How does one conclude that a function maps closed sets to closed sets?

Question

Let $f:(X,d_X)\to(Y,d_Y)$ be an equivalence (this means that $f$ is Lipschitz continuous, bijective and $f^{-1}$ is Lipschitz continuous). Let $K \subset X$. We want to prove that $K$ is closed and bounded iff $f(K)$ is closed and bounded.

Somewhere along the way in the proof, my instructor noted that, because $f$ is, in particular, continuous, bijective and $f^{-1}$ is continuous, then $f$ is a homeomorphism.

Then it is concluded that $K$ is closed in X iff $f(K)$ is closed in Y.

I'm unable to find this result in my notes or in the textbook I'm using for the class, moreover, when I looked it up, it is generally not true that a homeomorphic function always maps closed sets to closed sets.

So I think I made a mistake while taking the notes, there's something I missed, or the conclusion that $K$ is closed in $X$ iff $f(K)$ is closed in $Y$ comes from the fact that $f$ is an equivalence and not the fact that it is homeomorphic. However, I don't know which one is the correct option.

Answer 1

Let $f :X \to Y$. Here are three exercises that answer your question:

1.) Prove that $f$ is continuous if and only if, if $O$ is open in $Y$, then $f^{-1}(O)$ is open in $X$.

2.) Use properties of the preimage to conclude that the above is equivalent to preimages of closed sets being closed.

3.) Conclude homeomorphisms send closed sets to closed sets, both ways.

Answer 2

The preimage of a closed set under a continuous function is a closed set. If $f$ is a homeomorphism then in particular it is continuous and its inverse is continuous. Therefore $K\subseteq X$ is closed if and only if $f(K)\subseteq Y$ is closed.