Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 23 / 1

Let $M\times X$ have the diagonal right monoid action $(g,x)m=(gm,xm)$, and put $E={\mathrm{Hom}}_M(M\times X,Y)$. Define the right action on this set of equivariant maps of monoid sets byIt really stays in $E$: $(ek)(gm,xm)=e(kgm,xm)=e(kg,x)m$. Also $e1=e$ and $((ek)l)(g,x)=e(klg,x)=(e(kl))(g,x)$, so it satisfies the right-action law. Left multiplication in the first argument is intentional; no commutativity of $M$ has been assumed.

Define the evaluation map of an exponential objectIt is equivariant, becauseFor any right $M$-set $Z$ and equivariant map $h:Z\times X\to Y$, define its currying byFor fixed $z$, this is equivariant in the diagonal variables: $h(zgm,xm)=h(zg,x)m$. The map $z\mapsto \hat{h}(z)$ is also equivariant, sinceIt satisfies $\mathrm{ev}(\hat{h}(z),x)=h(z,x)$.

Conversely, given an equivariant $u:Z\to E$, set $h(z,x)=u(z)(1,x)$. Evaluation makes this equivariant, and currying recovers $u$:These constructions are inverse and natural in $Z$. Therefore they establish the exponential object universal propertyThis is the exponential of right monoid actions, not the set of ordinary maps $X\to Y$ with an arbitrarily guessed action.

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.