I want to prove that $H_0(X)\simeq\Bbb{Z}^k$, where $k$ is the number of path components of $X$.

What I tried Since $∂_0=0$, $$H_0(X)=\ker{∂_0}/\operatorname{Im}∂_1=C_0(X)/\operatorname{Im}∂_1=\left\{\sum_{f:\Delta_0\to X}{n_f f: n_f\in\Bbb{Z}}\right\}/\operatorname{Im}∂_1.$$ $\operatorname{Im}∂_1$ consists of elements like $g\restriction_{v_1}-g\restriction_{v_0}$, where $g:\Delta_1\to X$. Since $g$'s are continuous and $\Delta_1$ is path-connected, $g(\Delta_1)$ always lies in a path-component of $X$. By taking the quotient $C_0(X)/\operatorname{Im}∂_1$ we actually identify $f$'s whose images are in the same component. So we end up with $k$ many generators, which leads to $H_0(X)\simeq\Bbb{Z}^k$.

Questions How can I formalize my proof? Is there a way to write $\operatorname{Im}∂_1$ as a direct sum of $\Bbb{Z}$'s?


I think you've understood the idea of the result, which is an important issue.

One way to formalize it could be the following:

  • Trying to break down a problem into smaller pieces, first prove that, in general, if you have $X = \bigsqcup_i X_i $, where the $X_i$ are the path components of $X$, then

$$ H_*(X) = \bigoplus_i H_*(X_i) \ . $$

  • Then, prove that, if $X$ is path-connected,

$$ H_0(X) \cong \mathbb{Z} \ . $$

As for the first part, you've already said almost everything: if $\sigma : \Delta^p \longrightarrow X$ is a $p$-simplex, necessarily $\sigma (\Delta^p) \subset X_i$, for some path-component $X_i$. Also its boundary $\partial (\sigma) \in B_{p-1}(X_i)$. All in all, it's clear that you have

$$ C_*(X) = \bigoplus_i C_*(X_i) \qquad \text{and} \qquad H_*(X) \cong \bigoplus_i H_*(X_i) \ . $$

In particular,

$$ H_0(X) \cong \bigoplus_i H_0(X_i) \ . $$

As for the second part, a classical way to proceed is through the augmentation map:

$$ \varepsilon : C_0(X) \longrightarrow \mathbb{Z} \ , $$

which is defined as follows: you can identify each $0$-simplex as a formal linear combination of points of $X$ with integer coefficients

$$ \sigma = \sum_{x\in X} m_x x \ . $$

Then, you define

$$ \varepsilon \left(\sum_{x\in X} m_x x \right) := \sum_{x\in X} m_x \ . $$

And you verify just two things:

  • $\varepsilon$ is an epimorphism.
  • $\ker\varepsilon = B_0(X)$

From which you've got your result.

EDIT. Because of the first isomorphism theorem: if we've got a morphism of groups,

$$ \varepsilon : C_0(X) \longrightarrow \mathbb{Z} \ , $$

then, in general, you have an isomorphism

$$ C_0(X) /\!\ker\varepsilon \longrightarrow \mathrm{im}\ \varepsilon \ . $$

Since $\varepsilon$ is an epimorphism, $\mathrm{im}\ \varepsilon = \mathbb{Z}$ and we are done.

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  • $\begingroup$ The first isomorphism also holds without the assumption that $X_i$ is path-connected. $\endgroup$ – Martin Brandenburg Sep 6 '13 at 12:51
  • $\begingroup$ I don't really see how does $\ker{\epsilon}=B_0(X)$ solve our problem $\endgroup$ – Xena Sep 6 '13 at 13:44
  • $\begingroup$ I've edited my answer to include this last issue. $\endgroup$ – Agustí Roig Sep 6 '13 at 13:57
  • $\begingroup$ Aha! Thank you very much Agustí $\endgroup$ – Xena Sep 6 '13 at 14:00
  • $\begingroup$ You're welcome. :-) $\endgroup$ – Agustí Roig Sep 6 '13 at 14:41

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