What is $H_*(S^m \times S^n)$, without using Künneth theorem? As the question suggests, what is $H_*(S^m \times S^n)$ for $m \ge 1$ and $n \ge 1$? I would like to see a way without using the Künneth theorem...
 A: We can compute $H_i(S^n\times X)$ for any space $X$ and all $i$ without using Künneth's theorem as follows:
Step 1: There is a retraction $r:S^n\times X\to \{x_0\}\times X=X$ given by pinching $S^n$.
This means that $r\circ i=id_X$ where $i:X=\{x_0\}\times X\hookrightarrow S^n\times X$ is the inclusion. Then, by functoriality of $H_i$, the inclusion is injective on $H_i$ (since it has left inverse). Thus, considering the long exact sequence of the pair $(S^n\times X, X)$, we have split short exact sequences of the form:
$$0\to H_i(X)\to H_i(S^n\times X)\to H_i(S^n\times X, X)\to 0$$
and therefore $H_i(S^n\times X)\cong H_i(X)\oplus H_i(S^n\times X, X)$.
Step 2: Computation of $H_i(S^n\times X, X)$.
Let $D^n_+$ be an "$\epsilon$-thickened" upper hemisphere of $S^n$ and $D^n_{-}$ and "$\epsilon$-thickened" lower hemisphere of $S^n$, each containing $x_0$. Then $S^n\times X$ can be written as the union of the two open subsets $\underbrace{(D^n_+\times X)}_{A}$ and $\underbrace{(D^n_-\times X)}_{B}$ and further $\{x_0\}\times X = X\subset A\cap B$. Now let us consider the following chunk of the relative Mayer-Vietoris sequence:
$$H_i(A,X)\oplus H_i(B,X)\to H_i(S^n\times X, X)\to H_{i-1}(A\cap B, X)\to H_{i-1}(A,X)\oplus H_{i-1}(B,X)$$
We make the following observations:


*

*$(A,X)\simeq (X,X)\simeq (B,X)$  since $D^n_{\pm}$ is contractible

*$(A\cap B,X)\simeq (S^{n-1}\times X, X)$


Thus, the MV-sequence above gives isomorphisms $H_i(S^n\times X, X)\cong H_{i-1}(S^{n-1}\times X,X)\cong \cdots \cong H_{i-n}(X\sqcup X, X)\cong H_{i-n}(X)$.
Conclusion: Combining the two results, we get $H_i(S^n\times X)\cong H_i(X)\oplus H_{i-n}(X)$.
