# Serre Spectral sequence [Hatcher]

I'm having trouble understanding Serre Spectral sequence given in Hatcher. https://pi.math.cornell.edu/~hatcher/AT/ATch5.pdf

In the beginning of the section, (Page 526 of the book, or rather, Pg 9 of the pdf file) he writes "Since $$(B, B^p)$$ is $$p$$-connected, the HLP implies that $$(X, X_p)$$ is also $$p$$-connected..."

I'm unable to show this simple(?) statement - maybe I'm overthinking it, or I've missed something.

What I've tried :

We need to show that the relative homotopy groups are 0. So, let $$f \in [D^n, S^{n-1}; X, X_{p}]$$. (where $$n < p$$) Then $$\phi = \pi \circ f \in [D^n, S^{n-1}; B, B^{p}]$$. Since $$(B, B^p)$$ is $$p$$-connected, $$\phi$$ is homotopic to a constant map, say $$e_{b_0}$$. ($$e_{b_0}$$ is the constant map that sends everything to $$b_0 \in B$$.) Let the homotopy be $$F$$. Then, by the HLP, we get a map $$G:D^n \times I \rightarrow X$$ such that $$G\circ i = f$$ (or $$G(x,0) = f(x) \forall x \in D^n$$) and $$\pi \circ G = F$$, i.e. $$\pi \circ G(x,1) = F(x,1) = e_{b_0}$$.

(Basically I've used the fact that fibration => Serre's fibration)

Now I want to show that $$G$$ is the required homotopy between $$f$$ and the constant map, but I'm not able to see why.

Also, in Theorem 5.3, how is $$X$$ filtered if $$B$$ is not a CW complex?

• Hint: The restriction $X_p\rightarrow B_p$ is a pullback. What do you know about the fibres of the maps in a pullback square? – Tyrone May 8 at 16:55
• I haven't checked theorem 5.3, but the standard way to get a filtration in this case is to take a CW approximation. – Tyrone May 8 at 16:56