Calculating the cochain complex of $S^2 \times S^4. $

Question

Calculating the cochain complex of $S^2 \times S^4. $

My calculation:

$C^6 = \mathbb{Z}$

$C^5 = 0$

$C^4 = \mathbb{Z}$

$C^3 = 0 $

$C^2 = \mathbb{Z}$

$C^1 = 0$

$C^0 = \mathbb{Z}$

Am I correct?

Also, my procedure in the calculation was just using the homology groups of the sphere, is that correct? My professor told us the procedure in a fast way that I did realize it, could anyone tell me the general procedure of calculating the cochain complex?

Answer 1

If $C,C',D$ and $D'$ are chain complexes with $H_*(C) = H_*(C')$ and $H_*(D) = H_*(D')$ then $H_*(C \otimes D) = H_*(C' \otimes D')$.

(To prove this use the formula for the homology of tensor product of two chain complexes found on the last page here)

In particular if we have $X,Y$ be two $CW-$complexes with $C = C_*(X)$, $C'_n = H_n(X_n,X_{n-1})$, $D = C_*(Y)$ and $D'_n = H_n(Y_n, Y_{n-1})$

Then we have $H_*(C)=H_*(X) = H_*^{CW}(X) = H_*(C')$ and by the same token $H_*(D) = H_*(D')$ so we also know that $H_*(X \times Y) = H_*(C \otimes D) = H_*(C' \otimes D')$,

Basically what I'm saying is that the homology of the tensor product of the two $CW-$homology chain complexes actually gives you the singular homology of the product of the two spaces. I.e what you're doing is valid.