Definition of equivariant map if one of the actions is a right action Let $G$ be a group and $f:X\to Y$ a map between two $G$-sets which preserve the $G$ action. If $X$ has a left $G$ action and $Y$ right $G$ action then why do we define $f(g.x)=f(x).g^{-1}$ for all $ g\in G $ and all $x\in X?$
Is it to ensure that $f(x.gh)=f((x.g).h)$ for all $g,h \in G$ and $x \in X?$
Thanks! 
 A: The relevant fact here is that if a set $Y$ is equipped with a right $G$-action, then it is also automatically equipped with a left $G$-action by defining $g \cdot y := y \cdot g^{-1}$ (the axioms are easy to check, this is basically what's done in Michael Joyce's answer). Therefore, in your case we view $Y$ as a set with a left $G$-action, and then the definition is the usual one.
A: The key property you need for the definition to be well-defined is that $f((gh) \cdot x) = f(g \cdot (h \cdot x))$, since $(gh) \cdot x = g \cdot (h \cdot x)$ by the definition of a $G$-action on $X$.  Using the definition given, we see that
$$f(g \cdot (h \cdot x)) = f(h \cdot x) \cdot g^{-1} = (f(x) \cdot h^{-1}) \cdot g^{-1} \\ = f(x) \cdot (h^{-1} g^{-1}) = f(x) \cdot (gh)^{-1} = f((gh) \cdot x).$$
Note that we used the definition of a $G$-action on $Y$ in the derivation.
With a similar calculation, you'll see that trying to define equivariance by $f(g \cdot x) = f(x) \cdot g$ fails to be well-defined, i.e. $f((gh) \cdot x) \neq f(g \cdot (h \cdot x))$.
