Definition of singular n-simplex Hatcher's book says 'A singular n-simplex in a space is by definition just a map $\sigma:\Delta^n\rightarrow X$'. And $C_n(X)$ be the free abelian group with basis the set of singular n-simplices in $X$.
So does he mean $C_n(X)$ is the... of all singular n-simplices in $X$? 
And does he identify those 'same maps with different domain'? Because I see proposition $2.8$:

If $X$ is a single point, then $H_n(X)=0$ for $n>0$ and $H_0(X)=Z$.

The proof of this says: 

In this case there is a unique singular n-simplex $\sigma_n$ for each n, and $\partial(\sigma_n)=\sum(-1)^i\sigma_{n-1}$, which is $0$ for n odd and $\sigma_{n-1}$ for n even...

But by definition of the boundary map: $\partial_n(\sigma)= \sum(-1)^i \sigma|[v_0,v_1...\widehat{v_i},...v_n]$. But different i deleted gives different domain, although they all map to a single point. This is what I mean 'same maps with different domain'.
Thanks for help!

$C_n(X)$ is the set of all formal sums of $n-$simplices in $X$, an element of $C_n(X)$ is given by a finite sum $\sum n_i \phi_i$ for some $\phi_i : \Delta^n \rightarrow X$ and $n_i \in \mathbb Z$. This is just a formal sum. And yes he means that $C_n(X)$ is the free abelian group generated by the set of $\textbf{all}$ n-simplices. 
And that definition of the boundary map is really, really misleading, I could 100% see how you would think that deleting different $i$ give different domains. And according to that definition that is actually true. A better definition is $\partial \sigma = \sum (-1)^i \sigma \circ d_i$ where $\sigma : \Delta^n \rightarrow X$ and
$$d_i: \Delta^{n-1} \rightarrow \Delta^n : (x_0,...,x_{n-1}) \mapsto (x_0,...,x_{i-1}, 0,x_{i},...,x_{n-1})$$
then the resulting maps have the same domain and the resulting formal sum is in $C_{n-1}(X)$. 
The map $d_i$ is the same as the map that @bart described in his comment by the way.
