Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to get an intuition for why $(-)\otimes N$ is right-exact using its universal property involving bilinear maps, not by appealing to higher-level observations such as "left-adjoints preserve colimits". The argument below is the best I could do towards this goal, but clearly it is in need of rigorization (if indeed something along these lines is correct).

I have indicated two spots in the argument below that I would like to ask for detailed explanations of how to rigorize and/or fix.

This is an attempt to re-ask an earlier question of mine, which apparently was easy to misinterpret.

Basic idea:

Let $\mathcal{C}$ be a category. For any object $X$ of $\mathcal{C}$, let $h^X:\mathcal{C}\to\mathsf{Set}$ be the covariant hom functor: $$h^X(Y):=\mathrm{Mor}_{\mathcal{C}}(X,Y),\qquad h^X\left(Y\xrightarrow{\;f\;}Z\right)=\mathrm{Mor}_{\mathcal{C}}(X,Y)\xrightarrow{\;f\,\circ\, -\;}\mathrm{Mor}_{\mathcal{C}}(X,Z)$$ The Yoneda lemma implies that a natural transformation $\gamma:h^X\Rightarrow h^W$ must come from a morphism $g:W\to X$; that is, we must have that $\gamma_Y(k)=k\circ g$ for some such $g$.

If $\gamma_Y$ is injective for all objects $Y$ of $\mathcal{C}$, the corresponding $g$ is an epimorphism (by definition).

Let $A$ be a ring, and fix an $A$-module $N$.

If an $A$-module map $\psi:M_1\to M_2$ is surjective, then $(\psi,\mathrm{id}_N):M_1\times N\to M_2\times N$ is surjective, so that for all $A$-modules $P$, the map $$\mathrm{Hom}(M_2\otimes N,P)\underset{\text{natural}}{\cong}\mathrm{Bilin}(M_2,N;P)\xrightarrow{-\circ(\psi,\mathrm{id}_N)}\mathrm{Bilin}(M_1,N;P)\underset{\text{natural}}{\cong}\mathrm{Hom}(M_1\otimes N,P)$$ is injective. Therefore (?) the induced map $M_1\otimes_AN\to M_2\otimes_AN$ is an epimorphism, which is equivalent to being a surjection for $A$-modules.

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 $\psi:M_1\to\ker(\rho)$. Because the functor $(-)\otimes_AN$ "preserves surjectivity", it must therefore (?) be right-exact.

share|cite|improve this question
Would you happy with an argument that convinces you that $\text{Bilin}(M, N, P)$, as a functor of $M$, is representable? – Qiaochu Yuan Aug 10 '13 at 17:12
@Qiaochu: I'm not sure I understand how that would help show right-exactness, but I'm probably missing something obvious. But of course I'd be interested to see whatever you've got. – Zev Chonoles Aug 11 '13 at 1:51
I like very much the way tensor products are handled in Categories and Sheaves by Kashiwara and Schapira. – Pierre-Yves Gaillard Aug 11 '13 at 9:55
@Zev: it implies that the tensor product preserves all colimits by the Yoneda lemma. – Qiaochu Yuan Aug 11 '13 at 15:39

The first step is correct, but the second step is not: You cannot say anything about the kernel after tensoring. Also note that your second step is purely formal and would apply to every additive functor which preserves epis. But not every such functor is right exact.

Let $M_1 \to M_2 \to M_3 \to 0$ be an exact sequence. We want to show that, for every module $N$, the sequence $M_1 \otimes N \to M_2 \otimes N \to M_3 \otimes N \to 0$ is exact, i.e. that $M_2 \otimes N \to M_3 \otimes N$ is a cokernel of $M_1 \otimes N \to M_2 \otimes N$. This means, by the universal property of the cokernel, that for every "test" module $T$, the sequence $0 \to \hom(M_3 \otimes N,T) \to \hom(M_2 \otimes N,T) \to \hom(M_1 \otimes N,T)$ is exact (as abelian groups, but then also as modules). By definition of the tensor product, this sequence is isomorphic to the sequence $0 \to \mathrm{Bilin}(M_3,N;T) \to \mathrm{Bilin}(M_2,N;T) \to \mathrm{Bilin}(M_1,N;T)$. $(\star)$

Thus, the claim is actually equivalent to a statement about bilinear maps. And this can be checked now directly. I will leave out the trivial steps. For the only interesting one, let $\beta : M_2 \times N \to T$ be a bilinear map which vanishes on $M_1 \times N$. Define $\gamma : M_3 \times N \to T$ as follows: If $m_3 \in M_3$, $n \in N$, choose a preimage $m_2 \in M_2$ of $m_3$ and define $\gamma(m_3,n):=\beta(m_2,n)$. This is well-defined, because every other choice of $m_2$ is of the form $m_2+x$ for some $x$ coming from $M_1$, and then $\beta(m_2+x,n)=\beta(m_2,n)+\beta(x,n)=\beta(m_2,n)$. One sees directly that $\gamma$ is bilinear because $\beta$ is. And course $\gamma$ is the desired preimage in $\mathrm{Bilin}(M_3,N;T)$.

This is not the most conceptual proof. You have already mentioned the one using adjoint functors. But we can also choose an alternative ending for the proof above: The sequence $(\star)$ is isomorphic to $0 \to \hom(N,\hom(M_3,T)) \to \hom(N,\hom(M_2,T)) \to \hom(N,\hom(M_1,T))$, which is exact because $\hom(N,-)$ is left exact and $\hom(-,T)$ is right exact.

And yet another ending (which explains Qiaochu's comment): The isomorphism $\mathrm{Bilin}(-,N;T) \cong \hom(-,\hom(N,T))$ shows that this functor is representable and therefore right exact, hence $(\star)$ is exact.

share|cite|improve this answer
We didn't use the Yoneda Lemma at all, right? I don't see why we'd have to. – Eric Auld Nov 8 '15 at 18:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.