Can you express this easy theorem in fancy categorical terms?

Question

Here is a theorem (of homological algebra):

Given $A \rightarrow B \rightarrow C$ in an abelian category $\mathcal{A}$.

If for all $D \in \mathcal{A}$ we have that $Hom(D,A) \rightarrow Hom(D,B) \rightarrow Hom(D,C)$ is an exact sequence, then $A \rightarrow B \rightarrow C$ is an exact sequence.

This is a nice enough theorem that it feels like this can be expressed by saying that some certain representable functor is exact, or faithful, or a generator.

I tried using that an additive functor is faithful iff it sends nonexact sequences to nonexact sequences, but still haven't come up with anything.

Answer 1

I'm gonna have to go with @Malice on this one.

• Recall that a functor $F$ is “representable” precisely when $F ≅ \mathsf{Hom}(D,-)$ for some object $D$.

• Recall that a functor $F$ “reflects a property $P$” precisely when $P(X) \;⇐\; P(F\, X)$ for all $X$.

  ( This’ the converse of property preservation. )

• Recall that a family of functors consists of a functor $Fᵢ$ for each $i$ in some index set $I$.

• Recall that a family $Fᵢ$ “jointly reflects property $P$” precisely when $P(X) \;⇐\; (∀ i.\; P(Fᵢ \, x))$ for all $X$.

Now your phrase can be stated compactly,

“Representables jointly reflect exact ternary sequences”

Neato!