$H_0(X)\simeq\Bbb{Z}^k$, where $k$ is the number of path components 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?
 A: 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.
