# Understanding the right-exactness of the tensor product using *only* its universal property and the Yoneda lemma

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.

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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

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.