Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Good morning,

I have a question concerning the relative homotopy groups of a CW-pair as follows.

Let (X,A) be a CW-pair. What are results known for the relation between $\pi_{\ast}(X,A)$ and $\pi_{\ast}(X/A)$. If in addition, A is (n-1)-connected, do we have the equality $\pi_i(X,A) = \pi_i(X/A)$ for $i<n$? For $i\leq n$?

Any help is appreciated. Thanks in advance.

share|improve this question
4  
I believe what you want is called the "homotopy excision theorem". –  Justin Young Jan 25 '12 at 12:44
5  
See Proposition 4.28 in Hatcher for a reference. –  user17786 Jan 25 '12 at 14:00
    
Thank you very much. –  Đức Anh Jan 25 '12 at 14:04

1 Answer 1

up vote 1 down vote accepted

An answer is given by the Relative Hurewicz Theorem, for which a well known form can be stated as follows:

If $(X,A)$ is an $(n-1)$-connected pair, then the pair $(X \cup CA,CA)$ is $(n-1)$-connected and the morphism induced by inclusion

$$\pi_n(X,A) \to \pi_n(X \cup CA,CA) \cong \pi_n(X \cup CA)$$

is given by factoring out the action of $\pi_1(A)$ on $\pi_n(X,A)$.

Note that this implies $X \cup CA$ is $(n-1)$-connected and so the absolute Hurewicz Theorem implies $\pi_n(X \cup CA) \cong H_n(X \cup CA)$; and if $(X,A)$ has the HEP, then the map $X \cup CA \to X/A$ is a homotopy equivalence.

A proof of this form is given in

R. Brown and P.J. Higgins, ``Colimit theorems for relative homotopy groups'', J. Pure Appl. Algebra 22 (1981) 11-41,

and is shown to be a special case of a higher homotopy van Kampen Theorem, proved without using simplicial approximation or singular homology (but it uses a cubical higher homotopy groupoid!).

share|improve this answer
    
thank you very much –  Đức Anh Apr 26 '12 at 19:45

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.