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.

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

I think he's referring to the preimage, not the inverse. I haven't read the book, but I believe he's referring to the fact that for a continuous function, the preimage of an open set is open.


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