Homology relative to a point is reduced homology 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!
 A: The definition of the reduced homology group is as follows. Any space $X$ admits a unique map $X\rightarrow\ast$ to the one-point space and we set
$$\widetilde H_n(X)=\ker\left(H_n(X)\rightarrow H_n(\ast)\right).$$
Assuming that $X$ is nonempty, any choice of point $x\in X$ defines a map $x:\ast\rightarrow X$ which splits the surjection $X\rightarrow\ast$. Then by the functorality of homology, the induced map $x_*:H_n(\ast)\rightarrow H_n(X)$ is injective. 
Thus when we consider the long exact sequence 
$$\dots\rightarrow H_n(\ast)\xrightarrow{x_*} H_n(X)\rightarrow H_n(X,\ast)\rightarrow H_{n-1}(\ast)$$
we see by exactness that it splits in each degree to give
$$H_n(X)\cong H_n(X,\ast)\oplus H_n(\ast).$$
But under this isomorphism the map $H_n(X)\rightarrow H_n(\ast)$ becomes the projection onto the second factor. Hence
$$\ker\left(H_n(X)\rightarrow H_n(\ast)\right)\cong \ker\left(H_n(X,\ast)\oplus H_n(\ast)\xrightarrow{pr_2} H_n(\ast)\right)$$
and thus
$$\widetilde H_n(X)\cong H_n(X,\ast).$$
