Illustrating an Alternative Definition of Exactness Kashiwara and Schapira say in their book Categories and Sheaves that a functor $F:C \to D$ is right exact when the category $C_U$ is filtrant for every $U \in D$. Left exactness is defined dually, and exactness is defined to be both right and left exactness. I want to see what filtrant categories have to do with my intuitions from commutative and homological algebra, so I want to make some examples of things which are and are not exact. The tensor product seems like a familiar example, so I figured that's a good place to start.
Let's just think about abelian groups then, since in that setting $-\otimes \mathbb{Z}/2\mathbb{Z}$ is right exact and not left exact. So I should be able to try to check both conditions somewhat clearly here why one works and one does not, but I find these definitions very cumbersome to work with, because I either need to concoct a rather odd looking counterexample (I tried without success to prove that it's not left exact by using some standard counterexamples and trying to make them work in this setting, to no avail) or I struggle to construct the objects I need to prove it is filtrant on the right exact side.
In principle I had a course in group theory so this shouldn't be too hard, but something about these definitions just feels super opaque and unwieldy. Does anyone have any pointers? To be clear, I have read the proof that this notion is equivalent to preserving various limits, which I know in the case of $R-Mod$ is equivalent to the usual exactness. I just don't like abstract nonsense that I can't see or work through myself, and I'm looking for a more hands-on way to see what's going on with this notion.
 A: I don't think there's anything enlightening to be gleaned from this definition on its own.
It is one of those maximally general, maximally correct technical definitions that are invented for making theorems true.
Indeed, this particular definition of exactness is applicable to categories where finite (co)limits do not exist!
So it is not surprising that one has to do some work to connect it back to traditional definitions – specifically, one has to figure out what theorem is hiding behind the definition here.
Personally, I prefer to call functors that satisfy this condition corepresentably flat.
The reason is this:
Proposition.
Let $F : \mathcal{C} \to \mathcal{D}$ be a functor.
The following are equivalent:

*

*For every object $D$ in $\mathcal{D}$, the comma category $(F \downarrow D)$ is a filtered category.

*For every object $D$ in $\mathcal{D}$, the presheaf $\mathcal{D} (F -, D)$ is a filtered colimit of representable presheaves on $\mathcal{C}$.

*The functor $h^F \otimes_\mathcal{C} {-} : [\mathcal{C}, \textbf{Set}] \to [\mathcal{D}, \textbf{Set}]$, where $h^F : \mathcal{C}^\textrm{op} \times \mathcal{D} \to \textbf{Set}$ is given by $h^F (C, D) = \mathcal{D} (F C, D)$, preserves finite limits.

The point is that a general functor $\mathcal{C}^\textrm{op} \times \mathcal{D} \to \textbf{Set}$, i.e. a profunctor $\mathcal{D} ⇸ \mathcal{C}$, is an analogue of a bimodule.
(A functor $\mathcal{C} \to \textbf{Set}$, or profunctor $\mathcal{C} ⇸ \mathbf{1}$, is analogous to a left module.)
There is an operation on profunctors analogous to the tensor product of (bi)modules, and "tensoring on the left" by the profunctor $h^F : \mathcal{D} ⇸ \mathcal{C}$ preserves finite limits (i.e. is left exact in the traditional sense) if and only if $F : \mathcal{C} \to \mathcal{D}$ satisfies the above condition on the comma categories.
Thus, $h^F$ is analogous to a bimodule that is flat as a right module.
Morally, flatness should be defined as "tensoring is left exact", but unfortunately there are some set-theoretic issues (that I swept under the rug in this answer) that make it untenable as a rigorous definition.
More precisely, the "tensoring" operation is not even well defined if one does not impose some size restrictions.
On the other hand, even after imposing the size restrictions, the "comma categories are filtered" condition does not change, so it is this definition that we make "official".
You can think of it as inverting the analogue of Lazard's theorem that every flat module is a filtered colimit of free modules.
A: I did manage to find a way to capture the toy example from commutative algebra! It turns out that I was not carefully understanding the definitions, and so I want to record the argument for the sake of posterity. I'll limit the category slightly further from abelian groups to finitely generated abelian groups, just in case it makes a difference with some crazy large group - I don't think it matters. I apologize to myself and future readers for not having diagrams in this post - I don't think it's possible to include them.
Let $G$ be a fixed FGAG. An object of our category is a map $s:G \to H \otimes \mathbb{Z}/2\mathbb{Z}$, which we could call $H/2H$, since we may use the invariant factor decomposition of a FGAG together with the third(?) isomorphism theorem to componentwise simplify the decomposition. The free part of the decomposition is reduced to copies of $\mathbb{Z}/2\mathbb{Z}$, while the torsion part is either killed or scaled by a factor of $2$, depending on parity.
So now what is a morphism in this category? If the above object is called $(H,s)$, then an element of $\hom((H_1,s),(H_2,t))$ is a map $\phi: H_1 \to H_2$ such that $f \otimes \text{id}$ makes a commutative triangle. This is the important bit - the maps have to come from upstairs!
Now the cofiltrant condition says that whenever I have a pair of parallel morphisms $i \to j$, there needs to be an object $k$ and a morphism $k \to i \to j$ which plays the role of a kernel of the pair of morphisms, but it need not be universal.
For the pair of parallel morphisms, we take as the objects two surjections $G \twoheadrightarrow \mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}$, and for the morphisms, take the maps, say $3 \otimes 1$ and $1 \otimes 1$ upstairs, both descending to the identity map on $\mathbb{Z}/2\mathbb{Z}$. I claim there is no $X$ such that $X/2X$ can be placed into this diagram, where the maps $X \to \mathbb{Z} \to \mathbb{Z}$ end up equal upstairs.
Up to taking sums of other cyclic groups, $X$ can only be one of the three things mentioned above. It can't be $0$ because then the diagram downstairs does not commute - you would get a $0$ map and the maps $G \to \mathbb{Z}/2\mathbb{Z}$ are surjections and the maps between the copies are both the identity. It also can't be a $\mathbb{Z}/n\mathbb{Z}$ because the only map to $\mathbb{Z}$ is $0$ and we just said that's not allowed. But if $X \simeq \mathbb{Z}$, then it cannot equalize $1$ and $3$ unless the map is $0$, but again this is prohibited.
So there's no object that does the job.
