I'm trying to figure out the below sentence (in bold, starting with "since.." from Hatcher formally:
…This is done by defining the reduced homology groups $\tilde{H}_n(X)$ to be the homology groups of the augmented chain complex
$$
\cdots \to C_2(X) \overset{\partial_2}{\to} C_1(X) \overset{\partial_1}{\to} C_0 \overset{\epsilon}{\to} \mathbb{Z} \to 0
$$
[where $\epsilon(\sigma) = 1$ for all singular 0-simplices $\sigma$]…Since $\epsilon\partial_1 = 0$, $\epsilon$ vanishes on $\operatorname{Im}{\partial_1}$ and hence induces a map $H_0(X) \to \mathbb{Z}$ with kernel $\tilde{H}(X)$, so $H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}$.
I know that this sentence, and the isomorphism at then end, appear in quite a few questions (e.g. 1, 2), but either the answers seem to be too intuition-based or the responders seem to employ more advanced tools like splits, which I am not familiar with.
EDIT: I also find it slightly odd that Hatcher provides a section on "Exact sequences" on page 113, yet the aforementioned quote is from p. 110.
My main issue is with understand why the isomorphism holds, without going into splits (I am familiar, however, with the classic isomorphism theorems if that can help). Any advice here would be greatly appreciated!
Also – what is the importance of this discussion? Why should we care about this isomorphism?
Best Answer
Your question is not a precise duplicate of other questions, but I think it is useful to look at
Definition of $\tilde{H}_n(X,A)$
Suppose that $X$ is a topological space and $x_0\in X$. Prove that $\widetilde{H_n}(X)=H_n(X,x_0)$ for all $n\geq 0$.
Clarification about reduced Homology
Hatcher introdces both (ordinary) homology groups and reduced homology groups before he introduces the concept of exactness. Why should this be a problem? One has to define homology groups and induced homomorphisms (in the section "Homotopy Invariance" on p. 110 / 111) before one can study its properties.
I do not think that the classic isomorphism theorems will help you to understand why the exactness of $$0 \to \tilde{H}_0(X) \stackrel{\iota}{\to} H_0(X) \stackrel{\pi}{\to} \mathbb Z \to 0$$ implies $$H_0(X) \approx \tilde{H}_0(X) \oplus \mathbb Z . \tag{1} $$ So let us give a proof. It uses the concept of splitting a short exact sequence, but it avoids to use the word splitting.
Since $\pi$ is surjective, there exists $g \in H_0(X)$ such that $\pi(g) = 1$. Define $s : \mathbb Z \to H_0(X)$ by $s(n) = ng$. This is a group homomorphism such that $\pi \circ s = id$. Next define $$\phi : \tilde{H}_0(X) \oplus \mathbb Z \to H_0(X), \phi(a,n) = \iota(a) + s(n) .$$ This is a group homomorphism.
Let $\phi(a_1,n_1) = \phi(a_2,n_2)$. Noting that $\pi \circ \iota = 0$, we get $$n_i = \pi(s(n_i))) = \pi(\iota(a_i)) + \pi(s(n_i))) = \pi(\iota(a_i)) + \pi(s(n_i)) = \pi(\phi(a_i,n_i)) .$$ Thus $n_1 = n_2$ and therefore $\iota(a_1) = \iota(a_2)$. Since $\iota$ is injective, $a_1 = a_2$.
Let $h \in H_0(X)$. Set $n = \pi(h)$. Then $\pi(h - s(n)) = \pi(h) - \pi(s(n)) = n - n = 0$, thus $h - s(n) \in \ker \pi = \operatorname{im} \iota$, i.e. $h - s(n) = \iota(a)$ for some $a \in \tilde{H}_0(X)$. This shows $\phi(a,n) = h$.
This is philosophical question. Let us first observe that the above isomorphism $\phi$ depends on the choice of $g \in H_0(X)$ such that $\pi(g) = 1$. Thus $\phi$ is no canonical isomorphism. However, $(1)$ says that there is a transparent relationship between $H_0(X)$ and $\tilde{H}_0(X)$; this is not a priori obvious.
Intuitively it means that each $H_0(X)$ contains a trivial summand $\mathbb Z$ which comes from the group $Z = H_0(*) \approx \mathbb Z$ of the one-point space $*$. Note that $Z$ embeds into $H_0(X)$ via $\mu_* : H_0(*) \to H_0(X)$, where $\mu : * \to X$ is any map. Again this map involves the choice of a point $x_0 \in X$ with $\mu(*) = \{x_0\}$, and thus it is not a canonical map. The constant map $c : X \to *$ has the property $c \circ \mu = id$, thus $c_* \circ \mu_* = id$ which shows that $\mu_*$ is an injection.
Concerning the concept of augmentation map which is used in the definition of reduced homology also have a look at
The boundary of a $singular\; 0-simplex$
Definition of zeroth homology
Remark.
The fact that the isomorphism $\phi$ depends on the choice of $g \in \pi^{-1}(1)$ does not imply the non-existence of a canonical isomorphism (or more formally: natural isomorphism) $\tilde{H}_0(X) \oplus \mathbb Z \to H_0(X)$. The non-existence is proved in Isomorphism not natural in $X$?