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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 $n\geq 1$.

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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
up vote 3 down vote accepted

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.

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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

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