We wish to show that for a path connected space $X$, the zeroth singular homology group is just $\mathbb Z$. The intuition is that the zeroth homology group is given by consider $C[1] \xrightarrow{\partial_1} C[0]$, $C[0] \xrightarrow{\partial_0} 0$, and then taking $H[0] \equiv ker(\partial_0) / im(\partial_1) = C[0] / im(\partial_1)$. Recall that $C[0]$ is the abelian group generated by the direct sum of generators $\{ \Delta^0 \rightarrow X \}$, where $\Delta^0$ is the $0$-simplex, that is, a single point. So $C[0]$ is an abelian group generated by all points in $X$. Now, $C[1]$ contains all paths between all points $p, q \in X$. Thus the boundary of $C[1]$ will be of the form $q - p$. Quotienting by $C[1]$ identifies all points with each other in $C[0]$. That is, we get $H[0] \equiv \langle x \in X | y = z \forall y, z \in X \rangle$, which is isomorphic to $\mathbb Z$. Thus, the zeroth singular homology group is $\mathbb Z$.