Homology relative to a point I want to prove $$ \widetilde{ H_n}(X)\cong H_n(X, \ast)$$
The long exact sequence for pairs gives me $H_n(X) = \widetilde{H_n}(X) \cong H_n(X, \ast)$ for $n>0$
and $\widetilde{H_0}(X) \cong \widetilde{H_0}(X, \ast) $.
So I would like to show $\widetilde{H_0} (X, \ast) \cong H_0(X, \ast) $ to finish my proof.
Can anyone give me a hint on how to do this? Many thanks for your help.
 A: To elaborate on my third comment: The splitting defines an isomorphism $C(X) \cong C(X,*) \oplus C(*)$ with projections $\epsilon$ and $j$ and inclusion $i$. Using this identification we arrive at the following situation, where all the maps are the evident projections and inclusions:
$$\begin{array}{rcccl}
\vdots & & \vdots & & \vdots \\
\tilde C_1(X) & \stackrel{p_{1}}{\cong} & C_1(X,*) & \stackrel{q_{1}}{\cong} & \tilde C_1(X) \\
\downarrow & & \downarrow & & \downarrow \\
\tilde C_0(X) \cong C_0(X,*) \oplus \mathbb Z & \stackrel{p_{0}}{\to} & C_0(X,*) & \stackrel{q_{0}}{\to}  & C_0(X,*) \oplus \mathbb Z \cong \tilde C_0(X)\\
\downarrow & & \downarrow & & \downarrow \\
\mathbb Z & \stackrel{p_{-1}}{\to} & 0 & \stackrel{q_{-1}}{\to} & \mathbb Z

\end{array}$$
It should now be easy to see that $h: \tilde C(X) \to \tilde C(X)^{+1}$, given by the inclusion $h_{-1}: \mathbb Z \to C_0(X,*) \oplus \mathbb Z$ and $h_k = 0$ otherwise, defines a chain homotopy of the horizontal map $q_* \circ p_*$ in the diagram above and the identity map on $\tilde C(X)$. On the other hand we have $p_* \circ q_* = id_{C(X,*)}$, which proves $C(X,*)$ and $\tilde C(X)$ to be chain homotopy equivalent.
