I am struggling with the following question (2.2.36) from Hatcher for quite some time now: Show that $H_i(X\times S^n) \simeq H_i(X) \oplus H_{i-n}(X)$. I don't know how to use the hint given by Hatcher. I have been trying to use the Mayer-Vietoris sequence by covering $S^n$ by two discs of dimension $n-1$ and considering the cover obtained by taking product with $X$ but without any luck. Please help.

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    $\begingroup$ This is trivial by the Künneth formula. $\endgroup$ – J.R. Feb 9 '14 at 14:39
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    $\begingroup$ @TooOldForMath: I see, thanks for pointing it out. But this is after Section 2.2 in Hatcher, so there should be a simpler solution using material up to Section 2.2 in Hatcher, right? $\endgroup$ – Dave Feb 9 '14 at 14:54

Let us follow the hint given in Hatcher. Suppose that $x_0\in S^n$.

Claim 1: $$H_i(X\times S^n)\approx H_i(X)\oplus H_i(X\times S^n,X\times\{x_0\})$$

Proof: Since $X\times \{x_0\}$ is a retract of $X\times S^n$ (along the map $r=\mathrm{id}_X\times x_0$), the long exact relative homology sequence induced by $A\overset{i}{\hookrightarrow} X\rightarrow X/A$ breaks up into split short exact sequences

$$0\rightarrow H_i(X\times\{x_0\})\rightarrow H_i(X\times S^n)\rightarrow H_i(X\times S^n,X\times\{x_0\})\rightarrow 0$$

To see this note that $r\circ i=\mathrm{id}$ implies $r_*\circ i_*=\mathrm{id}$ which in turn means that the induced map $i_*:H_i(X\times \{x_0\})\rightarrow H_i(X\times S^n)$ is a split monomorphism. Now because the sequence splits, $$H_i(X\times S^n)\approx H_i(X\times \{x_0\})\oplus H_i(X\times S^n, X\times \{x_0\})\approx H_i(X)\oplus H_i(X\times S^n, X\times \{x_0\})$$ $\square$

Claim 2: $$H_i(X\times S^n,X\times \{x_0\})\approx H_{i-1}(X\times S^{n-1},X\times\{x_0\})$$

Proof: Pick some $\epsilon>0$ and decompose $S^n=U\cup V$ where $U=\{(x_0,\dots,x_n)\in S^n\,:\,x_0>-\epsilon\}$ and $V=\{(x_0,\dots,x_n)\in S^n\,:\,x_0<\epsilon\}$. Then $U,V$ are contractible and $U\cap V\approx S^{n-1}$. The claim now follows by the (relative) Mayer-Vietoris sequence. $\square$

Corollary 3: $$H_i(X\times S^n, X\times\{x_0\})\approx H_{i-n}(X)$$

Proof: Apply Claim 2 $n$ times and notice $H_{i-n}(X\times S^0, X\times\{x_0\})\approx H_{i-n}(X)$ (since $S^0$ is just two points). $\square$

Now Claim 1 and the Corollary to Claim 2 yield

$$H_i(X\times S^n)\approx H_i(X)\oplus H_{i-n}(X)$$

  • $\begingroup$ could you show how to use relative Mayer-vietoris sequence in details? $\endgroup$ – Bruce Jan 13 '18 at 10:05

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