Weibel - Left Derived Functor proof explanation Theorem 2.4.6

The full proof is freely available online page 46, Theorem 4.2.6 where Weibel proves that $$L_*F$$ is a a $$\delta$$ -functor. For an exact exact sequence $$0 \rightarrow A' \rightarrow A \rightarrow A'' \rightarrow 0,$$ we want to prove the naturality of the connecting homomoprhism $$\partial_i: L_iF(A'') \rightarrow L_{i-1}F(A').$$

Weibel states

I don't understand: what is "the connecting homomoprhism" and how one derives a long exact sequence from the given diagram.

You want to show that $$L_*F$$ defines a functor from the category of short exact sequences to that of long exact sequences. To do this, you want to prove that if you have a map $$t:S\to T$$ between exact sequences $$S:A''\to A\to A'$$ and $$T:B''\to B\to B'$$, the infinite "ladder diagram" between the LESs of $$S$$ and $$T$$ commute.
It is evident they commute everywhere except possibly at the square involving the connecting morphism of $$L_*F$$. Now Weibel observes that this ladder diagram of LESs is obtained by resolving the morhism $$t$$, that is, producing a morphism of SECs of projective resolutions, as you have written down, and using the LES on homology after applying $$F$$.
The connecting morphism in the LESs coming from the SECs of resolutions define that of the derived functors, so to prove that the connecting morphism of $$L_*F$$ is natural, it suffices to show that the connecting morphism for homology that defines a functor from SECs of complexes to LESs of abelian groups is natural.
• Sorry, what is SEC? My problem is precisely how "the ladder diagram of LESs is obtained by resolving the morphism $t$ " How do you resolve a morphism? – CL. Oct 26 '18 at 7:52