Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 23 / 1 / b / 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 isIndeed,so $e_f$ is an equivariant map. Conversely, if $e$ is equivariant, $(g,x)=(1,x{g}^{-1})g$ implies $e(g,x)=e(1,x{g}^{-1})g$, so $e=e_f$ for $f(x)=e(1,x)$.

Transport the right action from part (a) through this bijection. At $x$ its value isThus the exponential of right group actions isFor clarity, the right-action law holds even in a nonabelian group: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.