Continuous map on compact Hausdorff spaces is closed

Question

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?

No. You have to pick a closed $C \subset X$, and show that $f(C)$ is closed in $Y$. — Crostul, Apr 05 '16 at 08:39

score 4 · Answer 1 · edited Jun 12 '20 at 10:38

4

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:

Theorem: A closed subset of a compact topological space is compact.

Thoorem: A compact subset of a Hausdorff topological space is closed.

edited Jun 12 '20 at 10:38

Community

1

answered Apr 05 '16 at 08:39

Spenser

20,236

Don't we need $f$ to be surjective to say that $f(A)$ is compact in $Y$? – user1292919 Apr 05 '16 at 08:41
2

@user1292919 No, the image of a compact subset by a continuous map is always compact. – Spenser Apr 05 '16 at 08:42
Ok! that makes my life easier.. – user1292919 Apr 05 '16 at 08:44

Continuous map on compact Hausdorff spaces is closed

1 Answers1