If $X$ is a path-connected space, then $H_0(X) \cong \Bbb Z$.
I'm going over the proof of this and it relies on defining a map $\varepsilon : C_0(X) \to \Bbb Z$ such that $\sum_{i}n_i \sigma_i \mapsto \sum_{i}n_i$ in order to show that $\ker \varepsilon = \operatorname{Im}(\partial_1)$. This can be further used since $$H_0(X) \cong \frac{\ker(\partial_0)}{\operatorname{Im}(\partial_1)} = \frac{C_0(X)}{\operatorname{Im}(\partial_1)} = \frac{C_0(X)}{\ker(\varepsilon)} \cong \Bbb Z$$ since $\varepsilon$ is surjective.
I'm missing the intuition for why do we need to define this kind of map $\varepsilon$. What we have is that $$H_0(X) \cong \frac{\ker(\partial_0)}{\operatorname{Im}(\partial_1)}$$ and the denominator is the group of $1$-cycles i.e. "paths" in $X$, but I don't see a direct way of noticing that such an map $\varepsilon$ is reasonable to define. What might be the intuition here?
Best Answer
The map $\varepsilon$ is known as the augmentation map which is used to define the reduced homology group $\tilde H_0(X)$. See for example The boundary of a $singular\; 0-simplex$.
However, this does not explain why $\varepsilon$ is used to show that $H_0(X) \approx \mathbb Z$. As you say we have $$H_0(X) = \ker \partial_0 / \operatorname{im} \partial_1 = C_0(X) / \operatorname{im} \partial_1 . $$ So what is the intuition to identify $\operatorname{im} \partial_1$ with $\ker \varepsilon$?
$C_1(X)$ is generated by all singular $1$-simplices $\sigma^1 : \Delta^1 \to X$. Since we may identify $\Delta^1$ with $I = [0,1]$, $C_1(X)$ is generated by all paths $u : I \to X$.
$C_0(X)$ is generated by all singular $0$-simplices $\sigma^0 : \Delta^0 \to X$. We may identify singular $0$-simplices with points of $X$, i.e. we regard $C_0(X)$ as generated by the points of $X$.
Hence $\operatorname{im} \partial_1$ is generated by all difference chains $u(1) - u(0) \in C_0(X)$. If $X$ is path-connected, then the set of all $u(1) - u(0)$ agrees with the set of all difference chains $x - y \in C_0(X)$ with $x,y \in X$. Call them simple difference chains. Then we know that $\operatorname{im} \partial_1$ is generated by all simple difference chains. Trivially the sum of coefficients of any simple difference chain is $1 + (-1) = 0$. Therefore the same is true for all linear combinations of simple difference chains, that is, for all elements of $\operatorname{im} \partial_1$. This explains why $\varepsilon$ (which assigns to any singular $0$-chain the sum of all its coefficients) occurs in this context.
So far we know that $\operatorname{im} \partial_1 \subset \ker \varepsilon$.
Now consider $z = \sum_{i=1}^k m_iz_i \in \ker \varepsilon$. We can write $m_iz_i = \sum_{j=1}^{\lvert m_i \rvert} a_{ij} z_i$, where all $a_{ij} = +1$ or all $a_{ij} = -1$. Thus $$z = \sum_{j=1}^r b_j w_j$$ where all $w_j \in \{z_1,\ldots,z_k\}$ and all $b_j = \pm 1$. Since $\varepsilon(z) = \sum_{j=1}^r b_j = 0$ we see that the number of indices $j$ with $b_j = +1$ must be identical with the number of indices $j$ with $b_j = -1$. In particular $r = 2s$ and w.lo.g. we may assume that $b_j = +1$ for $j \le s$ and $b_j = -1$ for $j > s$. Thus $$z = \sum_{j=1}^s (w_j - w_{j+s}) \in \operatorname{im} \partial_1 .$$