Show that this function is an isomorphism (tensor product of homomorphisms)

Question

Proposition If $A, B, C$ are modules over a commutative ring $R$, then $Hom_R(A,Hom_R(B,C))\cong_R Hom_R(A\otimes B,C)$.

Proof

Let $f\in Hom_R(A,Hom_R(B,C))$.

We start by showing $$q_f:A\times B\to C, (x,y)\mapsto f(x)(y)$$ is bilinear, which is pretty simple.

Thus, there is a unique homomorphism $h_f:A\otimes B\to C$ such that $q_f=h_f\otimes$.

All there's left to do is prove that $$H:Hom_R(A,Hom_R(B,C))\to Hom_R(A\otimes B,C), f\mapsto h_f$$ is an isomorphism.

This last part is the one I'm having a hard time proving.

Any help with be appreciated.

Answer 1

You have actually done the hard direction here already. The easy part is taking $$g\in {\rm Hom}_R(A\otimes B,C)$$ and precomposing it with the natural bilinear map $$A\times B \to A\otimes B,$$ to get a bilinear map $$q_g\colon A\times B \to C.$$ Then let $I(g)\in Hom_R(A,Hom_R(B,C))$ be the map sending $a\mapsto q_g(a,\_)$.

You have already shown that for any $f\in Hom_R(A,Hom_R(B,C))$ there is a unique $g\in Hom_R(A\otimes B,C)$ such that $I(g)=f$ - namely $g=H(f)$.