Show that $H_q(S^n,x) \xrightarrow{\cong} H_q(S^n,E_n^+) \;\;\forall q$ I am new into the topic of algebraic topology and try to understand the argumentation of the following statement in the chapter of homology of CW-complexes. The goal is to show that:
$$H_q(S^n,x) \xrightarrow{\cong} H_q(S^n,E_n^+) \;\;\forall q$$
Where we used the following notations:  $E_n^+ := \{(x_1,\ldots,x_{n+1}) \in S^n \mid x_{n+1} \ge 0\}$.
The proof is given as follow: We observe the long exact sequence of $(S^n,E^+_n):$
$$H_q(E^+_n) \to H_q(S^n) \to H_q(S^n,E^+_n) \xrightarrow{\partial_q} \to H_{q-1}(E^+_n) \to \dots$$
Since $E^+_n$ is contractible, we have that $H_q(E^+_n)=0, \; \forall q>1$ and hence we have by exactness that $H_q(S^n) \to H_q(S^n,E^+_n)$ is an isomorphism for all $q>1$. Now we observe the sequence:
$$\dots \to 0= H_1(E^+_n) \to H_1(S^n) \to H_1(S^n,E^+_n) \xrightarrow{\partial_1} H_0(E^+_n) \to H_0(S^n) \to H_0(S^n,E^+_n) \to H_{-1}(E^+_n)=0 \to \dots$$
But from here on I am lost... How can I end now the argumentation? For example why do we have that $H_0(S^n) \to H_0(S^n,E^+_n)$ is surjective?
Many thanks for some help!
 A: This post does not answer your question about the details of the book's argument, but it answers the question of how to show the goal of the book (i.e. third line of your post). Apply the 5 lemma on the commutative diagram below and you are done.
$$\begin{array}{ccccccccc}
H_q(\{x\}) & \to & H_q(\mathbb{S}^n) & \to & H_q(\mathbb{S}^n,\{x\}) & \to & H_{q-1}(\{x\}) & \to & H_{q-1}(\mathbb{S}^n)\\\
\downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow\\\
H_q(E_n^+) & \to & H_q(\mathbb{S}^n) & \to & H_q(\mathbb{S}^n,E_n^+) & \to & H_{q-1}(E_n^+) & \to & H_{q-1}(\mathbb{S}^n)
\end{array}$$
Vertical morphisms $2,5$ (counting from left) are identity maps so isomorphisms. Vertical morphisms $1,4$ (counting from left) are isomorphisms since the $\{x\}$ is a deformation retract of $E_n^+$.
A: You can prove that for all contractible $C \subset S^n$ and all $q > 0$
$$H_q(S^n) \to H_q(S^n,C) \tag{1}$$
is an isomorphism. This shows that your claim is true for all $q > 0$.
For $q > 1$ you have already proved in your question that $(1)$ is an isomorphism. For $q = 1$ do it as follows:
Take any constant map $c : S^n  \to C$. Then $c i \simeq id_C$, where $i : C \to S^n$ denotes inclusion. Thus $c_*i_*$ is the identity on $H_0(C)$. This  implies that $i_*$ is injective, i.e. $\ker i_* = 0$. Hence $\operatorname{im} \partial_1 = 0$ which means that $\ker \partial_1 = H_1(S^n,C)$. By exactness we see that $H_1(S^n) \to H_1(S^n,C)$ is surjective. Since we already know that it is injective, it is an isomorphism.
It remains to consider $q = 0$. Here $(1)$ is not an isomorphism. We shall show that
$$H_0(S^n,C) \approx \ker (p_* : H_0(S^n) \to H_0(*))$$
where $*$ is a one-point space and $p : S^n \to *$ is the unique map onto $*$. This will prove your claim for $q = 0$. We have a short exact sequence
$$0 \to H_0(C) \stackrel{i_*}{\to} H_0(S^n) \to H_0(S^n,C) \to 0$$
(recall that $i_*$ is injective). With $q : C \to *$ we see that $q_* : H_0(C) \to H_0(*)$ is an isomorphism and we get a short exact sequence
$$0 \to H_0(*) \stackrel{i_*q_*^{-1}}{\to} H_0(S^n) \stackrel{j}{\to} H_0(S^n,C) \to 0$$
This sequence splits because $i_*q_*^{-1}$ has $p_*$ as a left inverse (note that $p i = q$, thus $p_*i_*q_*^{-1} = id$). Therefore
$$\phi :  H_0(S^n) \to H_0(*) \oplus H_0(S^n,C), \phi(g) = (p_*(g),j(g))$$
is an isomorphism. Obviously $H_0(S^n,C)$ is the kernel of the projection $\pi : H_0(*) \oplus H_0(S^n,C) \to H_0(*)$. Since $\phi$ is an isomorphism, the kernel of $\pi\phi$ is isomorphic to $H_0(S^n,C)$. But $\pi\phi = p_*$ which completes the proof.
