I have some difficulty understanding the following proof (source):
Claim: If $T: \mathcal{A} \rightarrow \mathcal{B}$ is a right exact functor of two Abelian categories, then $L_0 T$ and $T$ are canonically naturally equivalent.
Proof: Take a projective resolution $P_\bullet$ of some object $A\in \mathcal{A}$. Then we have an exact sequence $P_1 \rightarrow P_0 \rightarrow A \rightarrow 0$. As $T$ is assumed right exact, we have an exact sequence $TP_1 \rightarrow TP_0 \rightarrow TA \rightarrow 0$. Hence $H_0(TP_\bullet)\simeq TA$.
My difficulty is seeing this final conclusion. I could prove that this was indeed true for the more specific case of R-Modules, but there I used the first isomorphism theorem, which I'm not sure if exists in an arbitrary Abelian category. Could you please explain carefully how the above equivalence follows from all of this?
Edit A category C is abelian if it is 1) additive, 2) for every monomorphism $f: A \rightarrow B$ the pair $(A,f)$ is the kernel of the epimorphism $B \rightarrow coker f$, 3) For every epimorphism $f: A\rightarrow B$ the pair $(f,B)$ is a cokernel of the monomorphism $ker f \rightarrow A$.