What's wrong with my proof ?? This question was on my exam but apparently my answer was wrong as I only got half of the point:
Suppose $p:X \to Y$ is an open map, prove if $A$ is open in $X$ then $q:A \to p(A)$ obtained by restricting $p$ is also an open map.
My proof: Let $U$ be open in $A$. We have that $q(U) = p(U)$ is open in $Y$, hence it's open in $p(A)$. So q is an open map.
This sounded like a quick excercise but apparently my proof needs improvement. What's wrong about my proof?
 A: You are assuming that $p(U)$ is open. Why is that? The set $U$ is an open subset of $A$, but you should deduce from this that therefore $U$ is an open subset of $X$ (and this is where the fact that $A$ is open is used; it is not true in general). Then, yes, you can deduce that $p(U)$ is indeed an open subset of $Y$.
A: To be certain why you didn't get full credit, you should ask whomever graded the exercise. The most likely reason seems that you conclude that $p(U)$ is open in $p(A)$, without arguing why $U$ is open in $X$. Also, the codomain of $q$ is $Y$, not $p(A)$.
A: Let $A$ be open in $X$;
Then any open set $O$ in $A$ can be written as
$O=U \cap A$, where $U$ is open in $X;$
$O$ as the intersection of $U$ and $A$, which are both open in $X$, gives and open set $O$ in $X$.
Hence $f(O)$ is open in $X$.
A: Let $U$ be an open subset in $A$. Then for some open in $X$ like $O$, we have $U=O \cap A$.
Since $A$ is open in $X$ then $O  \cap A$ is also open in $X$. Hence, $q(U)=p(U)=p(O \cap A)$ is open in $Y$.
We conclude that $q(U)=p(U)$ is open in $p(A)$ since open sets in $p(A)$ are open sets in $Y$ contained in $p(A)$.
