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Currying transforms a morphism of two variables into a morphism taking values in an exponential object. The evaluation map of an exponential object recovers the original morphism. Its categorical universal property requires naturality and uniqueness, not just an underlying set bijection.

Give $M\times X$ the diagonal right monoid action. On its equivariant maps into $Y$, define $(ek)(g,x)=e(kg,x)$. The evaluation map of an exponential object is $e(1,x)$ and currying of $h:Z\times X\to Y$ is $\hat{h}(z)(g,x)=h(zg,x)$. Both maps are equivariant, and they are inverse under evaluation. The left multiplication $kg$ in this right action is essential in a noncommutative monoid.

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.