# is it possible to derive the künneth theorem from serre spectral sequences?

In case it is any easier, we can consider $\mathbb Q$ coefficients, so that there is no torsion.

In this case it should be enough to prove that in the second page of the spectral sequence, all differentials are zero.

Indeed, considering the fibration:

$$X \to X \times Y \to Y$$

we get (in the homological case, but I guess that the cohomological one is completely analogous) $$E_2^{p, q} = H_p(Y, H_q(X, \mathbb Q)) = H_p(Y, \mathbb Q) \otimes H_q(X, \mathbb Q)$$

so.. if this is the correct approach, how to prove that the differentials in $E^2$ are null, in this case?

• Try using the fact that there's a splitting $Y \to X \times Y$. – JHF Jan 30 '17 at 15:52
• sorry I don't get it.. – fritz Jan 30 '17 at 16:57

I will expand my comment a little. We'll work with the cohomological spectral sequence. Because of the multiplicative structure, it is enough to show that all the differentials $d_r: E_r^{0,q} \to E_r^{r,q-r+1}$ emanating from the zeroth column is zero.
The zeroth column is essentially a copy of $H^* X$. If there were nonzero differentials, the result on the $E_\infty$ page would be smaller, since we would be passing to strict subquotients.
However, this is not possible in the case of a trivial fibration. In our situation there is a splitting of $X \xrightarrow{i} X \times Y$ given by projection. This means that $i^*: H^*(X \times Y) \to H^*X$ is surjective, so the left hand side, which is reconstructed from the $E_\infty$ page of the spectral sequence, cannot be smaller than $H^* X$.