I was reading a paper which refrenced something called a "product functor". The was a functor mapping from $\bf{Sets}^{C^{op}}$ for some category $C$. I was wondering what this was?
Thanks for any help
I was reading a paper which refrenced something called a "product functor". The was a functor mapping from $\bf{Sets}^{C^{op}}$ for some category $C$. I was wondering what this was?
Thanks for any help
A profunctor is probably what you want, as Berci suggested in the comments. Given two categories $\mathcal{C,D}$, a profunctor from $\mathcal{C}\to\mathcal{D}$ is a functor $\mathcal{D}^{\text{op}}\otimes\mathcal{C}\to\mathbf{Set}$. Profunctors are adjunct to functors $\mathcal{C}\to\mathbf{PShv}_{\mathbf{Set}}(\mathcal{D})$, where $\mathbf{PShv}_{\mathbf{Set}}(\mathcal{D})$ is the category of $\mathbf{Set}$-valued presheaves on $\mathcal{D}$. It is also a discrete opfibration over $\mathcal{D}^{\text{op}}\otimes\mathcal{C}$. However, I do not know if you mean "profunctor" or "product functor".
If you do indeed mean the latter, then the following is the definition of a "product functor$^{1}$". Anyway, let $\mathcal{C}\times\mathcal{C}$ be the product category of $\mathcal{C}$ with itself. The product functor is the functor $\prod:\mathcal{C\times C\to C}$ such that
$\prod(X,Y):=X\times Y$ (the binary product);
$\prod(f,g):=f\times g$ (the product morphism).
The product functor is a functor, as this proof shows. It is the right adjoint to the diagonal functor, so $\left(\prod\dashv\Delta\right):\mathcal{C\times C\to C}$.
$^{1}$ I do not know how and/or where $\mathbf{Set}^{\mathcal{C}^{\text{op}}}$ comes into this picture.