= Solution
When $M=G$ is a <group>, evaluation at its identity gives a bijection from the previous equivariant-map set to all functions $X\to Y$. For an arbitrary function $f$, the inverse construction is
$$
e_f(g,x)=f(xg^{-1})g.
$$
Indeed,
$$
e_f(gk,xk)=f(xk(gk)^{-1})gk=f(xg^{-1})gk=e_f(g,x)k,
$$
so $e_f$ is an <equivariant map>. Conversely, if $e$ is equivariant, $(g,x)=(1,xg^{-1})g$ implies $e(g,x)=e(1,xg^{-1})g$, so $e=e_f$ for $f(x)=e(1,x)$.
Transport the right action from part (a) through this bijection. At $x$ its value is
$$
(ek)(1,x)=e(k,x)=f(xk^{-1})k.
$$
Thus the <exponential of right group actions> is
$$
\boxed{Y^X=\operatorname{Set}(X,Y),\qquad (fk)(x)=f(xk^{-1})k.}
$$
For clarity, the right-action law holds even in a nonabelian <group>:
$$
((fk)l)(x)=f(xl^{-1}k^{-1})kl=f(x(kl)^{-1})kl=(f(kl))(x).
$$
Evaluation is equivariant because $(fk)(xk)=f(x)k$. The underlying functions need not be equivariant; the fixed points of this exponential action are exactly the equivariant functions.
Back to article page
