Suppose given an exact category $\mathcal{C}$. The following question arises while proving the Gabriel-Quillen-Laumon embedding theorem following Laumon [1].

Laumon constructs an abelian category $\mathfrak{F}(\mathcal{C})$ as category of sheaves on $\mathcal{C}$, turned into a site by letting coverings be sieves that contain an admissible epimorphism.

So an object of $\mathfrak{F}(\mathcal{C})$ is a functor $F \colon \mathcal{C}^{\operatorname{op}} \to \mathbf{Z}\text{-Mod}$ such that for each short exact sequence


in $\mathcal{C}$, the sequence


is left-exact in $\mathbf{Z}\text{-Mod}$.

(The diagram


is a pullback.)

Call such a functor $F \colon \mathcal{C}^{\operatorname{op}} \to \mathbf{Z}\text{-Mod}$ quasi-left-exact.

If $F \colon \mathcal{C}^{\operatorname{op}} \to \mathbf{Z}\text{-Mod}$ is additive, then $F$ is quasi-left-exact if and only if $F$ is left-exact.

Do there exist (non-additive) functors that are quasi-left-exact?

Laumon [1,p.155] says that Quillen [2,p.92/100] works with additive functors but he does not follow him in that point.

Thomason and Trobaugh [3, p.400] use the term $\ll$"left exact"$\gg$ for what I call $\ll$quasi-left-exact$\gg$.

Bühler [4,p.54-55] says that a (pre)sheaf is not necessarily additive by definition but then he continues to work with additive functors and $\ll$(additive) sheaves$\gg$.

[1] Laumon, G. Sur la catégorie dérivée des $\mathcal{D}$-modules filtrés. In Algebraic geometry (Tokyo/Kyoto, 1982), vol. 1016 of Lecture Notes in Math. Springer, Berlin, 1983, pp. 151-237

[2] Quillen, D. Higher algebraic K-theory. I. In Algebraic K-theory, I: Higher K-theories (Proc. Conf., Battelle Memorial Inst., Seattle, Wash., 1972). Springer, Berlin, 1973, pp. 85-147. Lecture Notes in Math., Vol. 341.

[3] Thomason, R.W. and Trobaugh, T. Higher algebraic K-theory of schemes and of derived categories. In The Grothendieck Festschrift, Vol. III, vol. 88 of Progr. Math. Birkhäuser Boston, Boston, MA, 1990, pp. 247-435

[4] Bühler, T. Exact categories. Expo. Math. 28, 1 (2010), 1-69.


Take $\mathcal C$ to be the opposite of the category of abelian groups and $F(A)=A\otimes A$.

  • $\begingroup$ Thank you! A modification is necessary since the sequence $\mathbf{Z}/2 \xrightarrow{2} \mathbf{Z}/4 \xrightarrow{1} \mathbf{Z}/2$ is short exact in $\mathbf{Z}\operatorname{-Mod}$ but $\mathbf{Z}/2 \otimes_{\mathbf{Z}} \mathbf{Z}/2 \xrightarrow{2\otimes 2} \mathbf{Z}/4 \otimes_{\mathbf{Z}} \mathbf{Z}/4$ is not injective. If we take $\mathcal{C}=(k\operatorname{-mod})^{\operatorname{op}}$, where $k$ is a field, then $F \colon k\operatorname{-mod} \to \mathbf{Z}\operatorname{-Mod}$ with $F(A) = A \otimes_k A$ for a $k$-vector space $A$ is a non-additive quasi-left-exact functor. $\endgroup$ – Nico Stein Nov 28 '13 at 10:45
  • $\begingroup$ You're welcome. The modification is not really necessary, since it is the tensor square of $\mathbb Z/4\twoheadrightarrow \mathbb Z/2$ what must be surjective, if I haven't misunderstood your question. $\endgroup$ – Fernando Muro Nov 28 '13 at 15:25
  • $\begingroup$ The sequence $F(C) \xrightarrow{F(p)} F(B) \xrightarrow{...}F(A\oplus B)$ is required to be left-exact (in the usual sense) in $\mathbf{Z}\text{-Mod}$, so $F(p)$ has to be injective. $\endgroup$ – Nico Stein Nov 29 '13 at 21:59
  • $\begingroup$ Yeah, I think my functor is right exact instead :-( $\endgroup$ – Fernando Muro Nov 29 '13 at 22:04

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