0
$\begingroup$

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 "

  • 0
    Just a simple click on this link: http://books.google.ca/books?id=bhNxPQExK_MC&printsec=frontcover&hl=fr&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false ;page number 42012-09-11

1 Answers 1

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 $0\text{-cycles}/\text{boundaries} \cong G.$

  • 0
    @ Matt E, thanks so much. Using homomorphism above-mentioned is more persuasive for me.2012-09-12