Example of a continuous bijective function on and to the closure of the complex numbers, with an inverse that is not continuous?

Question

Note that by the closure of the complex number, I mean the union of the complex numbers and infinity.

I have been stumbling over this questions for a wile now, and I understand many examples of this for subsets of the closure of the complex numbers, but I wish to demonstrate it for the closure of the complex numbers.

I only know the very basics of topology, but is appear the Invariance of Domain Theorem implies that no such function would exist if this were for the complex numbers (rather than their closure). So my questions are as follows:

1. Am I mistaken in the claim I made about the Invariance of Domain Theorem; if not, can this idea be extended is some way to the closure of the complex numbers?
2. Does such a function exist; and if so, what is an example?

(Please comment if my question is unclear, I will monitor regularly, and clarify anything if requested.) (I apologize for not knowing how to typeset this.)

Answer 1

Such a function can't exist, and this is an easier result than the invariance of domain.

The extended complex numbers form a compact Hausdorff space $\hat{\mathbb{C}}$. In general, every closed subspace of a compact space is compact, every continuous image of a compact set is compact, and every compact subset of a Hausdorff space is closed. Therefore if $f: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ is continuous, then it maps closed sets to closed sets. If it is a bijection, then this means it also maps open sets to open sets, so $f^{-1}$ is continuous.

Answer 2

Another way: a continuous bijection between a compact space ($S^1$ as the compactification of $\mathbb C$) and a Hausdorff space-- $S^1$ again-- is automatically a homeomorphism, so that the inverse function is not just continuous, but an open and closed map.