rather "homomorphism" I hope.
Yes, of course.
... but don't ask me to give even a sketch of proof of tedious exercise.
No, thanks. Your answer is perfect.

Hello,

there are at least two ways to define Ext(A,B) of abelian groups A,B
namely by derived functor stuff and in an elementary way: Consider all
the short exact sequences
0 --> B --> E --> A --> 0
where E is an abelian group and define two such sequences (with E and
E' at the middle position) to be equivalent iff there is an
homeomorphism f:E --> E'.

This defines the SET Ext(A,B). Is there an
easy and elementary way to define the group structure on this?

Thanks,
S.

On 23 Apr., 10:24, Jannick Asmus <jannick.n...@web.de> wrote:> rather "homomorphism" I hope.
Yes, of course.
... but don't ask me to give even a sketch of proof of tedious exercise.

No, thanks. Your answer is perfect.

S.

there are at least two ways to define Ext(A,B)
[derived functor and exact sequences]
This defines the SET Ext(A,B). Is there an
easy and elementary way to define the group structure
on this?
The sum is called the Baer sum and described by Jannick

Is there any other group (non-isomorphic) than Z which
fits into the middle of the sequence
0 --> Z -*2-> Z -->> Z/2Z -->0

Ok, I suppose that I have at least learned the definition: If one
takes two representatives
0 --> B --a--> E --b--> A -->0
and
0 --> B --c--> F --d--> A -->0
of elements of Ext(A,B) then the two classes are mapped under + to the
class of the exact sequence
0 --> B --pg--> S/T --k--> A -->0
where the object S comes from the pullback
S --l--> E
| |
| b
| |
v v
F --d--> A
i.e. S={(e,f)|b(e)=d(f)} and T={(a(-b),c(b))|b in B} and
p:S --> S/T is the projection and g:B --> S the unique map induced by
a and c, i.e. g(b)=(a(b),c(b)) and k is the factorization of bl:S --
A over S/T, correct?

I was trying to find an easy example of two sequences to get an
example for the whole addition but I have failed: Is there any other
group (non-isomorphic) than Z which fits into the middle of the
sequence 0 --> Z -*2-> Z -->> Z/2Z -->0 or any other easy example?

Yes, just take the sum of this sequence with itself.

On 24 Apr., 20:10, Jack Schmidt
Jack.Schmidt.SciM...@gmail.com
wrote:
Yes, just take the sum of this sequence with
itself.

Oh, this is a bad day, I can't even calculate a
factor group.
S={(e,f) in ZxZ | e = f mod 2} and
T={(-2b,2b) in ZxZ}={(-b,b) in 2Zx2Z}
...?

S = < (2,0), (1,1) > since those two elements are in there
and (2,0)*x+(1,1)y = (2x+y,y) is the general form for such
an element. The only choice [x,y] to get 0 is [0,0], so
this is a free abelian group of rank 2 on basis { (2,0),
(1,1) }.

T = < (-2,2) >. (-2,2) = (2,0)*(-2) + (1,1)*2 = [-2,2]
(spooky), and so you want to determine the structure of
the group Z^2/<(-2,2)>.

Consider Z^2 = < (-1,1), (0,1) >, and in this basis
[-2,2] has coordinates 2 and 0. Hence S/T =~
Z^2/<(-2,2)> =~ Z^2/<(2,0)> =~ Z/2Z x Z.

Since Ext(Z_2,Z) = Z_2, there is exactly one

On 24 Apr., 20:12, Mariano Suárez-Alvarez
mariano.suarezalva...@gmail.com> wrote:

Since Ext(Z_2,Z) = Z_2, there is exactly one

How is it possible to compute that?

On 24 Apr., 20:12, Mariano Suárez-Alvarez
mariano.suarezalva...@gmail.com> wrote:

Since Ext(Z_2,Z) = Z_2, there is exactly one

How is it possible to compute that?

Thanks,
S.