Is a certain restriction of an open map open?

Question

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?

Answer 1

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 $S^1$.

Answer 2

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.