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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.

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    $\begingroup$ Dear Anton, As was noted in Buschi Sergios comment below my answer, there is a typo in the question. (You should replace "coker($F(f)$)" by "im($F(f)$)".) $\endgroup$
    – Emerton
    Commented Nov 15, 2010 at 4:15
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    $\begingroup$ @Emerton: Fixed. Thank you. I was so confused when I read that comment since it seemed inconceivable to me that anybody got the exactness condition wrong. How embarrassing! $\endgroup$
    – Anton Geraschenko
    Commented Nov 15, 2010 at 4:40

3 Answers 3

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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.

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    $\begingroup$ This is a very elegant example. Thanks. $\endgroup$
    – Anton Geraschenko
    Commented Nov 15, 2010 at 4:42
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    $\begingroup$ One thing to keep in mind is that the category of morphisms of vector spaces really wants to a 2-category, and this is '2-abelian', see M. Dupont's thesis, arxiv.org/abs/0809.1760 $\endgroup$
    – David Roberts ♦
    Commented Nov 15, 2010 at 4:47
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    $\begingroup$ An example which is similar in spirit: Let $F: Ab \to Ab$ be the functor $A \mapsto pA$ for a prime $p$. This functor preserves monomorphisms and epimorphisms, but it is not middle-exact. $\endgroup$
    – Tim Campion
    Commented Dec 31, 2020 at 20:24
  • $\begingroup$ See also this post for a generalisation: the image of a natural transformation $\eta \colon F \to G$ between exact functors $F, G \colon \mathscr A \to \mathscr B$ is exact in the middle if and only if its kernel is right exact (or its cokernel is left exact). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 26, 2022 at 16:06
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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).

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  • $\begingroup$ I think the condition is: $Ker(Fg)= Im(Ff)$ and no $Ker(Fg)= Cok(Ff)$ . $\endgroup$
    – Buschi Sergio
    Commented Nov 14, 2010 at 9:46
  • $\begingroup$ Yes, this is right. $\endgroup$
    – Emerton
    Commented Nov 14, 2010 at 15:29
  • $\begingroup$ Dear Anton, You're welcome. $\endgroup$
    – Emerton
    Commented Nov 15, 2010 at 4:20
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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.

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    $\begingroup$ For an explication of this example, you could look at Ex. 2.7.1 of the survey ams.org/journals/bull/2009-46-04/S0273-0979-09-01260-9 $\endgroup$
    – Sam Lichtenstein
    Commented Nov 14, 2010 at 17:53
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    $\begingroup$ Yes ... and the relation to Leonid's example is provided by MacPherson and Vilonen's "elementary construction of perverse sheaves", which shows that intermediate extension can be calculated at each step in the stratification by taking an image of a map of vector spaces. So in some sense these are the same example! $\endgroup$
    – Geordie Williamson
    Commented Nov 14, 2010 at 17:55

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