Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

while I am reading a proof in "Elements of homology theory" - V.V Prasolov, I felt there is something missing in the proof. precisely when he says " therefore a=0 ".

the proof that I am talking about is the following: " If K is a connected simplicial complex, then H_0(K,G)=G "

share|improve this question
1  
Can you post the proof? –  Billy Sep 11 '12 at 1:08
1  
Offhand without seeing the proof, you could prove this theorem with three steps: 1) Every connected simplicial complex is path connected. 2) Since it is path connected, the difference of any two points (as singular 0-chains) is a boundary. 3) Since every generator of $C_0(X,G) = X^G$ are equivalent up to boundary, $H_0(X,G)$ must just be $G$. Obviously you'd need to fill in details, but I think broken up like this you could prove it yourself. Maybe 1 would be troublesome, since you have to consider the infinite complex case. –  John Stalfos Sep 11 '12 at 1:19
    
for a finite case, it's easy, by the way, the author had given an example before he stated this theorem. –  Mopzer Moreena Sep 11 '12 at 1:31
    
@JohnStalfos, Did you mean that final step made by the author is superfluous? I don't think so John. –  Mopzer Moreena Sep 11 '12 at 1:43
    
you have to prove that there exists at least a nonboundary 0-chain, this is the aim of the final step, otherwise all 0-chains are boundaries. whence H_0(K,G) must be trivial, Is not? –  Mopzer Moreena Sep 11 '12 at 1:46

1 Answer 1

I just looked at the argument, and it can be rephrased as follows:

There is a homomorphism $$0\text{-cycles} \to G$$ given by taking the sum of the coefficients.

If we apply this map to a boundary, we get $0$ (pretty obviously). If we apply it to $a[m]$, we get $a$. Applying this homomorphism to either side of the equation $a[m] = \text{ a boundary},$ we get $a = 0$, as claimed.


To make this answer more self-contained for others who might read it:

The degree map shows that $$0\text{-cycles}/\text{boundaries}$$ admits a surjection onto $G$. On the other hand, anything in the kernel is of the form $\sum_{i} a_i[m_i]$ for some elements $a_i \in G$ and points $m_i$, with $\sum_i a_i = 0.$ Thus, if we fix a point $m$, we find that $\sum_i a_i[m_i] = \sum_i a_i([m_i] - [m]) = \sum_i a_i \partial[m_i,m],$ and thus every point in the kernel of the degree map is a boundary. (Here $[m_i,m]$ is some path joining $m_i$ to $m$.) So in fact $0\text{-cycles}/\text{boundaries} \cong G.$

share|improve this answer
    
@ Matt E, thanks so much. Using homomorphism above-mentioned is more persuasive for me. –  Mopzer Moreena Sep 12 '12 at 12:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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