compactness is preserved under a continuous map

Theorem [1, 2]Suppose $f:X\to Y$ is a continuous mapbetween topological spaces $X$ and $Y$.If $X$ is compact and $f$ is surjective, then $Y$ is compact.

The inclusion map $[0,1]\hookrightarrow [0,2)$ shows that the requirement for $f$ to be surjective cannot be omitted.If $X$ is compact and $f$ is continuous we can always conclude, however, that $f(X)$ is compact, since $f:X\to f(X)$ is continuous (http://planetmath.org/IfFcolonXtoYIsContinuousThenFcolonXtoFXIsContinuous).

Proof of theorem. (Following [1].)Suppose $\{V_{\alpha }\mid \alpha \in I\}$ is an arbitraryopen cover for $f(X)$. Since $f$ is continuous, it followsthat

$$\{f^{-1}(V_{\alpha })\mid \alpha \in I\}$$

is a collection of open sets in $X$.Since $A\subseteq f^{-1}f(A)$ for any $A\subseteq X$,and since the inverse commutes with unions(see this page (http://planetmath.org/InverseImage)),we have

$X$$\subseteq$$f^{-1}f(X)$

$=$$f^{-1}\left({\displaystyle \bigcup _{\alpha \in I}}(V_{\alpha })\right)$

$=$${\displaystyle \bigcup _{\alpha \in I}}{f}^{-1}(V_{\alpha }).$

Thus $\{f^{-1}(V_{\alpha })\mid \alpha \in I\}$ is an open cover for $X$.Since $X$ is compact, there exists a finite subset $J\subseteq I$such that $\{f^{-1}(V_{\alpha })\mid \alpha \in J\}$ is afinite open cover for $X$.Since $f$ is a surjection, we have $f{f}^{-1}(A)=A$ for any $A\subseteq Y$(see this page (http://planetmath.org/InverseImage)). Thus

$f(X)$$=$$f\left({\displaystyle \bigcup _{i\in J}}{f}^{-1}(V_{\alpha })\right)$

$=$$f{f}^{-1}{\displaystyle \bigcup _{i\in J}}{f}^{-1}(V_{\alpha })$

$=$${\displaystyle \bigcup _{i\in J}}{V}_{\alpha }.$

Thus $\{V_{\alpha }\mid \alpha \in J\}$ is an open cover for $f(X)$,and $f(X)$ is compact.$\mathrm{\square }$

A shorter proof can be given using thecharacterization of compactness by the finite intersectionproperty (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty):

Shorter proof.Suppose $\{A_i\mid i\in I\}$ is a collection of closedsubsets of $Y$ with the finite intersection property.Then $\{f^{-1}(A_i)\mid i\in I\}$ is a collection of closed subsets of $X$with the finite intersection property,because if $F\subseteq I$ is finite then

$$\bigcap _{i\in F}{f}^{-1}(A_i)=f^{-1}\left(\bigcap _{i\in F}{A}_i\right),$$

which is nonempty as $f$ is a surjection.As $X$ is compact, we have

$$f^{-1}\left(\bigcap _{i\in I}{A}_i\right)=\bigcap _{i\in I}{f}^{-1}(A_i)\ne \mathrm{\varnothing }$$

and so ${\bigcap }_{i\in I}{A}_i\ne \mathrm{\varnothing }$.Therefore $Y$ is compact.$\mathrm{\square }$