How should I prove that $H_i(X \times S^n) \cong H_i(X) \oplus H_{i-n}(X)$ for all $i,n$?
What was my approach. I am not allowed to use Kunneth's Formula because it would be too easy. Rather I think I can use Mayer-Vietoris Sequence.
Consider the open sets $U=X \times (S^n-N)$ and $V=X \times (S^n-S)$ then $U \cap V=X \times S^{n-1}$ and induction comes into play. But then I am not sure how to proceed. I am unable to see how, after using induction to use the maps $i^*$ and $j^*$ and complete.
Another Solution that use Hatcher's Hint. I found the solution here but cannot understand the following step. Can anyone explain without using splitting of sequences?
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, $$H_i(X\times S^n)\approx H_i(X\times \{x_0\})\oplus H_i(X\times S^n, X\times \{x_0\})\approx H_i(X)\oplus H_i(X\times S^n, X\times \{x_0\})$$ $\square$
