Prove that every continuous one-to-one mapping of a compact space is topological.
Does this problem statement refer to a mapping of a compact space to itself?
If so, suppose the mapping is $f:A\rightarrow A$, where $A$ is compact and $f$ is continuous and 1-1. We must show that $f$ is also onto (and hence bijective), and the inverse image $f^{-1}$ is continuous.
For any compact subset $B\subseteq A$, the set $f(B)$ is compact. Since closed subsets of a compact set are compact, for any closed subset $B\subseteq A$ , the set $f(B)$ is compact (and so, closed and bounded). How might that help?
Best Answer
This means that if $f:A\to B$ is continuous and one-to-one, and if $A$ is compact, then $f$ maps $A$ homeomorphically to $f[A]$. As written, though, it is incorrect. We need to know that $B$ has the property that compact subsets are closed. That's what will allow the proof to work.
It should be clear that such an $f$ maps $A$ bijectively and continuously to $f[A]$ (considered as a subspace of $B$). It remains to show, then, that $f$ maps closed subsets of $A$ to (relatively) closed subsets of $f[A],$ since this is equivalent (why?) to showing that $g:f[A]\to A$ given by $g(x)=f^{-1}(x)$ is a continuous function.
As a side note, "bounded" doesn't necessarily make sense for general spaces. It does suggest that you may be working with metric spaces, however, rather than arbitrary topological spaces. If that is the case, then there's no need to make the additional assumption that compact subsets of $B$ are closed, because this is always true in metric spaces.