# Is a certain restriction of an open map open?

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 9:45
• math.stackexchange.com/q/1513717/322548 I think my problem was solved here. – user322548 Nov 15 '17 at 2:55
• 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 4:46

Let X be the real line, Y its quotient by the integers and p be the quotient map. Let A=[0, $\infty)$ and consider the open set [0, 1) in A. Its image is not open in p(A) (which is the whole of Y).

• If $Y$ is a topological space, then surely $Y$ is open. – wckronholm Nov 14 '17 at 6:46
• The image of [0,1) is the whole space Y. Why would it not be open? – Mathemagical Nov 14 '17 at 6:54
• p(A) is the whole space Y, hence open. Look at p[0,a) where a is small. – Kabo Murphy Nov 14 '17 at 7:17

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 $S^1$.