Let there be an open surjective continuous function $f:X\to Y$, where $X$ and $Y$ are topological spaces. I am given to understand that this mapping need not be closed. But could you point out the flaw in the following proof?
Let the open set $A\subseteq X$ map to the open set $f(A)$. Also assume $f$ is surjective. Then the complement of $A$ or $A'$ maps to $f(A)'$. Both are closed as per the definition of closed sets.
I know counter-examples exist. I"m just looking to find the flaw.
Thanks in advance!