Although you do not mention it in your question, it it obvious that you are interested in the simplicial homology of $\Delta$-complexes.
As an $n$-simplex Hatcher understands an ordered $n$-simplex which is an $(n+1)$-tuple $[v_0,\ldots,v_n]$ of vertices $v_i$. This means that an $n$-simplex contains more information than the set $\{v_0,\ldots,v_n\}$ of its vertices - in fact, if we take different orderings of the set of vertices, then this yields different $n$-simplices. Do not confuse this with the concept of an oriented simplex which is usually defined as an equivalence class of ordered simplices, two ordered simplices being equivalent if they originate from each other by an even permutation of their vertrices (i.e. we have $[v_0,\ldots,v_n] \sim [v_{\pi(0)},\ldots,v_{\pi(n)}]$ for each even permutation $\pi$).
The boundary homomorphism $\partial_n : \Delta_n(X) \to \Delta_{n-1}(X)$ is defined on the generators $\sigma^n : [v_0,\ldots,v_n] \to X$ by
$$\partial_n(\sigma^n) = \sum_{i=0}^n (-1)^n \sigma^n \mid [v_0,\ldots,\hat{v}_i,\ldots,v_n] .$$
The ordered $(n-1)$-simplices $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$ are the faces of $[v_0,\ldots,v_n]$. More precisely, $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$ is the $i$-th face of $[v_0,\ldots,v_n]$. In the above formula it is essential that we associate the sign $(-1)^i$ to the $i$-th face $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$. Only these signs allow to show that $\partial_{n-1}\partial_n = 0$.
If you work with unordered $n$-simplices, i.e. with the sets $\{v_0,\ldots,v_n\}$, then we obtain of course a set of $n+1$ unordered $(n-1)$-simplices $\{v_0,\ldots,\hat{v}_i,\ldots,v_n\}$ which we may call the faces of $\{v_0,\ldots,v_n\}$, but we do not have any chance to reasonably define the notion of an $i$-th face of the set $\{v_0,\ldots,v_n\}$.
The geometric meaning of orderings of vertices if explained on p. 105.
The short answer is: the maps $\partial$ are defined on the free abelian groups generated by $n$-simplices, hence it makes sense to speak about sums of simplices in these groups.
I'll try to give a down to earth answer, for which we will need to detour first. Consider the set $\mathbb{Z}[X]$ of integer polynomials. An element there is an expression
$$
a_0 + a_1 X + \ldots + a_nX^n
$$
where each $a_i$ is an integer, and $X$ is an indeterminate. Even though we might be quite familiar with thinking of polynomials in this way, the terms 'expression' and 'indeterminate' are informal. A rigorous definition of $\mathbb{Z}[X]$ could be the set of sequences of integers that are eventually zero,
$$
\mathbb{Z}[X] = \{(a_n)_{n \geq 0} \in \mathbb{Z}^{\mathbb{N}_0} : \text{there exists $k \in \mathbb{N}_0$ such that $a_n = 0$ for all $k \geq n$}\}. \tag{1}
$$
Given integers $a_1,\ldots,a_n\in \mathbb{Z}$, we then define
$$
a_0 + a_1 X + \ldots + a_nX^n := (a_0,a_1,\ldots,a_n,0,0,\ldots) \tag{2}.
$$
All the operations on polynomials can be defined in terms of sequences as in $(1)$ and are compatible with the usual notation $(2)$. This is a way to formally capture the idea of having some unknown element $X$ that can be 'scaled' by integers, multiplied by itself, and such that we can make sense of the sums of expressions like this.
In the same spirit, suppose you have a set $X$ and you want to have another set $M$ such that:
- $X$ is, in some sense, contained in $M$,
- we can make sense of multiplying $x \in X$ by an integer $k \in \mathbb{Z}$,
- we can give meaning to an expression like $17x+28y+3z$ for some $x,y,z \in X$.
A way to do this is to define the free abelian group $\mathbb{Z}^{(X)}$ with basis $X$. This concept is closely related to bases of vector spaces. The definition (technically, one definition, there are 'many' equivalent ones) is as follows: as a set $\mathbb{Z}^{(X)}$ consists of functions $f \colon X \to \mathbb{Z}$ such that $f(x) \neq 0$ for finitely many $x \in X$. In other words, the elements of $\mathbb{Z}^{(X)}$ are finitely supported functions $\mathbb{Z} \to X$. We can make sense of sum and multiplication of elements here in the same way we do to define a vector space structure on $\mathbb{R}^X$. Namely:
- the sum of $f,g \in \mathbb{Z}^{(X)}$ is the function $(f+g)(x) := f(x)+g(x)$.
- if $k$ is an integer and $f \in \mathbb{Z}^{(X)}$, we define $(k\cdot f)(x) := kf(x)$.
You can check that these operations define once again elements of $\mathbb{Z}^{(X)}$.
Now, let's go back to the analogy with polynomials; in particular, to the relation between definition $(1)$ and notation $(2)$. Even though formally the free abelian group is defined as finitely supported functions, we want to think of it as "$\mathbb{Z}$-linear combinations of $X$". To do this, we write
$$
x := \chi_{\{x\}}, \quad \chi_{\{x\}}(y) = \begin{cases}1 &\text{if $x=y$}\\
0 &\text{otherwise}\end{cases}
$$
You can check that, following this notation, every element of $\mathbb{Z}^{(X)}$ can be written as a finite sum
$$
a_1 x_1 + \ldots +a_n x_n
$$
for some $x_i \in X$ and integers $a_i$. Moreover, two such expressions are equal if the elements of $X$ appearing are the same, and the coefficients accompanying each element coincide.
In this case, we consider $\mathbb{Z}^{(C_n)}$ with $C_n$ the set of $n$-simplices. In $\mathbb{Z}^{(C_n)}$ it makes perfect sense to speak of $3 \sigma + 22 \tau$ for some pair of simplices $\sigma,\tau$ or any expression of this sort; likewise the maps $\partial_k$ are well defined.
I hope this gives a little bit more context to the first sentence of this answer.
Best Answer
Hatcher is a bit misleading in the chapter 2.1. What he considers are not "raw" simplices. In fact the circle $S^1$ contains no "raw" simplices of dimension above $0$ (e.g. no straight line is a subset of a circle). Instead he considers a $\Delta$-complex structure on $X$ (which is defined at the begining of the chapter), that is a collection of maps $\Delta^n\to X$ which satisfies certain properties. This can be seen when he says:
He means one $0$-dimensional simplex $\Delta^0\to S^1$ and one $1$-dimensional simplex $\Delta^1\to S^1$ (the latter is a glueing of its endpoints). Both together form a $\Delta$-complex on $S^1$. Of course this depends on the choice of those maps, so these are fixed implicitly. Also Hatcher will use the word "simplex" or "vertex" or "edge" for such maps, and sometimes even for images of those maps as well. Yeah, it can be confusing.
He then builds $\Delta_n(X)$ chain complexes. Those complexes can be built on top of the $\Delta$-structure. Hatcher describes how it is done in the chapter as well.
No. $\Delta_0(S^1)$ is a free group generated by all $0$-simplexes. But not all $0$-simplexes taken from thin air. All $0$-simplexes taken from a fixed $\Delta$-complex structure. In our case this is the single $\Delta^0\to S^1$ map. Hence it is $\mathbb{Z}$. Analogously $\Delta_1(S^1)$ is a free group generated by all $1$-simplexes. In our case this is the single $\Delta^1\to S^1$ map. Thus again $\mathbb{Z}$.