# Prerequisite of understanding a topological construction in Spectral Sequences

This is a post on the construction of a spectral sequence. I am in fact lost in the first paragraph.

Let $$B$$ be a CW complex and $$\pi\colon X\to B$$ a Serre fibration. Put $$X^k=\pi^{-1}(B^k)$$. A cellular approximation~$$\Delta_B\colon B\to B\times B$$ of the diagonal can be lifted to an approximation $$\Delta\colon X\to X\times X$$ of the diagonal such that $$X^k\stackrel\Delta\longrightarrow\bigcup_{m+n=k}X^m\wedge X^n\;.$$

Few points.

(i) How does the diagonal map lift?

(ii) Why does the lifted map respects the cells?

(iii) Does $$X^k$$ give a cell structure to $$X$$?

Put the product CW structure on $$B\times B$$ and take a cellular approximation of the diagonal to get a homotopy $$F$$ from the diagonal $$\Delta_B$$ to a map $$\Delta_B':B\rightarrow B\times B$$ which satisfies

$$\Delta_B'(B^n)\subseteq\bigcup_k B^{n-k}\times B^k.$$

Since $$\pi$$ is a Serre fibration, so is the product $$\pi\times \pi:X\times X\rightarrow B\times B$$, and in particular has a certain homotopy lifting property. Since diagonal maps are natural, the composite $$F\circ(\pi\times 1)$$ is a homotopy from $$(\pi\times \pi)\circ\Delta_X=\Delta_B\circ\pi$$ to $$\Delta'\circ\pi:X\rightarrow B\times B$$ and thus lifts to a homotopy $$G$$ from $$\Delta_X:X\rightarrow X\times X$$ to a map $$\Delta'_X:X\rightarrow X\times X$$ satisfying $$(\pi\times \pi)\circ\Delta'_X=\Delta'_B\circ \pi$$

If we set $$X^k=\pi^{-1}(B^k)$$ then

$$(\pi\times \pi)\circ\Delta'_X(X^k)=\Delta'_B(B^k)\subseteq\bigcup_k B^{n-k}\times B^k$$

so that

$$\Delta'(X^k)\subseteq(\pi\times \pi)^{-1}\bigcup_k B^{n-k}\times B^k=\bigcup_k \pi^{-1}(B^{n-k})\times \pi^{-1}(B^k)=\bigcup_k X^{n-k}\times X^k.$$

Quotienting this to the smash returns the exact formula you have stated above.

This explains how the diagonal approximation lifts. Note that the stratification $$X^k$$ is not cellular, and is not a CW structure on $$X$$. The subset $$X^k$$ is simply the bit of $$X$$ lying over the $$k$$-skeleton of $$B_k$$. That said, as you have seen in your other post, the 'bit of $$X$$' that lies over a particular cell of $$B^k$$ has an easily understood structure, owing to the fact that the fibration is easily controlled over the contractible interior. If you have information about the structure of the typical fibre $$F$$, then you can piece things together to get a better idea of what each $$X^k$$ looks like.

If you want some examples to think about try: i) Take $$\pi:X\rightarrow\ast$$ the unique map. ii) The projection $$\pi=pr:X=X'\times B\rightarrow B$$. iii) Take $$\pi:X=S^n\rightarrow B=\mathbb{R}P^n$$ the covering projection. iv) Take $$\pi:EG\rightarrow BG$$ the milnor universal bundle for the compact Lie group $$G$$.

• Thanks a lot Tyrone, I will take some time to digest this! – CL. Apr 20 at 22:16