Suppose that $X$ is a topological space and $x_0\in X$. Prove that $\widetilde{H_n}(X)=H_n(X,x_0)$ for all $n\geq 0$. Suppose that $X$ is a topological space and $x_0\in X$. Prove that $\widetilde{H_n}(X)=H_n(X,x_0)$ for all $n\geq 0$.
We have a short exact succession
$$0\to S_*(x_0)\to S_*(X)\to S_*(X,x_0)\to 0$$
where $i_*:S_*(x_0)\to S_*(X)$ is the mapping induced by the inclusion and $p_*:S_*(X)\to S_*(X,x_0)$ is the mapping induced by the projection.
So by the Zig-Zag theorem we have a long exact sequence in homology
$$...\to H_n(x_0)\to H_n(X)\to H_n(X,x_0)\to H_{n-1}(x_0)\to...$$
From here it follows that if $n-1>0$ then $H_{n-1}(x_0)=H_n(x_0)=0$ and so $H_n(X,x_0)\cong H_n(X)=\widetilde{H_n}(X)$ if $n>1$.
If $n=1$ then I have
$$...\to H_1(x_0)\to H_1(X)\to H_1(X,x_0)\to H_{0}(x_0)\to H_0(X)...$$
where $H_0(x_0)\cong \mathbb{Z}$ and $H_1(x_0)=0$. 
But note that $H_{0}(x_0)\to H_0(X)$ is injective, so we can deduce that there is an exact sequence $0\to H_1(X)\to H_1(X,x_0)\to 0$ and so $H_1(X,x_0)\cong H_1(X)\cong \widetilde{H_1}(X)$but I do not know what to do in the case of $n=0$. Could someone help me please, thank you very much.
 A: We first have to understand reduced homology. I do not know how you define reduced homology groups $\tilde{H}_n(X)$, but in the context of singular homology they are usually defined as the homology groups of the augmented chain complex $\tilde{C}_*(X)$. See Hatcher p.110. It is immediate from the definition that $\tilde{H}_n(X) = H_n(X)$ for $n > 0$. Note that there is a pecularity: We have $\tilde{H}_{-1}(\emptyset) = \mathbb Z$ (and $\tilde{H}_n(\emptyset) = 0$ for all other $n$). For this reason one usually excludes $\emptyset$ when working with reduced homology. However, for any one-point space $P$ we have $\tilde{C}_n(P) = \mathbb Z$ for all $n \ge -1$; for the boundary maps see the proof of Hatcher Proposition 2.8 and note that $C_0(P) \to C_{-1}(P) = \mathbb Z$ is the identity. This shows that $\tilde{H}_n(P) = 0$ also for $n = 0$.
As you know, there is long exact sequence of unreduced homology groups for each pair $(X,A)$. This comes from the definition of $H_n(X,A)$ as the $n$-th homology group of the chain complex $C_*(X,A)$ which is defined by levelwise quotients $C_n(X,A) = C_n(X)/C_n(A)$. This construction yields a short exact exact sequence of chain compexes and chain maps
$$0 \to C_*(A) \to C_*(X) \to C_*(X,A) \to 0$$
We can do the same for augmented chain complexes (see Hatcher p.118). We define $\tilde{C}_*(X,A)$ of a pair $(X,A)$ again by levelwise quotients, i.e. $\tilde{C}_n(X,A) = C_n(X,A) = C_n(X)/C_n(A)$ for $n \ge 0$ and $\tilde{C}_{-1}(X,A) = \mathbb Z / \mathbb Z = 0$. Thus $\tilde{C}_*(X,A) = C_*(X,A)$. Therefore the reduced homology group $\tilde{H}_n(X,A)$ agrees with $H_n(X,A)$ for all $n$. Note, however, that $\tilde{C}_*(X,\emptyset)$ is not the same as $\tilde{C}_*(X)$. In fact, we have $\tilde{C}_*(X,\emptyset) = C_*(X)$. You see that $\emptyset$ again plays a peculiar role and usually only pairs $(X,A)$ with $A \ne \emptyset$ are considered when working with reduced homology.
Anyway, we get a short exact sequence of chain complexes
$$0 \to \tilde{C}_*(A) \to \tilde{C}_*(X) \to \tilde{C}_*(X,A) \to 0$$
which produces a long exact sequence of reduced homology groups. For pairs $(X,A)$ with $A \ne \emptyset$ it agrees with the long exact sequence of unreduced homology groups except at the end where it looks like
$$(*) \quad \quad H_1(X,A) \to \tilde{H}_0(A) \to \tilde{H}_0(X) \to H_0(X,A) \to 0$$
Now let us consider this sequence for the pair $(X, \{x_0\})$. Since $\tilde{H}_0(\{x_0\}) = 0$, it reduces to
$$0 \to \tilde{H}_0(X) \to H_0(X,\{x_0\}) \to 0$$
which completes the proof.
Remark 1.
Without knowledge of $\tilde{H}_0(P)$ it is impossible to prove the claim based only on the long exact sequences. Note that we have $(*)$ for reduced homology and
$$(**) \quad \quad H_1(X,A) \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0$$
for unreduced homology. This shows that it is essential to have information about the $0$-th (reduced) homology group of a one-point space.
Remark 2.
The long exact reduced sequence of the pair $(X,\emptyset)$ agrees with the long exact sequence of unreduced homology groups except at the end where it looks like
$$H_1(X,\emptyset) = H_1(X) \to \tilde{H}_0(\emptyset) = 0 \to  \tilde{H}_0(X) \to H_0(X,\emptyset) = H_0(X) \to \mathbb Z \to 0$$
The short exact sequence $0 \to \tilde{H}_0(X) \to H_0(X) \to \mathbb Z \to 0$ splits (because obviously $H_0(X) \to \mathbb Z$ has a left inverse) and we conclude $H_0(X) \approx \tilde{H}_0(X) \oplus \mathbb Z$. This can also be used to show that $\tilde{H}_0(P) = 0$.
Remark 3.
It is easy to see that $\tilde{H}_0(X)$ is naturally isomorphic to $\ker (p_* : H_0(X) \to H_0(P))$ where $p : X \to P$ is the unique map into a one-point space $P$. This comes from the fact that the augmentation map $\epsilon : C_0(X) \to \mathbb Z$ is nothing else than the induced $p_\# : C_0(X) \to C_0(P) \approx \mathbb Z$.
