Theorem [1,2] Suppose $f\colon X\to Y$ is a continuous map between 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 $f\colon X\to f(X)$ is continuous.

Proof of theorem. (Following [1].) Suppose $\{V_\alpha \mid \alpha \in I\}$ is an arbitrary open cover for $f(X)$ . Since $f$ is continuous, it follows that $$\{f^{-1} (V_\alpha) \mid \alpha \in I\}$$ 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), we have \begin{eqnarray*} X &\subseteq & f^{-1} f(X) \\ &= & f^{-1} \big( \bigcup_{\alpha \in I} (V_\alpha)\big) \\ &=& \bigcup_{\alpha \in I} f^{-1}(V_\alpha). \end{eqnarray*}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 a finite open cover for $X$ . Since $f$ is a surjection, we have $ff^{-1}(A)=A$ for any $A\subseteq Y$ (see this page). Thus \begin{eqnarray*} f(X) &= & f \big(\bigcup_{i\in J} f^{-1}(V_\alpha)\big)\\ &= & ff^{-1} \bigcup_{i\in J} f^{-1}(V_\alpha)\\ &=& \bigcup_{i\in J} V_\alpha. \end{eqnarray*}Thus $\{ V_\alpha \mid \alpha \in J\}$ is an open cover for $f(X)$ , and $f(X)$ is compact.

A shorter proof can be given using the characterization of compactness by the finite intersection property:

Shorter proof. Suppose $\{A_i\mid i\in I\}$ is a collection of closed subsets 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$$ f^{-1}\left(\,\bigcap_{i\in I} A_i\!\right) = \bigcap_{i\in I} f^{-1}(A_i) \neq \emptyset$$ and so $\bigcap_{i\in I} A_i\neq \emptyset$ . Therefore $Y$ is compact.

References

1

I.M. Singer, J.A.Thorpe, Lecture Notes on Elementary Topology and Geometry, Springer-Verlag, 1967.

2

J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.

3

G.J. Jameson, Topology and Normed Spaces, Chapman and Hall, 1974.