Homology of product of topological space and sphere is direct sum of homologies.

Question

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.

Answer 1

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.

Answer 2

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 $i = n$.