The boundary of a $singular\; 0-simplex$

Question

In Singular homology,

Given a $singular\; n-simplex$, $\varphi$, we define the $singular\; (n-1)-simplex$, $\partial_i{\varphi}$,

$$\partial_i{\varphi}(x_0,x_1,\dots,x_{n-1})=\varphi(x_0,x_1,\dots,x_{i-1},0,x_i,\dots,x_{n-1}), i=0,1,\dots n$$

and later on define the boundary operator:

$$\partial:=\partial_0-\partial_1+\cdots+(-1)^n\partial_n$$ Notice that $\partial:S_n(X)\to S_{n-1}(X)$, so, my question is how to compute or calculate or give the sense to "$\partial\varphi_0$" where $\varphi_0$ is a $singular\;0-simplex$, and then give the sense also to a $singular\; 0-chain$ (Lineal combination of $0-chains$). Because, when they compute the $0-th\; group \;of \;homology$ they say that $\partial \varphi_0=0$ for $\varphi_0$ a $singular\; 0-simplex$.

I'm take this from "Czes Kosniowski - A First Course in Algebraic Topology (1980, Cambridge University Press)"

Answer 1

Your question comes very close to Definition of zeroth homology. You should read my answer concerning the concept of a chain complex, especially the singular chain complex $S_*(X)$.

You can take it as a convention to define $S_k(X) = 0$ for $k < 0$, and this show that $\partial : S_0(X) \to S_{-1}(X) = 0$ is automatically the trivial homomorphism.

The above convention can be justified by the argument that there are no singular simplices of negative dimension, thus the free abelian groups $S_k(X)$ with $k < 0$ have the empty set as basis, i.e. are $0$.

However, you can also find an argument against the convention $S_{-1}(X) = 0$. In fact, singular $k$-simplices are continuous maps $\Delta^k \to X$ living on the standard (geometric) $k$-simplex. The simplex $\Delta^k$ has $k+1$ vertices and we can argue that $\Delta^{-1}$ is a "simplex with $0$ vertices" which is nothing else than the empty set. Therefore there exists exactly one singular $(-1)$-simplex (which is the unique function $\emptyset \to X$), thus $S_{-1}(X)$ is the free abelian group with the one generator $\emptyset \to X$. Thus $S_{-1}(X) \approx \mathbb Z$. With this interpretation we see that the boundary $\partial \varphi$ of any $0$-simplex $\varphi$ is nothing else than $\emptyset \to X$. Writing $S_{-1}(X) = \mathbb Z$ we therefore get $$\partial : S_0(X) \to \mathbb Z, \partial (\sum_i n_i\varphi_i) = \sum n_i .$$ This map is denoted as the augmentation map. The resulting chain complex is the reduced singular chain complex. Its homology groups are known as the reduced singular homology groups.