# Showing $q:A\to p(A)$ is an open map.

I'm working on this problem out of Munkres: Let $p:X\to Y$ be an open map. Show if $A$ is open in $X$, then the map $q:A\to p(A)$ obtained by restricting $p$ is an open map. $\newcommand{\uset}{\mathcal{U}}$ $\newcommand{\oset}{\mathcal{O}}$

This seems like a very straightforward problem. Let $U$ be an open set in $A$. Then $U=\oset\cap A$ for some open set $\oset$ of $X$. Then we would like to have $$q(U)=p(\oset\cap A)=p(\oset)\cap p(A)$$

which would be open in $p(A)$ because $p(\oset)$ is an open set in $Y$. However, they don't give me any more information about the map (like injectivity) so all I can guarantee is that $$q(U)=p(\oset\cap A)\subseteq p(\oset)\cap p(A)$$

and I'm not sure how I can be sure $q(U)$ is open. Is there something I'm overlooking?

• I assume here you intend $q$ to be an open map with respect to the subspace topology? – charlestoncrabb Nov 5 '15 at 0:06
• If $A$ is open in $X$ and $\mathcal{O}$ is open in $X$, then $A\cap \mathcal{O}$ is open in $X$. – Joe Johnson 126 Nov 5 '15 at 0:07
• @charlestoncrabb Yes – Alex Mathers Nov 5 '15 at 0:07
• Since $A$ is open in $X$ you're allowed to assume that $U=\mathcal O$. – Arthur Nov 5 '15 at 0:07
• @JoeJohnson126 I'm an idiot, thanks – Alex Mathers Nov 5 '15 at 0:07

Any open set $U$ in $A$ is also open in $X$. Then $q(U)=p(U)$ is open in $Y$ so it is also open in $p(A)$ since open set in $p(A)$ are just open set in $Y$ contained in $p(A)$.