The definition of (right-/left-) exact functors is that they preserve (right-/left-) exactness of SESs. However, for some certain nice functors, as $\def\Hom{\text{Hom}\,}\Hom (A,-)$ and $A\otimes-$ we have even more: namely that if a SES sequence $$\mathcal S:0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$$ of modules remains (right-/left-) exact after homing (or tensoring) with any module A, then the sequence $\mathcal S$ was exact to begin with. The proof I have of these two examples is made in a rather ad-hoc manner and I was wondering if there is a more general proof.

To illustrate this, let me give a sketch of how we show that if $\Hom(A,\mathcal S)$ is exact at $\Hom(A,M')$, then $\mathcal S$ is exact at $M'$. Suppose not; then consider the sequence $\Hom(N,\mathcal S)$, where $N$ is the kernel of $M'\rightarrow M$. It is not exact.

Another example that I'd like to add is that of taking the stalk of a sheaf . Since this is a right-exact functor, a SES in sheaves $$0\rightarrow\mathcal F\rightarrow \mathcal G \rightarrow 0 \rightarrow 0 \tag{*}$$ would yield a right-exact sequence $$0\rightarrow\mathcal F_p\rightarrow \mathcal G_p \rightarrow 0. \tag{**}$$

Now, I'm tempted to say that if (**) is exact for all $p$ then (*) was in fact exact to start with because this is very similar to the property that I alluded to above. This is in fact true, as I tried to show in my other post. My question is: is there a purely categorical proof that generalizes all these facts?

I'd even conjecture that, under some exactness properties, if a bifunctor $\mathcal B(-,-)$ is right-/left- exact adjoint in its second argument, then the functor $\prod_A \mathcal B(A,-)$ (i.e. $\mathcal B$ parametrized by a family) reflects exactness (i.e. it enjoys the property that if $\prod_A \mathcal B(A,\mathcal S)$ is exact, then $\mathcal S$ was exact).

  • $\begingroup$ Those sequences are not what are usually called short exact. They are a term too short. $\endgroup$ Oct 26, 2013 at 19:32
  • $\begingroup$ You can add an extra zero to the left of (*) if you like. But I think my point is still clear, right? $\endgroup$
    – Rodrigo
    Oct 26, 2013 at 19:51
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    $\begingroup$ In many categories (I won't say all as I am not familiar enough with general abelian categories to know how general this is), the sort of exact sequence you have here simply corresponds to an isomorphism. And all functors preserve isomorphisms. $\endgroup$ Oct 26, 2013 at 19:54
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    $\begingroup$ Also, if a short exact sequence is preserved by $\mathrm{Hom}(A, -)$ for all $A$ then it must be a split exact sequence. $\endgroup$
    – Zhen Lin
    Oct 26, 2013 at 20:02
  • $\begingroup$ @TobiasKildetoft: do you mean the sequence (*)? There, I wanted to say that (**) for all $p$ implies (*), so the fact that functors preserve isomorphisms doesn't help me. $\endgroup$
    – Rodrigo
    Oct 26, 2013 at 20:40


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