How do you prove that z->z^2 is a closed map ?

**Aki** · June 26th 2011, 03:18 AM

I want to prove that the mapping of the complex plane to the complex plane $z\mapsto z^2$ is a closed map.
Only way I can think of is to prove first that it is a proper map.
Is there a more straightforward way ?

**Jose27** · June 27th 2011, 09:09 PM

It's obvious that $(\cdot )^2 : \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$ is closed, and since it leaves $\infty$ fixed it behaves nice with respect to unbounded closed sets (an unbounded set is closed in the plane iff the set plus $\infty$ is closed in the sphere). ( $\hat{\mathbb{C}}$ denotes the Riemann sphere)

**Aki** · June 29th 2011, 06:07 PM

Thank you for your help.

From your reply I got the impression that the necessary and sufficient condition for a continuous mapping of $\mathbb{C}$ into $\mathbb{C}$ to be a closed map is to leave $\infty$ fixed.
Am I right ?

**Jose27** · June 29th 2011, 07:15 PM

Originally Posted by Aki

Thank you for your help.

From your reply I got the impression that the necessary and sufficient condition for a continuous mapping of $\mathbb{C}$ into $\mathbb{C}$ to be a closed map is to leave $\infty$ fixed.
Am I right ?

Not at all: Take $f,g: \mathbb{C} \rightarrow \mathbb{C}$ where $f(x+iy)=x$ the natural projection on the real axis, then $f$ fixes $\infty$ (in the sense that there is a mapping of the sphere to the sphere that coincides with $f$ via stereographic projection, ie. as usual), but it's not difficult to see that $f$ is not closed. On the other hand if $g$ is a constant, we can extend $g$ to the sphere and this is, trivially, a closed continous map that doesn't fix $\infty$ .

What I meant to say was that obviously your mapping takes compact sets to compact sets (both in the plane), so we only need to check whether it sends unbounded closed sets (say $F$ ) to closed sets ( $f(F)$ say), but by adding $\infty$ to $F$ we get a closed set $G$ in the sphere and thus $f(G)$ is closed in the sphere, but by the first post $f(G)\setminus \infty$ is closed in the plane, but this is exactly $f(F)$ and we're done.

**Aki** · July 1st 2011, 08:31 PM

Thank you for the detailed explanation.

Thread: How do you prove that z->z^2 is a closed map ?

LinkBack

Thread Tools

Display

How do you prove that z->z^2 is a closed map ?

Re: How do you prove that z->z^2 is a closed map ?

Re: How do you prove that z->z^2 is a closed map ?

Re: How do you prove that z->z^2 is a closed map ?

Re: How do you prove that z->z^2 is a closed map ?

Similar Math Help Forum Discussions

Prove that A is closed

prove that if T is closed then the inverse image is closed

Prove that S is a closed set.

prove a set is closed

prove that T is closed.