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This is an exercise problem from Hungerford's Algebra but first I'll state a result that forms the background to the problem. Let $R$ be a ring. Let $A,A'$ be right $R$-modules. Let $B,B'$ be left $R$-modules. Let $f:A\rightarrow A'$ and $g:B\rightarrow B'$ be $R$-module homomorphisms. Then there is a unique group homomorphism $f\otimes g:A\otimes_{R}B\rightarrow A'\otimes_{R}B'$ such that $a\otimes b\mapsto f(a)\otimes g(b)$ for all $a\in A,b\in B$.

What is the difference between the homomorphism $f\otimes g$ above and the element $f\otimes g$ of the tensor product of abelian groups $\mathrm{Hom}_{R}(A,A')\otimes\mathrm{Hom}_{R}(B,B')$?

The only answer I can think of is that the former is really a map but the latter is actually a coset. Is this right? If it is, can this answer be improved? Are there other answers?

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1 Answer 1

One interpretation is that there is a group homomorphism $$\hom_R(A, A') \otimes_{\mathbb Z} \hom_R(B, B') \to \hom_{\mathbb Z}(A \otimes_R B, A' \otimes_R B')$$ defined by sending $f \otimes g$ to the map $a \otimes b \mapsto f(a) \otimes g(b)$. So one is an element in the domain and the other is an element in the codomain of this homomorphism. In general this map is neither injective nor surjective, but in many circumstances it's a convenient way of inducing and specifying maps.

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