# Existence of $G$-map between finite $G$-simplicial complex

Let $X, Y$ be finite free $G$- simplicial complex. What kind of properties are necessary for existence a $G$-map,i.e, a continuous map which preserves $G$-action, from $X$ to $Y$? Does existence of such a map gives any information on Homotopy or Homology groups of $X$ and $Y$? For example, Dold's theorem says $Conn (X)+1\leq\dim (Y)$ where $conn(X)$ is the maximum $k$ such that all homotopy groups of $X$ vanish up to $k$.

I just want to add, it is very simple to see that the inverse of Dold's Theorem is not true! For example, one can consider $X=\mathbb{S}^2$ by the antipodal $\mathbb{Z}_2$-action, i.e, $x\to -x$, and $Y=\mathbb{S}^1\times\mathbb{S}^1$ with the same $\mathbb{Z}_2$-action, $(x,y)\to (-x, -y)$. We have $conn(X)=1$, and $\dim(Y)=2$. But, there is no $\mathbb{Z}_2$-map from $\mathbb{S}^2\to\mathbb{S}^1\times\mathbb{S}^1$. Since otherwise we have $\mathbb{Z}_2$-map from $\mathbb{S}^2\to\mathbb{S}^1$, as the projection on the first entry give a $\mathbb{Z}_2$-map from $\mathbb{S}^1\times\mathbb{S}^1$ to $\mathbb{S}^1$. This contradicts the Borsuck-Ulam theorem.

• Well, one obvious observation is that you can't have a map from a guy with a trivial action into a guy with a free action, since $g\cdot f(x) = f(g \cdot x) = f(x)$ would force a fixed point in the codomain. I guess you could play around with this to find some restrictions on the existence of a map based on the actions, but it doesn't seem to be what you're angling for in your question. Sorry, I don't know much about the connection to homology. – David White Dec 25 '17 at 15:35