For a fibration $F\to X\to B$, if we add some mild hypothesis, we would have a corresponding Serre Spectral Sequence.
It’s said that if the total space $X$ is contractible, then $E^\infty$ page of the spectral sequence is trivial, i.e. it has the only nonzero group $\mathbb{Z}$ on the $(0,0)$ position if we choose $\mathbb{Z}$ to be the coefficient group. I don’t know why this follow? Hope someone could help. Thanks!
Best Answer
So for each $n \geq 0$ you have a convergent filtration $\{L_{p, n-p}\}_{p \in \mathbb{Z}}$ of $H_n(X)$, i.e. $$0 = L_{-1, n+1} \subseteq L_{0, n} \subseteq L_{1, n-1} \subseteq \dots \subseteq L_{p, n-p} \subseteq \dots \subseteq H_n(X).$$
Since $X$ is contactible we have $H_n(X) = 0$ for $n > 1$ and $H_0(X) = \mathbb{Z}$ (if we choose $\mathbb{Z}$ to be the coefficient group).
Now remember the Serre Spectral Sequence is a first quadrant spectral sequence so $E^{\infty}_{p, q} = 0$ for any $p, q < 0$ automatically. Also we know that in general we have an isomorphism $$E^{\infty}_{p, q} = \frac{L_{p, q}}{L_{p-1, q+1}}$$ with respect to the filtration of $H_n(X)$ (note $p+q = n$). Since $H_n(X)=0$ for all $n>1$ the filtration is automatically trivial for any $p, q >0$ (remember $n = p+q$) and so we have from the above that $E^{\infty}_{p, q} = 0$ for any $p, q > 0$.
Finally to see that $E^{\infty}_{0, 0} = \mathbb{Z}$, remember that $E^2_{0, 0} = H_0(F; H_0(X))$ then apply universal coefficients and note that $F$ is simply connected (since we are dealing with a Serre spectral sequence) to expand this out and see that $E^2_{0, 0} = \mathbb{Z}$ and then compute and notice that $E^2_{0, 0} = E^3_{0, 0} = E^4_{0, 0} = \cdots$ and so on and thus you end up with $E^{\infty}_{0, 0} = \mathbb{Z}$.