This should be easy but I am unfortunately struggling
The setting
Given a topological space $X$, let $\epsilon_i: \Delta^{n-1} \rightarrow \Delta^n$ be the $i$-th face map, where $\Delta^n$ is the standard $n$-simplex.
Let $\sigma_n: \Delta^n \rightarrow X$ be a singular $n$-simplex in $X$. We call $S_n(X)$ the free abelian group generated by the singular $n$-simplices in $X$, with the convention that $S_{-1}(X)=0$.
The boundary of a $\sigma_k$ is defined as $$\delta_n(\sigma_k) = \sum_{i = 0}^n (-1)^i \sigma_k \circ \epsilon_i $$
($\delta_n(\sigma_k) : S_n(X) \rightarrow S_{n-1}(X)$) and the $n$-singular homology group is given by $\ker( \delta_n) / \text{img}(\delta_{n+1})$.
The question
I want to know why do singular $0$-simplices have zero homology except on the $0$-singular homology group.
My attempt
The $0$-singular homology group is $\ker( \delta_0) / \text{img}(\delta_{1})$, the $\ker( \delta_0)$ is all the singular $0$-simplices while $$\delta_1(\sigma_0) = \sigma_0 \circ\epsilon_0 - \sigma_0 \circ\epsilon_1 = ?$$ Ideally I would like $\sigma_0 \circ\epsilon_0 - \sigma_0 \circ\epsilon_1$ to cancel since then the homology group is not only zero but it would be formed by all the singular $0$-simplices ( I am assuming that "having zero homology" means that the homology group is just the zero element, correct?). But why would it cancel?
It seems to me that $\sigma_0 \circ\epsilon_1$ is not even well defined.
Moreover I would like to show than for the other homology groups the singular $0$-simplices have zero homology but since I am having so many problems understanding the zeroth case I will stop here.