0

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 "

Gerry Myerson
  • 179,216
Mopzer Moreena
  • 21
  • 1
    Can you post the proof? – Billy Sep 11 '12 at 01:08
  • 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 01:19
  • 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 01:31
  • @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 01:43
  • 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 01:46
  • Not all 0-chains are boundaries. Things like $x-y$ will be boundaries, for points $x,y$(in the former, we're actually looking at the zero chains not the points themselves) – John Stalfos Sep 11 '12 at 01:51
  • @JohnStalfos, Have you seen Prasolov's proof?. – Mopzer Moreena Sep 11 '12 at 01:53
  • No, I haven't. If you wanted informed discussion about the exact proof you're looking at I can't provide it unless you either write up the skeleton of it, or scan the page. – John Stalfos Sep 11 '12 at 01:54
  • 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 4 – Mopzer Moreena Sep 11 '12 at 01:56

1 Answers1

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

Matt E
  • 123,735