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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.

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  • $\begingroup$ Does this answer your question? math.stackexchange.com/questions/55399/reduced-homology?rq=1 $\endgroup$
    – User
    Sep 18 '20 at 18:29
  • $\begingroup$ There is always some argumentation when the issue of reduced homology comes up. While Hatcher is a good book, I recommend you not take his definition of reduced homology seriously. When you are a little more familiar with homology I recommend you look for an explanation about the point of reduced homology, the main being that it is basically only useful when you have a basepoint. $\endgroup$ Sep 18 '20 at 18:37
  • $\begingroup$ @0-thUser I saw that post, but it seemed like all the responses where just explaining the quotient relation $H_0/\tilde{H}_0 \cong \mathbb{Z}$ and not the direct sum. $\endgroup$
    – Nick A.
    Sep 19 '20 at 14:52
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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 $nb$.

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