Why is $H_0(X)\cong \tilde{H}_0(X)\oplus \mathbb{Z}$?

Question

Here is how hatcher proves this. Let $(C_•(X),\partial_•)$ be the reduced singular chain complex where $X$ is a nonempty topological space. Since $\partial_0:C_0(X)\rightarrow \mathbb{Z}$ is an epimorphism, we have $C_0(X)/\ker(\partial_0)\cong \mathbb{Z}$. Since $im(\partial_1)\leq \ker(\partial_0)\leq C_0(X)$, by the third isomorphism theorem, we have $H_0(X)/\tilde{H}_0(X)\cong C_0(X)/\ker(\partial_0)\cong \mathbb{Z}$. Hatcher wrote that this implies that $H_0(X)\cong \tilde{H}_0(X)\oplus\mathbb{Z}$, but I do not understand why. How do I show this? Thank you in advance.

Answer 1

We have the exact sequence

$$0 \to \widetilde{H_0}(X) \to H_0(X) \stackrel{r_*}{\to} H_0(P)\to 0,$$

where $P$ is a one-point space and $r:X \to P$ is the unique map from $X$ to $P$.

Any map $i: P \to X$ induces a splitting, since $Id_*=(r \circ i)_*=r_*\circ i_*$, from which follows that $$H_0(X) \cong \widetilde{H_0}(X) \oplus H_0(P) .$$ Note that the argument holds for any homology theory satisfying Eilenberg-Steenrod axioms ($H_0(P)$ does not need to be free, for example) , since the splitting is given by functoriality of homology.

Answer 2

There is a map $\mathbb{Z}\rightarrow H_0(X)$ inverse to $\partial$. Namely take sny point (or vertex) $v$ in X and define $n\mapsto n[v]$. The image of this map is isomorphic to $\mathbb{Z}$ and with the reduced group forms a direct sum.