Y')] = p[U] \cap Y'$? Because firstly, the image of the intersection is not always equal to the intersection of the images (consider $0 \mapsto 0$, $1 \mapsto 0$), and secondly, the image of the preimage is not always equal to the set you started with (if the map is not surjective). – Anakhand Sep 26 '23 at 22:16
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    @Anakhand Yep, but when you combine them like this, you get equality :) You can show it by starting from the expression $y \in p[U \cap p^{-1}(Y')]$ and manipulating it with (about six) equivalences to the form $y \in p[U] \cap Y'$. – kaba Sep 27 '23 at 01:05
  • Yeah, I realised later than you can just prove the reverse inclusion in an elementary way. If $y \in p[U] \cap Y'$, then $y = p(u)$ for some $u \in U$, and since $y \in Y'$, this means $u \in p^{-1}[Y']$ therefore $y \in p[U \cap p^{-1}[Y']]$. I'm leaving the comment though in case anyone else wonders about the same :) – Anakhand Sep 27 '23 at 07:55
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The result is not true. See here.

I am not flagging this as duplicate because of the second part of your question.

Also not true for quotient map. Take $p: \mathbb{R} \rightarrow S^1$ (usual topology on the reals, quotient by integers). Then let A be [0.1,0.2]. Then $p(A) $ is not open in $S^1$.

Mathemagical
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  • The given link provides an example where $p|(A \to Y)$ is not open. That counter-example is not valid for this question which asks for whether $p|(A \to p(A))$ is open. – kaba Jan 19 '22 at 23:34
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Unfortunately, restricting an open map on both of its domain and codomain does not always yield an open map.

$\pi : \mathbb{R} \to S^1$ that $\pi(x) = (\cos 2 \pi x, \sin 2 \pi x)$ can be taken as a counterexample, for $\pi$ is open but $\pi|_{(0, 1]} : (0, 1] \to S^1$ is not an open map ($(1/2, 1]$ is open in $(0, 1]$ but $\pi(1/2, 1]$ is not open in $S^1$).

A minor adjustment is required for correction:

Theorem Let $f : X \to Y$ be an open map, and $A$ be a saturated subspace of $X$, then $f|_A : A \to f(A)$ is also a open map.

Which is actually the same as kaba's answer.

HWC
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