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Question: Prove that any continuous function from a compact $T_2$-space onto a $T_2$-space is closed, that is, $f(F)$ is closed if $F$ is closed.

Is my general reasoning correct?

Any compact subset of a $T_2$ space is closed. Also, A subset of a compact $T_2$-space is compact iff it is closed.

Further, compactness is preserved by continuous functions. Since compactness is preserved by continuous functions and the codomain is a $T_2$-space, then the image is also closed. Thus, any continuous function from a compact $T_2$-space onto a $T_2$-space is closed.

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If S is a closed subset of a compact T2 space, then S is compact. Then, f(S) is compact since f is continuous. Since f(S) is compact, and the target space is Hausdorff, f(S) is closed.

Edit: The codomain (or target space) is always necessarily closed, when viewed as its own space, since its complement is the empty set, which is open. However, the image (or range) is closed in the above case.

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