# Stillwell — Preservation of Compactness under Continuous Functions

I’m working through Naive Lie Theory by John Stillwell. I have a specific question regarding his particular proof for preservation of compactness under continuous functions.

Theorem. If $$K$$ is compact and $$f$$ is a continuous function defined on $$K$$ then $$f(K)$$ is compact.

Proof. Given a collection of open sets $$O_i$$ that covers $$f(K)$$, we have to show that some finite subcollection $$O_1,O_2,...,O_n$$ also covers $$f(K)$$. Well, since $$f$$ is continuous and $$O_i$$ is open, we know that $$f^{−1}(O_i)$$ is open by Property (**) in Section 8.3. Also, the open sets $$f^{−1}(O_i)$$ cover $$K$$ because the $$O_i$$ cover $$f(K)$$. Therefore, by compactness of $$K$$, there is a finite subcollection $$f^{−1}(O_1), f^{−1}(O_2),..., f^{−1}(O_m)$$ that covers $$K$$. But then $$O_1,O_2,...,O_n$$ covers $$f(K)$$, as required.

There’s one part which bugs me about his proof. How does he know $$f^{-1}$$ exists? Obviously, the fact it’s continuous isn’t sufficient; it may not be injective.

Note: Property $$**$$ is only used to establish $$f^{-1}(O_i)$$ is open.

• He's not assuming $f$ is invertible. The notation $f^{-1}(O_i)$ denotes $\{x \in K \mid f(x) \in O_i \}$. Sep 5 '20 at 3:19