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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?

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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 $H_1(S^1, \{x, y \}) \cong \mathbb Z^2$...

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  • $\begingroup$ I posted the question* on another forum and I've got: "k = 0 is easy, since it's about path components. For n > 0, H0(X,A) = Z, and the map induced by f is the identity." (Where X=S^n and A={x,-x}). But as you said, I remember that homology of a contractible space is 0, so now I am confused about it Other than that, I understand all your points except the ones about reduced homology, since we haven't studied them yet (and the exam is on friday, so we won't), thank you very much. *reddit.com/r/learnmath/comments/8vagxn/… $\endgroup$ – Arekkusu Jul 2 '18 at 11:18
  • $\begingroup$ @AiYagami I don't think the answer on the other forum is correct! Since $X$ is path connected, any $0$-chain in $X$ is homologous to a $0$-chain in $A$. So $H_0(X, A) = 0$. $\endgroup$ – Kenny Wong Jul 2 '18 at 11:27
  • $\begingroup$ @AiYagami If you haven't seen reduced homology before, then forget about it. Instead of the LES in my answer, consider this one: $ \dots \to H_k(\{x, y\}) \to H_k(S^n, \{ x \}) \to H_k(\{x, y \}, \{x\}) \to \dots $. You should be able to get the same answers. (Essentially, the reduced homology group $\widetilde H_k(S^n)$ is the same thing as $H_k(S^n, \{ x \})$, where $x$ is any point on $S^n$.) $\endgroup$ – Kenny Wong Jul 2 '18 at 11:29

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