# Triangulated Categories: Abelianization of projective subcategory

[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!

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.
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.