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I can't prove this lemma in "Notes on motivic cohomology, Beilinson, Macpherson, Schechtman":

Lemma. A pure functor is exact.

Definitions: A mixed category $\mathcal{M}$ is a $\mathbb{Q}$-abelian category in which every object has a finite increasing filtration $W_\bullet$, such that morphisms are strictly compatible with $W_\bullet$, graded quotients $\mathrm{gr}_i^W M$ for every object $M$ and every integer $i$ are semisimple and $\mathrm{Hom}$'s are finite dimensional $\mathbb{Q}$-vector spaces. A pure functor between two mixed categories is an additive functor which sends every pure object (an object $M$ with at most one nonzero graded quotient w.r.t. weight filtration) to a pure object of the same weight.

A counterexample(?): Let $\mathcal{M}$ be the category of finite $\mathbb{Q}$-representations of the quiver: $\bullet\to\bullet$. Then $\mathcal{M}$ is a mixed category with the weight filtration of a representation $f:V_0\to V_1$ given by $W_{-1}=0\to 0 , W_0=0\to V_1, W_1 = V_0\stackrel{f}{\to}V_1$. It has two simple pure objects $\mathbb{Q}\to 0$ and $0\to\mathbb{Q}$ and only one other indecomposable non-pure object $X = \mathbb{Q}\stackrel{\mathrm{id}}{\to}\mathbb{Q}$. Then I think that there are pure endofunctors of $\mathcal{M}$ which act as the identity on pure objects and send $X$ to either of the simple pure objects and these functors are not exact. So I think purity is not sufficient for exactness.

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    $\begingroup$ I agree that, with the definitions as stated in the paper, your example works. Maybe the definition of a pure functor is meant to include the condition that it commutes with the functors $M\mapsto W_iM$ and $M\mapsto M/W_iM$ (which is not the case in your example). Would that make sense? $\endgroup$ Commented Jan 25, 2015 at 20:04
  • $\begingroup$ @JeremyRickard Yes, this is a reasonable condition and works. I was not confident on my example and so presented it here as a question. $\endgroup$ Commented Jan 25, 2015 at 20:26
  • $\begingroup$ To be explicit, a functor which is the identity on pure objects but is not exact is given by mapping $V \stackrel f \to W$ to $\ker(f) \stackrel 0 \to \mathrm{coker}(f)$. Another possible fix than Jeremy Rickard's suggestion might be to impose the condition that the functors $\mathrm{gr}^W_i$ are exact. (This is just a suggestion, I don't know if the lemma holds under either added hypothesis.) $\endgroup$ Commented Jan 25, 2015 at 20:55
  • $\begingroup$ @DanPetersen The graded quotient functors are exact by the definition of mixed categories ( morphisms are strict) $\endgroup$ Commented Jan 25, 2015 at 21:15
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    $\begingroup$ Possibly, you can benefit from Appendix D in users.unimi.it/~barbieri/der1mot.pdf $\endgroup$ Commented Jan 26, 2015 at 11:11

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