On Thu, May 9, 2019 at 1:50 PM Harley D. Eades III <harley.eades@gmail.com> wrote:
Hi, Mike.
Doh! That was an unfortunate typo. Actually, I just meant:
Hom(F(A (x) B),C) = Hom(A, F(B) => C)
The nearest thing that comes to mind is a cartesian closed category with an invertible tensorial strength for F. (A tensorial strength for F is a natural transformation b_{A,B}:A x F(B) -> F(A x B) satisfying some coherence laws.) Then given f: F(A (x) B) -> C we get curry(b_{A,B} o f): A -> F(B) => C; and given g: A -> F(B) => C we get b^{-1}_{A,B} o uncurry(g): F(A (x) B) -> C; so the two hom sets are isomorphic. -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com [For admin and other information see: http://www.mta.ca/~cat-dist/ ]