# Tensor product and its homomorphisms.

Given $f:A_R\rightarrow A'_R$, $g:B_R\rightarrow B'_R$ R-module homomorphism we can define $f\otimes g: A\otimes_R B\rightarrow A'\otimes_R B'$ such that $(f\otimes g)(a\otimes b)=f(a)\otimes g(b)$ this type of homormorphism is unique.

My question is: What is the relation between the homomorphism define above and the elements $f\otimes g$ of

$$\text{Hom }(A,A')\otimes \text{Hom }(B,B')$$

I can define a function $\phi:\text{Hom }(A,A')\times \text{Hom }(B,B')\rightarrow \text{Hom } (A\otimes_R B, A'\otimes_R B')$ as $(f,g)\mapsto f\otimes g$ . By the universal propierty of tensor product there exist a $\theta:\text{Hom }(A,A')\otimes \text{Hom }(B,B')\rightarrow \text{Hom } (A\otimes_R B, A'\otimes_R B')$ such that $\theta\circ \iota=\phi$. But I don know how the function $\theta$ is useful.

Note: The function $\iota:\text{Hom }(A,A')\times \text{Hom }(B,B')\rightarrow \text{Hom }(A,A')\otimes \text{Hom }(B,B')$ is the canonical epimorphism, i.e. $(f,g)\mapsto f\otimes g$.

Thanks a lot!

## 1 Answer

What you have stated here is that the tensor product $(\_ \otimes \_)$ is a covariant bifunctor from the category of bimodules to itself.

Thus it makes the category into a Monoidal Category.

If you are worried about isomorphisms, or the functor being fully faithful: as you know, this functor need not be right exact in either variable, in general case. But with suitable restrictions, such as in the category of finite dimensional vector spaces over a field, it will be.