Proof of $\tilde{H}_*(X) \cong H_*(X, x_0) $.

Solution 1:

Recall that $H_0(X)$ is the free abelian group generated by the path-connected components of $X$. That is, if $\{X_i\}_{i\in I}$ are the path-connected components of $X$, then $H_0(X) \cong \bigoplus_{i\in I}\mathbb{Z}$. The map $\{x_0\} \to X$ induces the map $H_0(\{x_0\}) \to H_0(X)$ which sends the generator of $H_0(\{x_0\}) \cong \mathbb{Z}$ to the element of $\bigoplus_{i\in I}\mathbb{Z}$ which has a $1$ in the index corresponding the path-connected component of $X$ containing $x_0$, and $0$ in every other index. This is the map $i$ and it is injective. In general, if $A \subset X$ and every path-connected component of $A$ is contained in a path-connected component of $X$, then the induced map $H_0(A) \to H_0(X)$ is injective. The only way the map can fail to be injective is if two path-connected components of $A$ are contained in the same path-connected component of $X$.

Solution 2:

You can make explicit the connecting morhpism $\delta$ following the steps in the proof of the long exact sequence lemma. Take an element that is a border $[\alpha]\in \frac{C_1(X)}{C_1(x_0)}$, this comes from $\alpha\in C_1(X)$ whose is such that $\partial(\alpha)\in C_0(x_0)$. Then $\partial(\alpha)=x_0-x_0=0\in C_0(X)$. Because the map $C_0(x_0)\to C_0(X)$ is injective this comes from $0$ in $C_0(x_0)$. This implies $\delta$ is the zero map.

Note that in $\partial(\alpha)=x_0-x_0=0\in C_0(X)$ I treated $\alpha$ like a generator and not a linear combination, but in any case $\partial(\alpha)=0$ holds.