I am trying to prove that $H_n(X, *) \cong \widetilde{H_n}(X)$ for all $n$. Hatcher uses the long exact sequence of reduced homology groups to prove this for all $n$ (example 2.18), but my professor has not mentioned this sequence. I am looking for guidance on the case $n=0$, as I am quite stuck. I would also appreciate feedback on my attempt for $n \geq 1$:
Let $n \geq 1$. Consider the long exact sequence of relative homology groups. Since $H_n(*) \cong 0$ and $\widetilde{H_n}(X) \cong H_n(X)$, we rewrite the sequence as $$ \cdots \xrightarrow{j_*} H_{n+1}(X, *) \xrightarrow{\partial_*} 0 \xrightarrow{i_*} \widetilde{H_n}(X) \xrightarrow{j_*} H_n(X, *) \xrightarrow{\partial_*} 0 \xrightarrow{i_*}\cdots .$$ By exactness, $\text{Ker } j_* = \text{Im } i_* = 0$ and $\text{Im }j_* = \text{Ker }\partial_* = H_n(X, *)$. Thus $j_* : \widetilde{H_n}(X) \to H_n(X, *)$ is an isomorphism.
This seems almost correct, but I think there is an issue in the case $n = 1$, since $H_0(*) = \mathbb{Z} \neq 0$, and thus $\text{Ker }\partial_* \neq H_1(X, *)$. How I can I fix this?
Thank you!