Let $G$ be a compact Lie group. Suppose that $M$ and $N$ are smooth manifolds on which $G$ acts smoothly. If $f:M\longrightarrow N$ is a $G$equivariant proper continuous map, my question is: can we always find a smooth and proper $G$equivariant map $g:M\longrightarrow N$ which is homotopic to $f$ ? In particular, $M$ can not be compact.

$\begingroup$ Where do you use the assumption that $M$ and $N$ are $G$equivariant ? $\endgroup$ – Damian Rössler Jul 15 '13 at 11:58

$\begingroup$ Sorry, in fact I want to say that we need a proper smooth Gequivariant map $g$ which is homotopic to $f$. $\endgroup$ – yangyang Jul 15 '13 at 14:02
Proper maps form an open set in $C^0(M,N)$ in the Whitney $C^0$topology, and smooth maps are dense. If two maps are near enough in the Whitney topology, they are homotopic. What do you want with the $G$action?
Edit:To make all this equivariant, choose first a smooth equivariant embedding of $N$ into a finite dimensional representation space $V$ of $G$. This exists if there are only finitely many $G$orbit types, if I remember correctly. Then you approximate $f$ by a smooth $g_1:M\to V$, which you can integrate now over $G$ to make it equivariant. Since $f$ was equivariant and $g_1$ was near $f$, the integrated smooth map $g_2$ is still near $f$. Then use a $G$invariant inner product on $V$ to describe a smooth equivariant tubular neighborhood $U$ of $N$ in $V$ with equivariant projection $p$ onto $N$. Without loss we may assume that $g_2:M\to U$. Then $p\circ g_2:M\to N$ is smooth, equivariant, and still near $f$, so it is homotopic to $f$.

$\begingroup$ Sorry, in fact I want to say that $f$ and $g$ are all $G$equivariant. $\endgroup$ – yangyang Jul 15 '13 at 14:11