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

Question

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.

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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.

Answer 1

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.