1
$\begingroup$

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 enter image description here

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

$\endgroup$
1
$\begingroup$

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.

This last claim is achieved by a straightforward but perhaps slightly tedious arrow chasing, which I think Weibel does in his book.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – CL. Oct 26 '18 at 7:52
  • $\begingroup$ @CL. My bad, I meant SES (short exact sequence). SEC is the acronym in Spanish. What I mean is that you resolve the diagram of SESs by resolving each of the six objects but also doing this carefully (two uses of the horseshoe lemma and some uses of lifting morphisms to projective resolutions) to get a diagram of SESs of resolutions. $\endgroup$ – Pedro Tamaroff Oct 26 '18 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.