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Let $\mathcal{A}$, $\mathcal{B}$ be two triangulated categories and $F: \mathcal{A}\to \mathcal{B}$ be a triangulated functor between them. Then $F$ induces an homomorphism between their Grothendieck groups $$ \bar{F}: K_0(\mathcal{A})\to K_0(\mathcal{B}). $$

Now assume $F$ is also fully-faithful, it is true that $\bar{F}$ is always an injection? If not, is their any counter-examples?

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    $\begingroup$ I believe the answer is no. Take $R$ a ring, and let $\mathcal{B}$ be the derived category of complexes of $R$-modules (no restriction whatsoever on ranks of modules or size of complexes). Then take $\mathcal{A}$ the full subcategory of bounded complexes of finite rank $R$-modules. For $R$ suitably complicated, the $K_0(\mathcal{A})$ is non-trivial, but $K_0(\mathcal{B})$ is trivial by Eilenberg swindle. The functor is also fully faithful. If I didn't miss something... $\endgroup$
    – Matthias Wendt
    Commented Apr 16, 2015 at 17:19
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    $\begingroup$ @MatthiasWendt suitably complicated may be $R=\mathbb Z$ and, in this case, if you impose complexes in $\mathcal B$ to consist of countable abelian groups, then $\mathcal B$ is essentially small. This concerns Adeel's comment below. $\endgroup$
    – Fernando Muro
    Commented Apr 16, 2015 at 18:27

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One should really only talk about $K_0$ of small categories, otherwise one runs into the difficulty explained in Matthias's comment. Assuming $\mathcal{B}$ is small, let me identify $\mathcal{A}$ with its essential image by $F$, which will be a (strictly) full triangulated category of $\mathcal{B}$. For a positive answer to your question, it is reasonable to impose the condition that $\mathcal{A}$ is dense in $\mathcal{B}$, i.e. every object of $\mathcal{B}$ is a direct summand of an object of $\mathcal{A}$. Then the claim is indeed true and follows for example from Thomason's classification theorem, which says that there is a bijection between dense strictly full triangulated subcategories $\mathcal{A} \subset \mathcal{B}$ and subgroups of the Grothendieck group $K_0(\mathcal{B})$. See Corllary 2.3 in [Thomason, The classification of triangulated subcategories, Comp. Math. 105 (1997), pp. 1-27].

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    $\begingroup$ Density is a very strong assumption! $\endgroup$
    – Sasha
    Commented Apr 16, 2015 at 18:36
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    $\begingroup$ A counterexamle in the general case (generalizing the one of Matthias Wendt): if you embedd your category $A$ into $B$ that is closed with respect to countable coproducts (or products) then you kill the whole $K_0(A)$ since $K_0(B)=0$. $\endgroup$
    – Mikhail Bondarko
    Commented Apr 16, 2015 at 19:26

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