# The spectral sequence of the path fibration of $S^2$

Bott and Tu use the fibration $\Omega S^2 \to PS^2 \to S^2$ to compute the cohomology of $\Omega S^2$. They do this by looking at the (Serre?) spectral sequence. Then $E_2^{p,q} = H^p(S^2,H^q(\Omega S^2))$.

They claim that by the universal coefficient theorem, all columns in $E_2$ except when $p=0$ or $p=2$ are zero, but I don't see how this is being applied. Could someone give a more detailed explanation?

• By the way, yes, it's the Serre spectral sequence. – Najib Idrissi Mar 19 '15 at 10:52

Recall that $H_p(S^2;\mathbb{Z})$ is $\mathbb{Z}$ when $p=0,2$ and trivial otherwise.

The universal coefficient theorem gives a short exact sequence

$$0 \to \text{Ext}(H_{p-1}(S^2;\mathbb{Z}),G)\to H^p(S^2;G)\to \text{Hom}(H_p(S^2;\mathbb{Z}),G)\to 0$$

Whatever $G=H^q(\Omega S^2;\mathbb{Z})$ may be:

• The $\text{Ext}$ term always disappears since $H_{p-1}(S^2;\mathbb{Z})$ is always free (in particular, projective).
• $\text{Hom}(H_p(S^2;\mathbb{Z}),G)=0$ for $p\not= 0,2$ since $\text{Hom}(0,G)=0$.
• $\text{Hom}(\mathbb{Z},G)\cong G$.

The exactness of the sequence then implies that

$$E_2^{p,q}=H^p(S^2;G)\cong \begin{cases} \text{Hom}(\mathbb{Z},G)=G,& \text{if } p = 0,2\\ \text{Hom}(0,G)=0, & \text{otherwise} \end{cases}$$

Therefore the only possible non-zero terms are located at the $E_2^{0,*}$ or the $E_2^{2,*}$ columns.