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I have been doing some review with the goal of trying to understand as much as I can via universal properties and category theory (already feeling comfortable with the mundane way of doing things). What follows is what I wrote down in thinking about the operation of tensoring with a fixed module $(-)\otimes_AN$; in particular,

  • why it should be a functor, and
  • why it is right-exact.

My questions are at the end, and more generally I would welcome any feedback.

Everything here will be commutative and unital.

Let $A$ be a ring, and let $M$ and $N$ be $A$-modules. The functor $\mathbf{B}_{M,N}:\mathsf{Mod}_A\to\mathsf{Mod}_A$ defined by $$\begin{align*} \mathbf{B}_{M,N}(P)\;&=\;\mathrm{Bilin}(M,N;P)\\[0.05in] \mathbf{B}_{M,N}(\phi:P\to Q)\;&=\;\mathrm{Bilin}(M,N;P)\xrightarrow{\;\phi\,\circ\;}\mathrm{Bilin}(M,N;Q) \end{align*}$$ is (famously) representable. The representing object, unique up to unique isomorphism, is denoted by $M\otimes_AN$, and is referred to as the tensor product of $M$ and $N$.

Fix the module $N$, and consider the contravariant functor $\mathbf{T}_N:\mathsf{Mod}_A\to\mathsf{Mod}_A^{\mathsf{Mod}_A}$ defined by $$\begin{align*} \mathbf{T}_N(M)\;&=\;\mathbf{B}_{M,N}\\[0.05in] \mathbf{T}_N(\psi:M_1\to M_2) \;&= \; \widetilde{\psi}:\mathbf{B}_{M_2,N}\Rightarrow\mathbf{B}_{M_1,N} \end{align*}$$ where the natural transformation $\widetilde{\psi}$ is defined by $$\widetilde{\psi}_P:\mathrm{Bilin}(M_2,N;P)\xrightarrow{\;\circ \,(\psi,\,\mathrm{id}_N)\;}\mathrm{Bilin}(M_1,N;P)$$ Thus, a map $\psi:M_1\to M_2$ also induces a natural transformation $\widetilde{\psi}:h^{M_2\otimes_AN}\Rightarrow h^{M_1\otimes_AN}$, and by the enriched Yoneda embedding (??), we can see that such data is equivalent to a map $M_1\otimes_AN\to M_2\otimes_AN$. Thus, using only the universal property, $(-)\otimes_AN$ can be seen to be a (covariant) functor because a map $\psi:M_1\to M_2$ acts contravariantly on the spaces of bilinear maps, and hence covariantly on the representing objects.

First, I'd like to ask: am I correct that there is an "enriched Yoneda lemma" that works the way I cited above? Can someone provide a reference that explains it at a relatively basic level? To what extent do I need to apologize for referring to $\mathsf{Mod}_A^{\mathsf{Mod}_A}$?

Now for the question in the title, I would like to argue as follows for why $(-)\otimes_AN$ is right-exact:

If the map $\psi:M_1\to M_2$ is surjective, then it acts injectively on the spaces of bilinear maps (that is, two elements of $\mathrm{Bilin}(M_2,N;P)$ pull back by $(\psi,\mathrm{id}_N)$ to the same element of $\mathrm{Bilin}(M_1,N;P)$ if and only if they were the same to start with - (bi)linearity isn't even necessary here, just thinking about these as maps of sets shows this), and hence surjectively on the representing objects; that is, the induced map $M_1\otimes_AN\to M_2\otimes_AN$ is surjective. A short exact sequence $$M_1\xrightarrow{\;\psi\;}M_2\xrightarrow{\;\rho\;} M_3\longrightarrow 0$$ is equivalent to having a surjective map $\rho:M_2\to M_3$ and a surjective map $\psi:M_1\to\ker(\rho)$, and since $(-)\otimes_AN$ "preserves surjectivity" everything's good.

Of course, this argument is not rigorous as written; some sentence sounding approximately like

the enriched Yoneda embedding preserves epimorphisms

would have to be true. What is the (true) sentence I need?

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I assume that you mean $Mod_{A}^{Mod_{A}}$ is the functor category (i.e. category of functors $Mod_{A}\rightarrow Mod_{A}$). Is it clear why this category is small? –  DBS Jul 28 '13 at 4:10
    
@DBS: Yes, I mean the functor category; since $\mathsf{Mod}_A$ itself is not small, as far as I understand it the category $\mathsf{Mod}_A^{\mathsf{Mod}_A}$ shouldn't even exist, technically. That was part of what I was hoping to get rigorized about my argument :) –  Zev Chonoles Jul 28 '13 at 4:16
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I think the enriched version of Yoneda that you refer to is here ncatlab.org/nlab/sho/enriched+Yoneda+lemma –  DBS Jul 28 '13 at 4:44
    
In your case the functor category is not just a category but it is a $Mod_{A}$ enriched category (i.e. the Hom sets are not just sets but also A-Modules. –  DBS Jul 28 '13 at 4:48
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1 Answer

QUOTE:My questions are at the end, and more generally I would welcome any feedback.

$(-)\otimes_AN$ is a left adjoint functor of Hom$_\mathbb{Z}(N, -$)(see my answer to this question and the comments for it). Hence it preserves colimits(see, for example, Mac Lane's Categories for the Working Mathematician). Hence it is right exact.

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I agree that this is a correct argument for why $(-)\otimes_AN$ is right-exact, but I don't see how this is the rigorization of the argument I gave in my post. –  Zev Chonoles Jul 28 '13 at 1:09
    
@ZevChonoles I think this proof explains very clearly why the tensor product is right exact. You said you would welcome any feedback. Sorry if this is not what you meant by a feedback. –  Makoto Kato Jul 28 '13 at 1:26
    
Sorry, I misunderstood this as an answer to one of the questions I asked. Thanks for your feedback. –  Zev Chonoles Jul 28 '13 at 1:44
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