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 $G$.

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

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