Why is this homology group is zero?

[kakarotyjn]

For example,K is a triangulation of S^2;H_1 (K ) = Z_1 (K )/B_1 (K ).And Z_1 (K ) = B_1 (K ).Then I think H_1(K)=[z],z is any element of Z_1(K),[z] is the equivalent class of z .But why is it zero?Thank you!

 

[lavinia]

For example,K is a triangulation of S^2;H_1 (K ) = Z_1 (K )/B_1 (K ).And Z_1 (K ) = B_1 (K ).Then I think H_1(K)=[z],z is any element of Z_1(K),[z] is the equivalent class of z .But why is it zero?Thank you!

Your question is a little vague because you are not describing the triangulation of the 2 sphere that you are using and you are not describing the cycle,z.

Be assured that the first homology is zero because the sphere is simply connected.

 

[Tinyboss]

Maybe the question is why we say "zero" for the group consisting of a single equivalence class? In that case, the answer is because homology groups are abelian, so the identity is written as zero, and the group consisting of only the identity is also called "zero".

 

[lavinia]

Maybe the question is why we say "zero" for the group consisting of a single equivalence class? In that case, the answer is because homology groups are abelian, so the identity is written as zero, and the group consisting of only the identity is also called "zero".

yes. zero is the convention for the identity in abelian groups.

 

[Arif Syed]

K would be the same as k K=kMaybe the question is why we say "zero" for the group consisting of a single equivalence class? In that case, the answer is because homology groups are abelian, so the identity is written as zero, and the group consisting of only the identity is also called "zero"
Re: Why is this homology group is zero?
Originally Posted by Tinyboss View Post

Maybe the question is why we say "zero" for the group consisting of a single equivalence class? In that case, the answer is because homology groups are abelian, so the identity is written as zero, and the group consisting of only the identity is also called "zero".

yes. zero is the convention for the identity in abelian groups. Your question is a little vague because you are not describing the triangulation of the 2 sphere that you are using and you are not describing the cycle,z.

Be assured that the first homology is zero because the sphere is simply connected.
and that my friendf is the... answer

 

[kakarotyjn]

Oh,thank you all! Now I'm clear about it,it is truly consisting one equivalent class.