# Construction of $G$-invariant map between manifolds

Let $$M,N$$ be two closed differential manifolds and let $$G$$ be a compact Lie group. Assume that $$G$$ acts on both manifolds $$M,N$$ nicely (i.e. free and proper so that $$M/G$$ and $$N/G$$ have the structure of manifolds, but I don't think that this is important for my question). Furthermore, assume that there exists open neighborhoods $$U_1\subset M$$ and $$V_1 \subset N$$ and a map $$s_1:U_1\rightarrow V_1.$$

Question: Is it possible to construct $$G$$-invariant open subset $$U\subset M$$ and $$V\subset N$$ and a map $$s:U\rightarrow V$$ such that $$\forall g\in G$$ and $$\forall p \in U$$ we have $$s(g\bullet_{M} p)=g\bullet_{N}s(p),$$ where $$\bullet_{M}$$ and $$\bullet_{N}$$ denote the $$G$$-action in both $$M$$ and $$N$$, respectively and such that $$U_{1} \subset U$$, $$V_{1} \subset V$$ and there exists some subset $$A\subset U_{1}$$ such that $$s|_{A} \equiv s_{1}$$. If this is in general not possible, then under what conditions on $$s_1$$ can this be realized?

For an answer I would be thankful!

• I forgot to mention that all open subsets here are supposed to be connected. But I don't know if this is important for the question. – Stan Jun 14 '19 at 9:13
• How does $s_1$ relate to $s$? Are you aware of the slice theorem? This seems to give what you want – Thomas Rot Jun 14 '19 at 9:16
• I am not sure how they relate. I am thinking that one can "push" $s_1$ in direction of the group action in order to define $s$. But I do not know how to make this precise. – Stan Jun 14 '19 at 9:19
• But as you formulate it $s_1$ does not seem to be doing anything (You can always define a map from $M$ to $N$, e.g. a constant map). – Thomas Rot Jun 14 '19 at 9:20
• Yes you are right. I wish that $U_1 \subset U$ and $V_1 \subset V$ and that $s|_{U_{1}} \equiv s_1$. I will change this in the question. Thanks I forgot to mention. – Stan Jun 14 '19 at 9:24

If $$A$$ is just one point $$p$$ then you can define $$s$$ by the required equivarianace property on orbit of $$p$$ and then you can extend it to whole $$M$$ since you said that $$M/G$$ is a manifold.
On the other hand, if $$g \in G$$ is sufficiently close to identity, then $$g\cdot p$$ will be close to $$p$$ and your $$s_1$$ must itself satisfy equivariance with respect to some small neighborhood of identity element of $$G$$. So if you want bigger $$A$$ you need some assumptions on $$s_1$$.