# Is there a trick to prove injectivity for maps out of tensor products?

For a ring $R$, an $R$-right-module $M$, an $R$-left-module $N$, and an abelian group $P$, one can use the universal property of the tensor product to construct maps $$M\otimes_R N\to P.$$ It concrete cases, it is often easy to see that the constructed map is surjective by just writing down pre-images.

It seems to be harder to verify that the map is injective, because then one has to consider general sums of elementary tensors in $M\otimes_R N$.

Are there any tricks to avoid this, and to achieve injectivity more elegantly?

More concretely, an example I have in mind is the following: I have a pre-additive category $C$ with finitely many objects. I consider modules over this category (that is, functors from $C$ to Abelian groups; or, equivalently, modules over the category ring of $C$). Now, I consider the "free $C$-right-module $Q_Y$ over an object $Y$ in $C$" which is the hom functor $C(\bullet,Y)$. I want to show that for an arbitrary left-module $M$: $$Q_Y\otimes_C M\cong M(Y)$$ as abelian groups. Using the module structure of $M$, I can accomplish a natural epimorphism $Q_Y\otimes_C M\to M(Y)$ which should be injective.

Currently I think that the formula $$M(Y)\to Q_Y\otimes_C M;\quad m\mapsto\mathrm{id}_Y\otimes m$$ gives indeed a (two-sided) inverse. Is this correct?

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 Can you give an example? I don't really know what you mean. – Qiaochu Yuan Oct 28 '10 at 15:28 please edit your question to include the example, so that the questyion body contains all the relevant details about what you want to know (and so that we don't have to read the slightly amusingly small font of the comments...) – Mariano Suárez-Alvarez♦ Oct 28 '10 at 21:27 (Isn' t your example application just Yoneda's lemma?) – Mariano Suárez-Alvarez♦ Oct 28 '10 at 21:29 @Mariano: I think you are right. Yoneda's lemma gives $M(Y)\cong\mathrm{Hom}_{C-Mod}(P_Y,M)$, where $P_Y$ is the free left-module over $C$ (the covariant hom-functor). Now I have to understand why $\mathrm{Hom}_{C-Mod}(P_Y,M)\cong Q_Y\otimes_C M$, so that in some sence $Q_Y$ is dual to $P_Y$, right? – Rasmus Oct 28 '10 at 21:45