Reduced homology

1. Nov 19, 2005

homology

Two questions:
(1) What can you mod Z(+)Z by and get Z?
Certainly modding out by Z works, but does anything else?

(2) I've just started reading about reduced homology (in particular the reduced zeroth homology (singular). So as a refresher: We define a homomorphism f (called the augmentation) between the zero chains C0 and the integers Z, and what f does is to add all the coefficients of the 0-chain together.

Its clear to me that all the zero boundaries are in the kernel of f, but why wouldn't it always be equal to it?

Another way of asking this is to say: why doesn't the reduced zeroth homology just have one less Z than the usual zeroth homology?

Thanks,
Kevin

2. Nov 19, 2005

mathwonk

the only thing you can mod out by woould be a submodule of Z+Z. now all submodules of Z+Z are isomorphic either to {0}, to Z, or to Z+Z.

there are how ever many ways to embed Z and even Z+Z into Z+Z, and the embedding affects the nature of the quotient.

any embedding of Z+Z will have rank 2 however and hence the quotient will be finite, so not isomorphic to Z.

hence only Z can be embedded so as to have quotient isomorphic to Z.

there are however many different embeddings of Z which have quotient Z, not just the obvious ones taking n to (0,n) or to (n,0).

Any embedding is given by a 2 x 1 matrix of integers of rank 1.

then one can "diagonalize" this matrix by row and column operations, and afterwards, the quotient would appear to be isomorphic to Z if and only if the diagonalized matrix has a zero in it, so

3. Nov 19, 2005

homology

So does this mean if you have an epimorphism from Z+Z to Z that the kernel of the epimorphism must be isomorphic to Z? (Just want to clarify).

4. Nov 22, 2005

mathwonk

yes. all submodules of a free finitely generated module over a pid are free of rank less than or equal to the rank of the top module.

and rank is additive over exact sequences, so the rank of a quotient equals the difference of the rank of the top and bottom modules.