Is the inverse of a continuous monotonic function defined on a compact also continuous?

Question

Let $K \subset \mathbb{R}$ be a compact set, and $f:K \rightarrow \mathbb{R}$ a continuous and strictly monotonic function. Is it true that its inverse function, $f^{-1}: f(K) \rightarrow K$, is also continuous? I know that this result is true if $K$ is an interval, so if $K=[a,b]$ this result holds, but does it still hold if $K$ is an arbitrary compact set?

Answer 1

You don't need monotone, except to guarantee the existence of an inverse. Any bijective continuous map $f:K \to M$ where $M$ is Hausdorff and $K$ compact is open, so its inverse is continuous. In particular $f:K \to f(K)$ is in fact a bijection, and $f(K) \subset \mathbb R$ is clearly Hausdorff, so $f^{-1}$ is continuous