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In my class of category theory, my professor stated (without prove it) that the existence of tensor products between modules over commutative rings follows from the following result: a category $\mathcal{C}$ contains a final object, the equalizer of any pair of arrows and the direct product of any pair of objects iff all finite limits exist in $\mathcal{C}$.

Of course, since the tensor product is an initial object, one should use the dual version of the above result. More precisely, fixed a commutative ring $A$, $M_1$ and $M_2$ modules over $A$, we consider $\mathcal{D}$ as the category of $A$-bilinear morphisms of the form $M_1\times M_2\stackrel{f}{\rightarrow} X$, where a morphism between $M_1\times M_2\stackrel{f}{\rightarrow} X$ and $M_1\times M_2\stackrel{g}{\rightarrow} X'$ is a morphism $h:X\rightarrow X'$ such that $hf=g$. Then, an initial object of $\mathcal{D}$ is the usual tensor product $M_1\times M_2\rightarrow M_1\otimes_A M_2$.

So, my question is: how to use the result above in order to prove the existence of tensor product (avoiding its usual construction)?

Thank you.

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    $\begingroup$ Are you willing to accept the existence of the tensor product of Abelian groups? In that case, $M ⊗_A N$ is just the coequalizer of the two maps $M ⊗ A ⊗ N → M ⊗ N$ given by the $A$-action on $M$ and $N$. I don't know if the product over $ℤ$ can be constructed as a colimit in Ab. $\endgroup$ – user54748 Aug 30 '15 at 18:20
  • $\begingroup$ That result is more or less irrelevant. What you need is an adjoint functor theorem. $\endgroup$ – Zhen Lin Aug 30 '15 at 18:50
  • $\begingroup$ @ZhenLin I am aware of the adjoint functor theorem, the point is that it was stated to me that this irrelevant result can be used to prove the existence of tensor products, I don't know if it is really true. $\endgroup$ – Renan Maneli Mezabarba Aug 30 '15 at 19:05
  • $\begingroup$ @user54748 If I can prove this, then I can accept this :) Anyway, I'll ask my professor. Thanks $\endgroup$ – Renan Maneli Mezabarba Aug 30 '15 at 19:08
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    $\begingroup$ I'm sure you can prove it, it's just a rewording of the usual construction. $\endgroup$ – user54748 Aug 30 '15 at 19:33

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