TITLE: how is this a functor?
QUESTION [1 upvotes]: This is from Page 78 of Rotman's homological algebra book:
Give an (R,S)-bimodule A and left R-module B, then Hom(A, B) is a left S-module, where sf takes a to f(as) and Hom (A, ) is a functor from left R-Mod to left S-Mod.
I've proved everything except the following part:
Suppose f is R-map from B to C, then Hom(A, f) is a S-map from Hom(A, B) to Hom (A, C).
I got stuck here: Hom(A, f)(sh)(a) = f(sh)(a) = fh(as) = Hom(A, f)h(as) for h in Hom(A,B) and s in S.
I am not sure how to copy with this s. Any help would be appreciated!

REPLY [1 votes]: So you want to check that $\hom(A,f)$ is indeed an $S$-map, right ?
If that is so, then $\hom(A,f)(sh) = f\circ (sh)$.
Now if you evaluate that on $a$, you get $f((sh)(a)) = f(h(as)) = f\circ h(as)= s(f\circ h)(a)$, so $f\circ (sh) = s(f\circ h)$, and so $\hom(A,f)(sh) = s\hom(A,f)(h)$
{"set_name": "stack_exchange", "score": 1, "question_id": 3725987}