MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

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

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 $\pi_{1}(f(\Omega))$?

-
Clearly not, the image could be a circle for instance (for $n=1$, say). – Torsten Ekedahl Jan 11 2011 at 10:59
f either injective or constant, perhaps? – Ketil Tveiten Jan 11 2011 at 12:23
This question is much too general. One could state a million conditions, but without knowing where this question comes from, they are almost certain to be useless. – Igor Rivin Jan 11 2011 at 16:02

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.

-
 Dear Thomas Rot: In order to apply invariance of domain,$\Omega$ must be a domain in $R^{2}$.However,it is not so in my question. – xiao Jan 11 2011 at 14:04 What's more,if $n>2$,I think $f$ can not be injective in intuitution.And I assume $f$ to be smooth,because I think differential topology may help. – xiao Jan 11 2011 at 14:14 I think Brouwers invariance of domain holds in all $\mathbb{R}^n$?, see en.wikipedia.org/wiki/Invariance_of_domain . In many cases even if $f$ is not injective, we will have that $\pi_1(\Omega)\sim \pi_1(f(\Omega)$, this is true. I don't think that there is an easy criterion. Maybe the complement of $f(\Omega)$ characterizes this. What if $f(\Omega)^c$ is path connected? Why do you think smoothness helps? Can you give some more context of the problem? – Thomas Rot Jan 11 2011 at 14:22 Dear Thomas Rot: I know invariance of domain holds in all dimensions.However,it is of crucial importance that $\Omega$ and $f(\Omega)$ are contained in Euclidean space of the same dimension.But in my question,$f(\Omega) \subset \mathbb{C}$. – xiao Jan 11 2011 at 14:33 What I was considering is the image of simply connectted space.And I think the above is the simplest and non-trivial case. – xiao Jan 11 2011 at 14:42
show 1 more comment