If $(X,\tau)$ and $(Y,\tau_1)$ are compact Hausdorff spaces and $f:(X,\tau) \rightarrow (Y,\tau_1)$ is a continuous mapping, prove that $f$ is a closed mapping.
My outline of the proof is that Let $A \in Y$ be a closed set, then $f^{-1}(A)$ is closed in X since $f$ is continuous. I have to show that $f(f^{-1}(A)) = A$ and therefore $f$ is close mapping.
Am I doing the right thing?
Best Answer
You have to show that if $A\subseteq X$ is closed, then $f(A)\subseteq Y$ is closed. Because $X$ is compact, the closed set $A\subseteq X$ is compact. Since $f$ is continuous, $f(A)$ is compact in $Y$. Since $Y$ is Hausdorff and $f(A)$ is compact in $Y$, then $f(A)$ is closed in $Y$.
I am using these two standard results: