Question

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.

Answer 1

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.

Answer 2

Composing two random "exact-in-the-middle" functors should give a counterexample.

E.g. let's consider the functor from $\mathbb Z$-modules to itself given by $$ M \mapsto Hom(\mathbb Z/p\mathbb Z, M/p^2 M),$$ for some fixed prime $p$. Applying this to the short exact sequence $$0 \to \mathbb Z/p^2 \mathbb Z\to \mathbb Z/p^3\mathbb Z \to \mathbb Z/p \mathbb Z \to 0$$ (the first non-trivial arrow being mult. by $p$ and the second being the natural projection) gives the sequence $$ \mathbb Z/p\mathbb Z \to \mathbb Z/p\mathbb Z \to \mathbb Z/p\mathbb Z,$$ with all the arrows being 0 (assuming I haven't miscalculated).

Answer 3

As far as I remember, there is an important example of a functor that transforms mono- and epimorphisms into mono- and epimorphisms, respectively, but is not half-exact; this is the functor of intermediate extension of perverse sheaves.