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Suppose I have an additive category $\mathcal{C}$ and a pair of composable arrows: $$A \longrightarrow B \longrightarrow C.$$ It makes no sense to ask if this sequence is exact at $B$ since the category $\mathcal{C}$ doesn't have kernels or images. However: the Yoneda embedding produces a sequence of functors $$\mbox{Hom}(A,-) \longleftarrow \mbox{Hom}(B,-) \longleftarrow \mbox{Hom}(C,-)$$ and it does make sense to ask if this sequence is exact since the functor category $\[\mathcal{C},\mbox{Ab}\]$ is abelian.

Similarly we may ask if $$\mbox{Hom}(-,A) \longrightarrow \mbox{Hom}(-,B) \longrightarrow \mbox{Hom}(-,C)$$ is an exact sequence. So here are my questions:

  1. What are criteria for determining if a given composable pair of arrows become exact under one of the two Yoneda embeddings?
  2. What properties of $\mathcal{C}$ guarantee that the co- and contravariant Yoneda embeddings agree on which sequences become exact?
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You don't need kernels or images to talk about exactness. Exactness is equivalent to a certain square constructed from $A, B, C$ being both a pullback square and a pushout square, and constructing this square only requires the existence of a zero object. –  Qiaochu Yuan Jan 23 '13 at 18:40
    
In abelian categories exactness of the sequences involving Hom$(--,A)$ is measured by the derivatives of the Hom-functors which are traditionally called Ext. –  Jochen Wengenroth Jan 24 '13 at 10:16
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Here is an example: Let $A \to B \to C \to 0$ be a sequence in $\mathcal{C}$. Then it becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ (where $\widehat{\mathcal{C}} := [\mathcal{C}^{\mathrm{op}},\mathsf{Ab}]$) after applying $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}, X \mapsto \hom(X,-)$ if and only if $B \to C$ is a cokernel of $A \to B$ (immediate). On the other hand, it becomes exact in $\widehat{\mathcal{C}}$ after applying $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}}, X \mapsto \hom(-,X)$ if and only if it is isomorphic to a sequence of the form $F \oplus E \xrightarrow{p} C \oplus E \xrightarrow{q} C \to 0$, where $E,F,C \in \mathcal{C}$, $q$ is the projection and $p$ is the projection $F \oplus E \to E$ followed by the inclusion $E \to C \oplus E$ (exercise). Thus we get a split exact complex. Therefore, exactness in $\widehat{\mathcal{C}}$ applies the one in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$, but not vice versa.

Similarily, $B \to C \to 0$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if $B \to C$ is an epimorphism, and it becomes exact in $\widehat{\mathcal{C}}$ if and only if $B \to C$ is a split epimorphism.

By duality, $0 \to A \to B \to C$ becomes exact in $\widehat{\mathcal{C}}$ if and only if $A \to B$ is a kernel of $B \to C$, and it becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if it is a split exact complex.

Combining these results, we see that $0 \to A \to B \to C \to 0$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if it becomes exact in $\widehat{\mathcal{C}}$ if and only if it is a split exact complex.

Even for abelian $\mathcal{C}$, this is not the "correct" notion of exactness. I doubt that there is anything interesting which can be said about sequences of the form $A \to B \to C$.

Edit: For abelian $\mathcal{C}$ the sequence $A \to B \to C$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ iff $A \to B \to C$ is exact and the image of $B \to C$ is a direct summand of $C$. See mathunderflow 360234.

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