MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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($F(f)$)".) – Emerton Nov 15 '10 at 4:15
@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! – Anton Geraschenko Nov 15 '10 at 4:40
up vote 21 down vote accepted

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.

share|cite|improve this answer
This is a very elegant example. Thanks. – Anton Geraschenko Nov 15 '10 at 4:42
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, – David Roberts Nov 15 '10 at 4:47

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).

share|cite|improve this answer
I think the condition is: $Ker(Fg)= Im(Ff)$ and no $Ker(Fg)= Cok(Ff)$ . – Buschi Sergio Nov 14 '10 at 9:46
Yes, this is right. – Emerton Nov 14 '10 at 15:29
Very nice example! Thank you. – Anton Geraschenko Nov 15 '10 at 4:13
Dear Anton, You're welcome. – Emerton Nov 15 '10 at 4:20

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.

share|cite|improve this answer
For an explication of this example, you could look at Ex. 2.7.1 of the survey – Sam Lichtenstein Nov 14 '10 at 17:53
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! – Geordie Williamson Nov 14 '10 at 17:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.