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I've written the following "proof" that if $X$ is path-connected then $H_0 (X) = 0$. I know that that's not the case, yet I can't find the mistake in my "proof". Can you please point it out to me? Here is my "proof":

$X$ path-connected $\implies $ any 2 singular 0-simplexes (constant maps) $x,y$ in $X$ are the boundary of a 1-simplex (path) $\sigma$

$\iff \partial \sigma = x - y$

$\implies x,y$ differ by a boundary

$\implies x,y$ are in the same homology class $\{ c \} \in H_0(X)$

$\implies H_0 (X) = \{ c \} = 0$.

Thanks for your help, it's much appreciated.

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Why is the class of $c$ zero? – Bonanza Aug 10 '11 at 11:54
Because it's a group and it only has one element, so that element has got to be neutral element otherwise it's not a group. – Rudy the Reindeer Aug 10 '11 at 11:57
Remember, the chain group consists of formal sums of 0-simplices, not just of 0-simplices themselves. – Chris Eagle Aug 10 '11 at 11:58
Oh!! $\{ c\}$, $\{c + c\}$, $\dots$ are all different elements of $H_0 (X)$. That's what you're saying? – Rudy the Reindeer Aug 10 '11 at 12:07
That's it, Matt :-) – Eivind Dahl Aug 10 '11 at 12:52
up vote 4 down vote accepted

As you point out, in a path-connected space, all 0-simplices (i.e. points) are homologous. But the 0th chain group consists not of 0-simplices, but of formal sums of 0-simplices. So, assuming our space is also nonempty, and so contains a point $p$, the 0th chain group contains chains $0$, $p$, $-p$, $p+p$, $-p-p$, and so on. Since every point is homologous to $p$, every chain is homologous to one of these chains. Moreover, since every boundary of a 1-chain is a sum of points with total weight 0, there are no non-trivial homology relations among these chains, and so the 0th homology group is $\mathbb{Z}$.

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