Is the category of hypergraphs cartesian-closed?

If $$H_i = (V_i, E_i)$$ for $$i=1,2$$ are hypergraphs then a map $$f:V_1\to V_2$$ is said to be a hypergraph homomorphism if $$f(e_1)\in E_2$$ for all $$e_1\in E_1$$. Hypergraphs together with hypergraph homomorphisms form a category. Is this category cartesian closed?

• Do you have a hypergraph structure on $Hom(G_1,G_2)$ in mind? Maybe something drawn from category theory (since graphs are kinda like categories)? Commented Jan 26, 2019 at 16:26
• Good point. How is it done in the category of graphs? Maybe for the category of graphs we could say that $f,g\in \text{Hom}(G,H)$ form an edge if $\{f(v),g(v)\}\in E(H)$ for all $v\in V(G)$. Analogously for hypergraphs, how about ${\cal S}\subseteq\text{Hom}(H_1, H_2)$ is a hyperedge in $\text{Hom}(H_1,H_2)$, iff $\{f(v): f\in {\cal S}\} \in E(H_2)$ for all $v\in V(H_1)$? Commented Jan 26, 2019 at 21:27
• In this link it says Cat is cartesian closed, with functor categories as the internal hom. It says directed graphs are cartesian closed, probably using the same idea. en.wikipedia.org/wiki/Cartesian_closed_category Commented Jan 27, 2019 at 15:29
• I take that previous link to suggest that the cartesian product on directed graphs makes it cartesian closed. There's also a cartesian product on graphs (link below) but I do not know if hom-tensor duality is satisfied. There's also a tensor product on graphs (link below): en.wikipedia.org/wiki/Cartesian_product_of_graphs, math.stackexchange.com/questions/302147/… Commented Jan 27, 2019 at 15:31
• I remember from a course in grad school that there are 4 monoidal products on Graph, but I don't remember all of them. Commented Jan 27, 2019 at 17:36

First let's describe the cartesian product. I believe the category $$\rm HyGph$$ is a topological concrete category over $$\rm Set$$, in the following way. Suppose $$X$$ is a set and $$\{S_i\}$$ is a family of hypergraphs, with functions $$f_i : X \to V_{S_i}$$ to their vertex sets. Then there is a "largest hypergraph" on $$X$$ rendering the $$f_i$$ homomorphisms, where $$E_X$$ consists of all subsets $$e\subseteq X$$ such that $$f_i(e)\in E_{S_i}$$ for all $$i$$. In particular, given hypergraphs $$Y,Z$$, if we take $$X = V_Y \times V_Z$$ with the $$f$$'s the projections, we obtain the product hypergraph structure on $$V_Y\times V_Z = V_{Y\times Z}$$: thus $$E_{Y\times Z}$$ consists of all subsets $$e\subseteq V_Y\times V_Z$$ such that $$\pi_1(e)\in E_Y$$ and $$\pi_2(e)\in E_Z$$. Put differently, each edge of $$Y\times Z$$ is obtained by choosing an edge $$e_Y\in E_Y$$ and an edge $$e_Z\in E_Z$$, and then choosing a subset of the rectangle $$e_Y \times e_Z \subseteq V_Y \times V_Z$$ that projects onto both $$e_Y$$ and $$e_Z$$, i.e. contains at least one ordered pair with each possible first coordinate and at least one ordered pair with each possible second component.
Now there is a standard way to discover what an internal-hom must look like if it exists, by mapping out of small objects. To start with, let $$I$$ be the hypergraph with one vertex and no edges. Then for any hypergraph $$X=(V_X,E_X)$$, a homomorphism $$I\to X$$ simply picks out a vertex of $$X$$. Thus, if there is an internal-hom $$Y^X$$, the vertices of $$Y^X$$ are in bijection with the homomorphisms $$I\to Y^X$$, which must in turn be in bijection with the homomorphisms $$I\times X\to Y$$. But since $$I$$ has no edges, the above description of the product means that $$I\times X$$ has no edges either (in fact we could see this simply because any edge of $$I\times X$$ must project to an edge of $$I$$). Thus every set-function from $$V_{I\times X} \cong V_X$$ to $$V_Y$$ determines a vertex of $$Y^X$$, whether or not it is a homomorphism.
The information about homomorphisms is instead carried by the unary edges (edges containing only one vertex). The terminal hypergraph $$1$$ is a one-element set with the unique possible (unary) edge. A homomorphism $$1\to X$$ picks out a vertex of $$X$$ that belong to a unary edge. Thus, a homomorphism $$1\to Y^X$$ picks out such a vertex of $$Y^X$$, but such homomorphisms must be bijective to homomorphisms $$1\times X \to Y$$, i.e. $$X\to Y$$ since $$1\times X \cong X$$. So the vertices of $$Y^X$$ are the arbitrary set-functions $$V_X\to V_Y$$, while the homomorphisms are the set-functions that belong to unary edges. This is similar to some other examples of cartesian closed categories, such as $$G\text{-}\rm Set$$ in which the elements of $$Y^X$$ are arbitrary functions $$X\to Y$$, with $$G$$ acting by conjugation, so that the actual $$G$$-equivariant maps $$X\to Y$$ (the "real morphisms" in the category) are instead the $$G$$-fixed-points of the $$G$$-set $$Y^X$$.
We can detect all the other edges of $$Y^X$$ in a similar way. For any nonempty set $$A$$, let $$J_A$$ be the hypergraph with vertices $$A$$ and the only edge being the whole set $$A$$. Then a homomorphism $$J_A\to X$$ is a function $$A\to X$$ whose image is an edge. Thus, an edge of $$Y^X$$ must be uniquely determined by picking a subset $$A\subseteq V_{Y^X}$$ of functions $$V_X\to V_Y$$ such that the corresponding function $$A\times V_X \to V_Y$$ is a homomorphism $$J_A \times X\to Y$$, which is to say such that for any edge $$e_X\in E_X$$, if a subset $$e\subseteq A\times e_X$$ projects onto both $$A$$ and $$e_X$$, then $$\{ f(x) \mid (f,x)\in e \}$$ is an edge of $$Y$$.
This defines a hypergraph $$Y^X$$ which must be the cartesian-closed exponential if such exists. It remains to prove that it actually is, but I think this is fairly straightforward. The evaluation map $$Y^X\times X\to Y$$ is a homomorphism, essentially by definition of the edges of $$Y^X$$. And if $$Z\times X\to Y$$ is a homomorphism, then there is a unique induced function $$V_Z\to V_{Y^X} = (V_Y)^{V_X}$$, which is a homomorphism again essentially by definition of the edges of $$Y^X$$.