On $\pi_{1}(f(\Omega))$ with $\Omega$ convex - MathOverflow

On $\pi_{1}(f(\Omega))$ with $\Omega$ convex

2011-01-11T10:20:56Z

Suppose $\Omega\subset R^{n}$ is an open,convex and bounded set,$f:\Omega\to\mathbb{C}$ is a smooth map.

My question:

1)when $\pi_{1}(f(\Omega))=\lbrace 1 \rbrace$? Or in order to make $\pi_{1}(f(\Omega))=\lbrace 1 \rbrace$, whether there is some non-trivial restrictions on $f$?

What's more,if we do not need $\pi_{1}(f(\Omega))=\lbrace 1 \rbrace$,then comes the following:

2)How does $f$ affect $\pi_{1}(f(\Omega))$?

Answer by Thomas Rot for On $\pi_{1}(f(\Omega))$ with $\Omega$ convex

2011-01-11T12:40:28Z

In general it is not true. We have the following however.

$\Omega$ is open. If $f$ is injective, $\Omega$ is homeomorphic to it's image $f(\Omega)$ via $f$ by Brouwers invariance of domain. The induced map $f_*:\pi_1(\Omega)\rightarrow\pi_1(f(\Omega))$ is hence an isomorphism. If $\Omega$ is convex, $\pi_1(\Omega)={1}$, hence $\pi_1(f(\Omega))={1}$.

We don't need smoothness of $f$ at all, only continuity.