[Bel2011]: Beligiannis, 2011, Relative Homological Algebra.

Let $T$ be a triangulated category.
Regard an ideal $I\subset T$. Suppose it is also saturated.

Consider the full subcategory of projective objects $\operatorname{proj}(I)\subseteq T$, $$P\in\operatorname{proj}(I),x\in I:\quad \hom(P,x)=0.$$

In [Bel2011, prop 4.19] it is then considered $\operatorname{ab}(\operatorname{proj}I)$.
The $\operatorname{ab}(..)$ is meant as the abelianization functor.
As its construction it is taken in that article the one by Peter Freyd, see [Bel2011, sec 3.1]:

ab(...) := fin pres additive functors: (...) –> abelian groups, i.e. $$(\_,B)\to(\_,A)\to F(\_)\to0.$$

For triangulated categories this is fine. However for the class of projectives it is not clear to me why this defines an abelian category, more precisely why $\operatorname{ab}(\operatorname{proj}I)$ has weak kernels. (All the other properties are clear.)

I doubt it, but it would be sufficient to know that the class of projectives is itself triangulated: $$P,Q\in\operatorname{proj}(I):\quad P\to X\to Q\to SP\implies X\in\operatorname{proj}(I)?$$

Do you have some idea whether $\operatorname{ab}(\operatorname{proj}I)$ is abelian?

Thank you very much in advance!


1 Answer 1


If $\mathcal{C}$ is an additive category, then for Freyd's category of finitely presented functors to have kernels, it is sufficient for $\mathcal{C}$ to have weak kernels.

In Beligiannis' Prop. 4.19, there is an added hypothesis: "enough projectives".

Your notation and terminology is a bit different from that of Beligiannis, but if I'm understanding correctly how the two correspond, then in your terms I think that this means that, for any object $Z$ of $T$, there is a triangle $$W\to P\to Z\xrightarrow{\alpha}\Sigma W$$ where $P$ is projective and $\alpha\in I$.

If $X\to Y$ is any map between projectives fitting into a triangle $$Z\to X\to Y\to \Sigma Z$$ in $T$, then the composition $$P\to Z\to X$$ will be a weak kernel of $X\to Y$ in the subcategory of projectives.

  • $\begingroup$ Very nice, thank you very much Jeremy Rickard! :) $\endgroup$ Sep 13, 2020 at 12:26

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