Suppose $F: C\to D$ is an additive functor between abelian categories and that


$$0\to X\xrightarrow f Y\xrightarrow g Z\to 0$$


is and exact sequence in $C$. Does it follow that $F(X)\xrightarrow{F(f)} F(Y)\xrightarrow{F(g)} F(Z)$ is exact in $D$? In other words, is $\ker(F(g))=\mathrm{im}(F(f))$?


**Remark 1:** If the answer is no, a counterexample must use a *non-split* short exact sequence. This is because additive functors send split exact sequences to split exact sequences. A splitting is a pair $s:Y\to X$ and $r:Z\to Y$ so that $id\_Y=f\circ s+r\circ g$, $id\_X=s\circ f$, and $id\_Z=g\circ r$. An additive functor preserves these properties, so $F(s)$ and $F(r)$ will split the sequence in $D$.


**Remark 2:** You probably know you know lots of left exact and right exact additive functors, but you also know lots of *exact in the middle* additive functors. $H^i$ and $H\_i$ for any (co)homology theory are neither left or right exact, but they are exact in the middle by the long exact sequence in (co)homology.



Is there an additive functor between abelian categories which isn't exact in the middle?
Consider the abelian category of morphisms of vector spaces, i.e., the objects are linear maps $f:U\to V$, and the morphisms are commutative squares. Let the functor $Im$ assign to a morphism $f$ its image $Im(f)$. Consider the short exact sequence of morphisms $(0\to V)\to (U\to V)\to (U\to 0)$. The functor $Im$ transforms it to the sequence $0\to Im(f)\to 0$, i.e. $Im$ is not exact in the middle.


On the other hand, notice that $Im$ is epimorphic and monomorphic, i.e., transforms epimorphisms to epimorphisms and monomorphisms to monomorphisms.


