I have to prove this:
Let $p:X \rightarrow X'$ an identification map (I mean continuous, surjective and open or closed) and let $A \subset X$ and $A'\subset X'$ subspaces. If $A'$ is open in $X'$ and $A=p^{-1}(A')$ then $p_{|A}:A \rightarrow X'$ is an identification map.
I have this:
If $p:X \rightarrow X'$ is an identification map and $A \subset X$ is open and p-saturated ($A=p^{-1}(p(A))$) then the restriction is also an identification map.
It's easy to see that $A$ is open ($p$ is continuous, $A'$ is open and $A=p^{-1}(A')$) so I just need to show that $A=p^{-1}(p(A))$. I tried this:
$p(p^{-1}(A'))=A'$ because p is surjective. Then using $A=p^{-1}(A')$ we have $p(A)=A' \Rightarrow p^{-1}(p(A))=p^{-1}(A')=A $ cqd.
Is this correct?