On $\pi_{1}(f(\Omega))$ with $\Omega$ convex - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:32:37Z http://mathoverflow.net/feeds/question/51747 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51747/on-pi-1f-omega-with-omega-convex On $\pi_{1}(f(\Omega))$ with $\Omega$ convex xiao 2011-01-11T10:20:56Z 2011-01-11T16:01:16Z <p>Suppose $\Omega\subset R^{n}$ is an open,convex and bounded set,$f:\Omega\to\mathbb{C}$ is a smooth map.</p> <p>My question:</p> <blockquote> <p>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 $f$?</p> </blockquote> <p>What's more,if we do not need $\pi_{1}(f(\Omega))=\lbrace 1 \rbrace$,then comes the following:</p> <blockquote> <p>2)How does $f$ affect $\pi_{1}(f(\Omega))$?</p> </blockquote> http://mathoverflow.net/questions/51747/on-pi-1f-omega-with-omega-convex/51756#51756 Answer by Thomas Rot for On $\pi_{1}(f(\Omega))$ with $\Omega$ convex Thomas Rot 2011-01-11T12:40:28Z 2011-01-11T12:40:28Z <p>In general it is not true. We have the following however. </p> <p>$\Omega$ 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 $\pi_1(f(\Omega))={1}$.</p> <p>We don't need smoothness of $f$ at all, only continuity.</p>