(Theorem) A continuous function from a compact space to a Hausdorff space is a homeomorphism

Theorem (Munkres 26.5 & 26.6): Continuous functions and compactness

• (i) If $f:X\to Y$ is continuous and $X$ is compact, then $f(X)$ is compact.
• (ii) Let $f:X\to Y$ be a continuous bijection. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.

Proof of (i) from Modern Analysis I.

Suppose $K\subseteq X$ is a compact subset and $f:X\to Y$ is continuous. Then let $\{U_{\alpha }\}$ be an open cover of $f(K)$. By continuity of $f$, the preimages $f^{-1}(U_{\alpha })$ are also open and cover $K$. Since $K$ is compact, we may find a finite subcover $\{f^{-1}(U{)}_i{\}}_{i=1}^n$. Then the union of the images of this collection is a finite subcover of $f(K)$, and we conclude that $f(K)$ is compact.

Sketch of (ii) from Modern Analysis I.

Suppose $(X,d_X)$ and $(Y,d_Y)$ are metric spaces. To show $f^{-1}$ is continuous, it suffices to show that $f(A)$ is open for any $A\subseteq X$.

1. If $A\subseteq X$ is open, then $A^C$ is closed and therefore compact by Heine-Borel.
2. $A^C$ compact $⟹$ $f(A^C)$ compact, hence closed by Heine-Borel again.
3. Conclude that $f(A)$ open.

Sketch of (ii) from Topology.

To show $f^{-1}$ is continuous, we want to show that images of closed sets under $f$ are also closed. This can be done by applying the earlier lemmas.

1. Suppose $A\subseteq X$ closed. Since $X$ is compact, $A$ is also compact (Munkres 26.2).
2. Since $A\subseteq X$ is compact, its image under a continuous function $f(A)\subseteq Y$ is also compact (above; Munkres 26.5).
3. By (Theorem) Compact subspaces of Hausdorff spaces are closed, since $f(A)\subseteq Y$ is compact and $Y$ is Hausdorff, $f(A)$ is also closed.