Thanks to the several people who pointed out that a) (for no good reason) I switched left with right as I typed out my query; and b) the Cartesian closed categories provide the correct context. David PS In case anyone has further comments, the corrected query follows.
I would appreciate help with something probably very simple that I just can't see right now:
In any category with products and coproducts one gets a map
n:AxC+BxC ---> (A+B)C
In the category of sets this turns out an isomorphism, so we have our usual distributive law for cardinal arithmetic.
Surely this remains valid in an arbitrary topos, right? But how to prove it. One needs *something* because while one similarly has a natural map
AB+C ---> (A+C)(B+C)
this map almost never turns out iso.
My first stumbling block occurs trying to think of a map
(A+B)C ---> AxC + BxC
as natural maps either *to* coproducts or *from* products seem hard to come by. But wait, starting from the map n above, I get the following natural map by composition (I write 2 for the subobject classifier):
2^n: 2^((A+B)C) ---> 2^(AxC + BxC)
or (more or less) equivalently
2^n: 2^(A+B)^C ---> (2^A)^C x (2^B)^C.
But surely I have natural isos
2^(A+B) ---> 2^A x 2^B and (2^A x 2^B)^C ---> (2^A)^C x (2^B)^C.
At the end of the day, this shows at best an isomorphism between 2^(AxC + BxC) and 2^((A+B)C). In an arbitrary topos one can't just "take the log," right?
So I don't know how to proceed, or alternatively, the right way to start over from scratch.
More generally, what conditions which happen to hold in Sets suffice to make n an isomorphism?