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I am trying to prove that

$$H_p(B_{n+1},S_n;\mathbb{A}) \cong \left\{\begin{array}{ll} H_{p-1}(S_n,\mathbb{A}) & \text{if } p\geq2\\\ 0&\text{if } p=1, n\geq 1\\ \mathbb{A} &\text{if } p=1, n=0\\0 & \text{if } p=0 \end{array}\right.$$

For $p=0$ that's ok but I don't understand how to prove the case when $n=0$ or $n\geq 1$.

Dan Rust
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Vrouvrou
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    Have you tried writing out the long exact sequence in homology for a pair for the different cases of $n$? – Dan Rust Nov 10 '13 at 17:55

1 Answers1

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Recall that for a pair $(X,A)$ with $A$ a subspace of $X$, we get a long exact sequence in homology $$\cdots\to H_p(A)\to H_p(X)\to H_p(X,A)\to H_{p-1}(A)\to H_{p-1}(X)\to\cdots$$ over the given coefficient ring.

Next, recall that $B^{n+1}$ is contractible for all $n$ and so $H_p(B^{n+1})=0$ for all $p\geq 1$. Also, $H_0(X)=\oplus_{i=1}^k\mathbb{A}$ for all $X$ with $k$ path components.

Also, remember that if $A_1\to A_2\stackrel{f}{\to} A_3\to A_4$ is an exact sequence of groups, then $A_1=0=A_4$ if and only if $f$ is an isomorphism.

Finally, recall that for the pair $(X,A)$, if there exists a neighbourhood $U\subset X$ of $A$ such that $A$ is a deformation retract of $U$, then $H_p(X,A)\cong \tilde{H}_p(X/A)$ where $\tilde{H}$ denotes reduced homology.

Dan Rust
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  • you dont answer my question, why we have a difference when n=0 and $n\geq 1$ ? – Vrouvrou Nov 10 '13 at 19:32
  • How many path components does $S^0$ have? Everything I've said in the above answer, together with some basic group theory, should pretty quickly give you your answer. – Dan Rust Nov 10 '13 at 20:20
  • i think that $S^0$ has one path compoments no ? – Vrouvrou Nov 10 '13 at 20:23
  • $S^0$ is the boundary of the closed $1$-ball (ie an interval). Are you sure $S^0$ has one path component? – Dan Rust Nov 10 '13 at 20:24
  • $S^0={-1,1}$ if we talk about unit sphere so it has 2 not 1 – Vrouvrou Nov 10 '13 at 20:26
  • So what's $H_0(S^0)$? – Dan Rust Nov 10 '13 at 20:28
  • $H_0(S^0)=\mathbb{A}\oplus \mathbb{A}$ – Vrouvrou Nov 10 '13 at 20:29
  • Now, what's $H_0(B^1)$? Can you see what the map induced between $H_0(S^0)$ and $H_0(B^1)$ is? Using exactness, can you then work out what the group $H_1(B^1,S^0)$ must be? – Dan Rust Nov 10 '13 at 20:35
  • $H_0(B^1)=\mathbb{A}$,what it means the map induced between $H_0(S^0)$ and $H_0(B^1)$ ? – Vrouvrou Nov 10 '13 at 20:39
  • If $i\colon S^0\to B^1$ is inclusion, then what is the induced map $i_*\colon H_0(S^0)\to H_0(B^1)$? (Check the definition given in your text) – Dan Rust Nov 10 '13 at 20:46
  • $i_*: C_0(S^0)\rightarrow C_0(B^1) $ ? – Vrouvrou Nov 10 '13 at 21:04
  • It's true that a continuous map $f\colon X\to Y$ induces a chain map $f_{#}\colon C_p(X)\rightarrow C_p(Y)$ because $f_{#}\partial=\partial f_{#}$, but then chain maps send cycles to cycles and boundaries to boundaries, hence induce homomorphisms on the level of homology. The map is given by, if $[x]\in H_p(X)$, then $f_*([x])=[f(x)]$. – Dan Rust Nov 10 '13 at 21:15
  • yes , i'm ok with you but what is the relation with $H_1(B^1,S^0)$ ? – Vrouvrou Nov 10 '13 at 21:17
  • You should have written down at some point the exact sequence $$\cdots\to H_1(B^1)\to H_1(B^1,S^0)\stackrel{\delta}{\to} H_0(S^0)\stackrel{f_}{\to} H_0(B^1).$$ You know $H_1(B^1)$ is trivial, and so $\delta$ is injective into $\mathbb{A}\oplus\mathbb{A}$. You should find that $\ker f_=\langle(x,-x)\rangle$. That means, by exactness, that $\delta$ has image equal to $\langle(x,-x)\rangle$. – Dan Rust Nov 10 '13 at 21:25
  • ok, thank you, i'm confuse with the case when $p\geq 2$ – Vrouvrou Nov 11 '13 at 08:11
  • What does the exact sequence look like when $p\geq 2$? You really should be writing this stuff down and showing me what you've tried. – Dan Rust Nov 11 '13 at 09:50
  • $\cdots\to H_p(B^{n+1})\to H_p(B^{n+1},S^n)\stackrel{\delta}{\to} H_p(S^n)\stackrel{f_*}{\to} H_{p+1}(B^{n+1})$ the sequence is exact and $H_{p+1}(B^{n+1})$and $H_p(B^{n+1})$ are 0 then we obtain the result thank you for all – Vrouvrou Nov 11 '13 at 14:14
  • The indices for the homology groups in your sequence are slightly wrong, but yes, you get an isomorphism from the usual 'trapped between two zeros' rule. – Dan Rust Nov 11 '13 at 14:16
  • can you please see this and tel me what i must prove exactly http://math.stackexchange.com/questions/551575/question-on-relative-homology/ – Vrouvrou Nov 11 '13 at 17:50
  • for 2) please ... – Vrouvrou Nov 11 '13 at 17:51