Understanding homologies of pairs and induced maps Consider the map of $(S^n$,{$x,-x$}$)$ into itself $(x\in S^1)$ induced by the antipodal map of $S^n$. Describe the induced homology of pair map.
I guess that first I have to compute the homology group, but I don't know if I'm doing this right. Let's consider the long exact sequence
$$H_k(\text{{$x,-x$}}) \longrightarrow H_k(S^n) \longrightarrow H_k(S^n,\text{{$x,-x$}}) \longrightarrow H_{k-1}(\text{{$x,-x$}}) \longrightarrow 
 ... $$
Then, since $H_k(\text{{$x,-x$}})=0$ for all $k\neq 0,1$, I have: 
$$0 \longrightarrow H_k(S^n) \longrightarrow H_k(S^n,\text{{$x,-x$}}) \longrightarrow 0$$
And so $H_k(S^n) \cong H_k(S^n,\text{{$x,-x$}})$. But I don't know how to continue from this point. 
Also, if $k=0$ the homology group is just $\mathbb{Z}$ (because of connected components).
If $k=1$ I have the sequence
$$0\longrightarrow H_1(\text{{$x,-x$}}) \longrightarrow H_1(S^n) \longrightarrow H_1(S^n,\text{{$x,-x$}}) \longrightarrow H_0(\text{{$x,-x$}}) \cong \mathbb{Z}^2 \longrightarrow H_0(S^n)\cong \mathbb{Z} \longrightarrow H_0(S^n,(\text{{$x,-x$}})\cong \mathbb{Z} \longrightarrow 0$$
How do I go from here to know $H_1(S^n,\text{{$x,-x$}})$?
And, after all, what is the role of the antipodal map in all this problem? I don't understand its role. How does it define $H_k(f)$? And what would be the explicit map? 
 A: I'll use $\{x, y\}$ instead of $\{x, -x\}$ as my notation for the antipodal points. A few comments:


*

*$H_0(S^n, \{x, y \} )$ is not $\mathbb Z$ -  it is zero!. The reason is that, since $S^n$ is path connected, any point in $S^n$ is homologous to $x$ (and also to $y$, for that matter...).

*To compute $H_1(S^n , \{ x, y\})$, we can use the reduced homology. The LES for reduced homology is
$$ \dots \longrightarrow  H_1(\{x, y\}) \cong 0 \longrightarrow H_1(S^n) \longrightarrow H_1(S^n, \{x, y\}) \longrightarrow \widetilde H_0(\{ x, y \})\cong \mathbb Z \longrightarrow \widetilde H_0(S^n) \cong 0\longrightarrow \dots$$

*Just for fun, an alternative way of computing $H_k(S^n \{ x, y \})$ for $k \geq 1$ is to note that $(S^n , \{ x, y \})$ is a good pair, so $$H_k(S^n , \{ x, y \}) \cong H_k(S^n / \{ x, y\}) \cong H_k(S^n \vee S^1) \cong H_k(S^n) \oplus H_k(S^1)$$ for $k \geq 1$, where the middle equality is by homotopy equivalence.

*Since the antipodal map sends the set $\{x, y \}$ into itself, the antipodal map defines a map of pairs $$f : (S^n, \{ x, y \}) \to (S^n, \{x, y \})$$ This induces maps
$$ C_k(f) : C_k(S^n, \{ x, y \}) \cong \frac{C_k(S^n)}{C_k(\{ x, y \})} \to  C_k(S^n, \{ x, y \})  \cong  \frac{C_k(S^n)}{C_k(\{ x, y \})}$$
on the groups of relative chains. The map $$H_k(f) : H_k(S^n, \{ x, y \}) \to H_k(S^n, \{x, y \})$$ is then defined as the map induced by $C_k(f)$ on the relative homology groups . The question is asking you to characterise this map $H_k(f)$, for each $k$.

*In the case $n \geq 2$, we have $H_n(S^n) \cong H_n(S^n , \{ x, y \}) \cong \mathbb Z$. We can characterise the map $H_n(f) $ by examining the commutative diagram:
$\require{AMScd}$
\begin{CD}
H_n(S^n) @>{\cong}>> H_n(S^n, \{ x, y \})\\
@V{H_n(f)}VV @V{H_n(f)}VV\\
H_n(S^n) @>{\cong}>> H_n(S^n, \{ x, y \})
\end{CD}
[Remember that $H_n(f) : H_n(S^n) \to H_n(S^n)$ maps a generator $\alpha$ of $ H_n(S^n)$ to $(-1)^{n+1} \alpha$.]

*In the case $n \geq 2$, we have $H_1(S^n , \{ x, y \}) \cong \widetilde H_0(\{ x, y \}) \cong \mathbb Z$. We can characterise the map $H_1(f)$ by examining the commutative diagram:
$\require{AMScd}$
\begin{CD}
H_1(S^n, \{ x, y\}) @>{\cong}>> \widetilde H_0( \{ x, y\})\\
@V{H_1(f)}VV @V{\widetilde H_0(f)}VV\\
H_1(S^n, \{ x, y\}) @>{\cong}>> \widetilde H_0(\{ x, y \})
\end{CD}
[Remember, a generator of $\widetilde H_0(\{ x, y \})$ is $x - y$, and the image of this under the antipodal map is $y - x$.]

*You will also need to handle the  case $n = 1$, where $H_1(S^1, \{x, y \}) \cong \mathbb Z^2$...
