Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been stuck in the following small detail which is part of the calculation of the $E^2$ term of the Serre Spectral Sequence.

Let $p: E\to B$ be a fibration where $B$ is a CW-complex. Denote by $E^p=p^{-1}(B^p)$ where $B^p$ is the p-skeleton of $B$. Let $\phi_{\alpha}: D^p \to B^p$ be the characteristic map of the p-cell $e^p_{\alpha}$. Let $p_{\alpha}: E_{\alpha}=\phi_{\alpha}^*(E) \to D^p$ be the pullback fibration and let $\partial E_{\alpha}=\partial \phi_{\alpha}^*(E) \to \partial D^p$ be the restriction. In his notes, Hatcher claims that by an excision argument we have an isomorphism \begin{eqnarray} \bigoplus_{\alpha} H_n(E_{\alpha},\partial E_{\alpha}) \to H_n(E^p,E^{p-1}). \end{eqnarray}

Can I have some help in seeing how the excision theorem is applied? It is probably an easy application but its not clear to me. Thanks!

share|cite|improve this question
I think I wrote this out in some detail here (look at Lecture 9, p. 3): – Dan Ramras Jun 24 '12 at 23:14
Thank you Dan! I have one more question, at the end of the calculation (page 5 of your notes) the simply connected hypothesis is used in the fact that there is a canonical isomorphism $H_n(\pi^{-1}(b_1)) \to H_n(\pi^{-1}(b_0))$. Why do we need these isomorphisms to be canonical? It looks like it is enough to have ANY isomorphism between the homology of the fibers to conclude that $E^1_{p,q}\cong \bigoplus_{\alpha}H_q(\pi^{-1}(b_{\alpha})) \cong \bigoplus_{\alpha}H_q(F)$ where $F$ is the fiber over a fixed pt. We know we always have such isomorphisms, but they may depend on the choice of path. – Manuel Jun 25 '12 at 4:21
The issue, which I didn't write out in the notes (but is addressed in some detail by Hatcher) is that you need to identify the $d_1$ differential with the cellular boundary map so that the $E^2$ page gives the ordinary homology groups. This is where the identifications start to really matter. – Dan Ramras Jun 25 '12 at 7:04
Right, some identifications are made but still it is not clear why we need a canonical map between the homology of the fibers. It is not even clear to me what canonical really means in this context and how it is used here. – Manuel Jun 25 '12 at 16:46
I'm not sure what to say, other than that these identifications are used in determining the differential. Canonical doesn't really have any precise meaning here, maybe, but if you tried to modify Hatcher's determination of the differential so as not to use these identifications it would at best get very confusing. – Dan Ramras Jun 26 '12 at 16:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.