Question

Suppose I define a multicategory $M=(Ob(M),Hom_M)$ to be simply closed if

• for every sequence $S=(b_1,\ldots,b_n;x)$ of $n+1$ objects in $M$, we provide an object $Exp(S)\in Ob(M)$, and
• for every sequence $A=(a_1,\ldots,a_m)$ of $m$ objects in $M$ we provide a bijection $$\phi_S^A\colon Hom_M(a_1,\ldots,a_m,b_1,\ldots,b_n;x)\to Hom_M(a_1,\ldots,a_m;Exp(b_1,\ldots,b_n;x)),$$

subject to the condition below, for any object $x\in Ob(M)$. Fix such an object $x\in Ob(M)$ and let $X:=Exp(;x)\in Ob(M)$. (Apologies: fixing $x$ and introducing $X$ is done here only for typographical reasons: mathjax was having trouble rendering.) We name a "constant element" $c_x\in Hom_M(x;X)$ by setting $A=(x)$ and $S=(;x)$, so $\phi^A_S\colon Hom_M(x;x)\to Hom_M(x;X)$, and putting $$c_x:=\phi^{(x)}_{(;x)}(id_x)\in Hom_M(x;X).$$ We also name an "evaluation element" by setting $A=(X)$ and $S=(;x)$, so $(\phi^A_S)^{-1}\colon Hom(X;X)\to Hom(X;x)$, and putting $$e_x:=\left(\phi^{(X)}_{(;x)}\right)^{-1} (id_X)\in Hom_M(X;x).$$ Then our condition is that the constant element and the evaluation element for $x$ are mutually inverse in the sense that $$c_x\circ e_x=id_X \;\;\;\;\;\;\text{ and }\;\;\;\;\;\; e_x\circ c_x= id_x$$ in $M$.

As alluded to above, we obtain an ``evaluation map" as an element $ev\in Hom_M(Exp(A;x),A;x)$, coming as $\phi^{-1}(id)$, where $id$ is the identity element $id\in Hom_M(Exp(A;x),Exp(A;x))$.

The multicategory of sets is simply closed in this sense.

What problems exist for, and/or what niceties are missing from, this notion of "simply closed multicategory"?

Answer 1

In fact, if you include naturality, then your second condition becomes automatic, and moreover it suffices to consider $n=1$. See e.g. section 3 of this paper.

Answer 2

What an interesting question! I have never thought about such exponents before.

Nevertheless, I think you miss more coherence conditions to make exponents behave well. Take for example a category with finite products. It corresponds to a multicategory with the same objects and with arrows $a_1, \dotsc, a_n \rightarrow x$ interpreted as arrows $a_1 \times \cdots \times a_n \rightarrow x$. The bijections: $$\phi_S^A\colon \hom(a_1,\dotsc,a_m,b_1,\dotsc,b_n;x)\to \hom(a_1,\dotsc,a_m;\mathit{Exp}(b_1,\dotsc,b_n;x))$$ turn into bijections: $$\hom(a_1 \times \dotsc \times a_m \times b_1 \times \cdots \times b_n;x)\to \hom(a_1 \times\cdots \times a_m;x^{b_1 \times \cdots \times b_n})$$ A category is cartesian closed precisely when the above bjiections are natural in $a_i$'s. I guess this naturality condition is really what you miss. Your "constant" elements $c_x\in \hom(x;\mathit{Exp}(;x)))$ correspond to isomorphisms $x \rightarrow x^1$ with evaluations $x^1 \approx x^1 \times 1 \rightarrow x$ being their inverses - but this does not help much.

Here is an explicit example what can go wrong --- consider the category of $\omega$-sets (or Hyland's effective topos if you wish) and a classical set $X$ as an object in the category. Almost by definition $X \not\approx X \sqcup 1$. However, if $X$ is infinite, then $\hom(A, X) \approx \hom(A, X \sqcup 1)$ for every $\omega$-set $A$. This means that according to the above definition both $X \times X$ and $(X \times X) \sqcup 1$ are candidates for an exponent of $1 \sqcup 1$ with the classical infinite set $X$, though are not isomorphic.

Perhaps a more elementary reformulation of simple exponents would be "by universal arrows". However, I do not see how one could give an external (that is 2-categorical, or 2-multicategorical) characterisation of such exponents!