Is a continuous bijection $f:\mathbb C \to \mathbb C$ a homeomorphism?

Asked Sep 27 '22 at 13:11

Active Sep 27 '22 at 13:14

Viewed 82 times

Let $X$ be an open subset of $\mathbb K$, $f: X \to \mathbb K$, and $Y:=f(X)$.

Theorem: If $\mathbb K = \mathbb R$, and $f$ is injective and continous, then $f^{-1}:Y \to X$ is continuous.

The proof relies on the order of $\mathbb R$. Does above theorem hold if we replace $\mathbb R$ by $\mathbb C$?

edited Sep 27 '22 at 13:14

asked Sep 27 '22 at 13:11

Analyst

What about $f(z) = e^{z}$? Or $z^2$? – preferred_anon Sep 27 '22 at 13:17
@preferred_anon The exponential is not injective. – mathcounterexamples.net Sep 27 '22 at 13:19
2

@preferred_anon You forgot the word bijection. – azif00 Sep 27 '22 at 13:22
5

It's true, see invariance of domain. You may want to study it from a textbook. – Sassatelli Giulio Sep 27 '22 at 13:24
@SassatelliGiulio Thank you so much for your help! Do you have any suggestion of textbook containing this theorem? – Analyst Sep 28 '22 at 08:14
1

@Analyst You can find it as theorem 2B.3 in Allen Hatcher's Algebraic Topology, which is notoriously free to download from the author's website. – Sassatelli Giulio Sep 28 '22 at 08:51

0 Answers0