Re: Ext^1 Question.

From: Robin Chapman (rjc_at_ivorynospamtower.freeserve.co.uk)
Date: 03/22/05 

Date: Tue, 22 Mar 2005 19:01:20 +0000

sirix wrote:

> Robin Chapman wrote:
>>>>>With this definition averything goes smoothly. But, Charles A. Weibel
>>>>>acts with Hom(A,__) on extension xi: 0->B->X->A->0 and gets
>>>>>0 -> Hom(A,B) -> Hom(A,X) -> Hom(A,A) -> Ext^1(A,B)
>>>>
>>>>
>>>>If you do this you should use an injective resolution of B :-)
>>>>
>>>
>>>And does it give rise to the same bijection or some other?
>>
>> Maybe the same, maybe the negative :-)
>
> Well, few hours ago I've told people (on student's seminary) that "I
> don't see any reason why it should be the same bijection", so please,
> tell me how to prove I'm nasty liar :-)

Use both a projective resolution of A and an injective one of B
to get a double complex whose cohomology is Ext^*(A,B) (and do a lot
of diagram chasing to show that this cohomology is the same
as that got by just doing either one of the resolutions). Then do
more diagram chasing to show that the two possible definitions
of the characteristic class of a short exact sequence coincide.
(I'm sure the key will be to set up your sign conventions
in the double complex cunningly :-) ) 

-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
  "Elegance is an algorithm"
    Iain M. Banks, _The Algebraist_