I'm having some trouble with Hatcher's introduction of reduced homology on p. 110 of his Algebraic Topology:
…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 understand everything except the last claim that $H_0$ is a direct sum. All I see from the rest of the discussion is that we have an exact sequence $0 \to \tilde{H_0} \to H_0 \to \mathbb{Z} \to 0$, but I can't figure out why this sequence splits.
Best Answer
Actually, since $\mathbb{Z}$ is a projective abelian group, any exact sequence of the form $$ 0 \to A \to B \to \mathbb{Z} \to 0 $$ splits, although not canonically. If you haven't seen this argument before, just note that for any $b \in B$ mapping to $1$, we can define a map $\mathbb{Z} \to B$ splitting the sequence by sending $n$ to $nb$.