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A covariant set-valued functor $F$ is a decidable object precisely when every transition map $F(u)$ is injective. The pointwise complement of its diagonal consists of unequal pairs; it is a subfunctor exactly when transition maps preserve inequality. For left M-sets, this says that every action map is injective.

An equivariant map $h:A\to B$ between left M-sets obeys $h(m\cdot a)=m\cdot h(a)$. These are the natural transformations of the corresponding functors from the one-object monoid category.

For left M-sets, $B^A$ consists of equivariant maps of monoid sets $M\times A\to B$ for the diagonal action on the domain. Its action is $(m\cdot f)(w,a)=f(wm,a)$ and evaluation is $f(1,a)$. The curry of an equivariant $h:C\times A\to B$ is $\hat{h}(c)(w,a)=h(w\cdot c,a)$. This construction can have noninjective action maps even when both $A$ and $B$ are decidable.

A left monoid action assigns each $m\in M$ a map $a\mapsto m\cdot a$ with $1\cdot a=a$ and $(mn)\cdot a=m\cdot (n\cdot a)$. It need not be invertible or injective. A left M-set is equivalently a covariant functor from the one-object category determined by $M$.

If each $m$ admits $p,q$ with $pmq=p$, then $B^A$ is decidable whenever the left M-set $B$ is. For equivariant $f,g$, equality of all values at $(1,a)$ implies equality at $(m,a)$: apply $p$ and use $f(pm,pa)=pm\cdot f(1,qa)$, then cancel the injective action of $p$ on $B$. Equality after the exponential action of $m$ then gives equal traces by cancelling its action on $B$.

The two introductory requests can be settled before the lettered applications. In the functor category $[C,\mathbf{Set}]$, finite limits and finite unions of subobjects are pointwise. The component of the diagonal at $c$ is ordinary equality on $F(c)$. Its only possible complement isThis forms a subfunctor exactly when each $F(u)$ sends unequal elements to unequal elements, equivalently when every $F(u)$ is injective. In that case $D$ and the diagonal are disjoint and their union is $F^2$ at every component. Conversely, a diagonal complement must have these components and be stable under every transition map. Hence $F$ is decidable exactly when every transition map is injective.

Regard a monoid $M$ as a one-object category; a covariant set-valued functor is a left M-set. Give $M\times A$ the diagonal left action $t\cdot (w,a)=(tw,t\cdot a)$. LetRight multiplication on the first coordinate commutes with the diagonal left action, so $m\cdot f$ remains equivariant. The formula obeys $m\cdot (n\cdot f)=(mn)\cdot f$ and $1\cdot f=f$. Evaluation isIt is equivariant because $\mathrm{ev}(m\cdot f,m\cdot a)=f(m,m\cdot a)=m\cdot f(1,a)$.

For an equivariant $h:C\times A\to B$, defineThis is equivariant in $(w,a)$, and $\hat{h}(m\cdot c)=m\cdot \hat{h}(c)$. Evaluation recovers $h$. Conversely, currying the evaluation of a map $C\to E$ recovers that map by its equivariance. This proves the exponential universal property and the natural identificationwith exactly the stated action. In particular, decidability in a set-valued functor category says that a left M-set is decidable if and only if each of its action maps is injective.