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Show that for $i > n \in\mathbb{N}$: $$H_{i}\left(X \times \mathbb{S}^{n}\right) \simeq H_{i}\left(X\right) \oplus H_{i - n}\left(X\right).$$

My first idea motivated by $n=0$ case (which is obvious) was to try induction but I cannot see how to perform next step. However question seems to be very neat so I decided to share it.

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    $\begingroup$ Are you familiar with the Kunneth theorem? If not, try to use Mayer-Vietoris. $\endgroup$ Jun 14, 2014 at 17:32
  • $\begingroup$ No I'm not. I'll try to show my attempt (unsuccessful) in Mayer-Vietoris usage. $\endgroup$ Jun 14, 2014 at 17:34

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Ok so here is what I've got so far:

Lemma 1. $$H_i(X\times S^n) \simeq H_i(X \times \{s\})\oplus H_i(X\times S^n,X\times\{s\})$$

Let's write seuquence for pair $(X \times S^n, X \times \{s\})$:

$$H_{i + 1}(X \times S^n, X \times \{s\})\to H_i(X \times \{s\}) \to H_i(X \times S^n) \to H_i(X \times S^n, X \times \{s\}) \to H_{i-1}(X \times \{s\})$$ second arrow is generated by inclusion (monomorphism) so first and last arrows are equal to $0$ (and therefore third arrow is epimorphism) and moreover we can take arrow in opposite direction i.e. $H_i(X \times S^n) \to H_i(X \times \{s\})$ by taking retraction then their composition will be identity on $X \times \{s\}$ thus also on $H_i(X \times \{s\})$. But we know that for short exact sequence $A \to B \to C$ where first arrow is mono and second epi and there exists opposite arrow $B \to A$ which composited with $A \to B$ gives identity we have $B = A \oplus C$.

Lemma 2. $$H_i(X \times S^n, X \times \{s\}) \simeq H_{i-1}(X \times S^{n-1}, X \times \{s\})$$

Let's take $A = X \times S^n \setminus\{s\}$, $B = X \times S^{n} \setminus\{n\}$, $C = D = X \times \{s\}$. We have $H_i(X \times \{s\}, X \times \{s\}) \simeq 0$, $H_i(A, X \times \{s\}) \simeq H_i(B, X \times \{s\}) \simeq H_i(X \times \{s\}, X \times \{s\})$ and from relative Mayer-Vietoris sequence for $(A, C), (B, D)$: $$0 \oplus 0 \to H_i( X \times S^n, X \times \{s\}) \to H_{i - 1}(X \times S^{n-1}, X \times \{s\}) \to 0 \oplus 0.$$ Hence the lemma.

Difficulty. After applying lemma 2 for $n$ times in lemma 1 we get: $$H_i(X\times S^n) \simeq H_i(X \times \{s\})\oplus H_{i - n}(X\times S^0,X\times\{s\}).$$ $X\times \{s\} \to X\times S^0$ is cofibration and hence we have $$H_{i - n}(X\times S^0,X\times\{s\}) \simeq H_{i - n}((X\times S^0)/ (X\times\{s\})) \simeq H_{i - n}(X\times\{n\} \cup \{x\}\times\{s\}) \simeq H_{i - n}(X \times \{n\}).$$ But the last equality holds only for $i - n > 0$. (I hope that the meaning of $n$ is clear from the context ;))

I'd be grateful for checking, simplifying this solution and fixing case $i = n$.

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    $\begingroup$ In your proof of the first lemma you state "second arrow is generated by inclusion (monomorphism) so first and last arrows are equal to 0 (and therefore third arrow is epimorphism)". It seems to me you are under the impression that if a map $i:X\to Y$ is a monomorphism in the category of spaces, then it induces a monomorphism in homology, but this is false, for instance the inclusion of the circle in the plane $S^1\hookrightarrow\Bbb R^2$ couldn't induce a monomorphism in homology because the first homology group of the plane is zero, but that of the circle is nonzero. Am I missing something? $\endgroup$ Jun 14, 2014 at 23:10
  • $\begingroup$ I'm afraid that you are right there is a serious lack but I don't know how to fix it so far. (Lemma surely holds - it was hint for this problem. Maybe you have an idea how to fix it?) $\endgroup$ Jun 15, 2014 at 19:33
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    $\begingroup$ Late to the game, but I've been working on this problem myself so I thought I'd comment. $X\times\{s\}$ is a retract of $X \times S^n$. Hence, the inclusion induces an injection. $\endgroup$
    – SFSH
    Oct 23, 2020 at 23:01
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The result follows directly from the Kuenneth formula, since $H_p(S^n) = {\mathbb{Z}}$ for $p = 0, n$ and vanishes in all other dimensions.

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  • $\begingroup$ It is surely elegant answer but Kunneth formula is machinery I'm not familiar to so far. I'd be grateful for other attempts. $\endgroup$ Jun 14, 2014 at 17:37
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    $\begingroup$ Sure, that's reasonable. (Unfortunately, the Kuenneth formula is not easy to derive separately; the two standard proofs I'm aware of are (a) showing that any functor satisfying certain axioms must agree with the standard (say, simplicial) homology $H_n(\cdot)$; or (b) deriving it through spectral sequences. Neither method is short or easy. $\endgroup$
    – anomaly
    Jun 14, 2014 at 17:58
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    $\begingroup$ @anomaly: probably this is an exercise in a section of an algebraic topology textbook before Kunneth is introduced. $\endgroup$ Jun 14, 2014 at 17:59
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    $\begingroup$ @Qiaochu Yuan: Sure, that makes sense. I'll try to elaborate on the Mayer-Vietoris argument you mentioned above, then. $\endgroup$
    – anomaly
    Jun 14, 2014 at 18:01
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    $\begingroup$ @StephenDedalus: Try using $X \times S_+$ and $X \times S_-$ in the Mayer-Vietoris decomposition, where $S_\pm$ are the upper- and lower-hemispheres of $S^n$ . Each of those terms contracts to $X$ itself, and their intersection is $X \times S^{n-1}$. (You may need to consider the actual maps in the sequence, rather than just relying on homological algebra itself, to prove the result.) $\endgroup$
    – anomaly
    Jun 14, 2014 at 18:05

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