9
$\begingroup$

Let $p:E \rightarrow B$ be a fibration with fiber $F$ . Associated to this we have a long exact sequence $$\cdots \rightarrow \pi_n(F) \rightarrow \pi_n(E) \rightarrow \pi_n(B) \rightarrow \pi_{n-1}(F) \rightarrow \cdots.$$

I am trying to show that the image of $\pi_2(B) $ in $\pi_1(F)$ is in the center of $\pi_1(F)$, but with no luck. Any help, solution or reference is welcome!

$\endgroup$
2
$\begingroup$

There is a general result due to Quillen that for a (pointed) fibration $F \to E \to B$ the map $\pi_1(F) \to \pi_1(E)$ may be given the structure of crossed module. This is a variation of the fact due to J.H.C. Whitehead that for a pointed pair of spaces $(X,A)$ the boundary map $\partial: \pi_2(X,A) \to \pi_1(A)$ may be given the structure of crossed module. Recall that a crossed module is a morphism of groups $\mu: M \to P$ together with an action of the group $P$ on say the right of the group $M$ written $(m,p) \mapsto m^p$ satisfying the two rules

  1. $\mu(m^p)= p^{-1} \mu(m) p$;

  2. $n^{-1}mn= m^{\mu n}$

for all $m,n \in M, p \in P$. A standard property of such a crossed module is that the kernel of $\mu$ lies in the center of $M$.

For more see the paper

Loday, J.-L. "Spaces with finitely many nontrivial homotopy groups". J. Pure Appl. Algebra 24 (1982) 179--202. and also Section 2.6 of the book Nonabelian Algebraic Topology, EMS Tract Vol 15, (2011). (Loday uses left actions. Beware that Mac Lane's book CFTWM, second edition, in its last section, omits the second axiom for a crossed module.)

Another way to think of this is to use the fibration property to show that there is an action of $\Omega E$ on $F$ satisfying the crossed module axioms up to homotopy, but I do not have a reference for that.

$\endgroup$
  • $\begingroup$ Thank you for your answer. I looked at your book and the pages suggested, there it is claimed that the action of $\pi_1(E)$ on $\pi_(F)$, given by, for $\alpha \in \pi_1(F)$ and $\beta \in \pi_1(E)$ , by choosing a lift $\tilde{H}$ of a nullhomotopy H of $p(\beta^{-1}\alpha\beta)$ and finally, defining the action $\alpha \cdot \beta := \tilde{H}_1$. I managed to verify that this is independent of lifts, homotopies ofthe loops, but I can't seem to see why this is independent on the choice of nullhomotop H. If you hadthe time to explain further, I would be very grateful. $\endgroup$ – user161954 Jul 16 '14 at 18:33
  • $\begingroup$ So, sorry for being so unclear but what I am asking is basically: When defining the action of $\pi_1(E)$ on $\pi_1(F)$ as on pg. 52 in your book - is the action independen of the choice of $H$, the nullhomotopy of $p(\alpha \mu \alpha^{-1})$? $\endgroup$ – user161954 Jul 16 '14 at 20:35
  • $\begingroup$ @user161954: I do not have an immediate answer to your question which is a good one, partly as I see this area more from the double groupoid viewpoint, which is pursued in Chapter 6 of the book. You might also like to look at [129] on my publication list, but that does not work at the space level. I'll think on it. $\endgroup$ – Ronnie Brown Jul 16 '14 at 20:40
  • $\begingroup$ Wrt my last comment, I should say "is not set up at the space level" rather than "does not work". $\endgroup$ – Ronnie Brown Jul 16 '14 at 20:50
  • $\begingroup$ @user161954: Actually what you ask is not quite so direct, and is verified further on in that section using the notion of cat$^1$-group. Crossed modules are equivalent to cat$^1$-groups, and also to (edge symmetric) double groupoids with connections and one vertex, see Chapter 6. It is useful to understand all of these to see clear proofs. The "narrowest" concept, that of crossed module, is often not the easiest in which to write down proofs. $\endgroup$ – Ronnie Brown Jul 17 '14 at 20:50
1
$\begingroup$

If $\alpha \colon S^2 \to B$, then the image $[\psi]$of $[\alpha]$ under the connecting homomorphism $\pi_2(B) \to \pi_1(F)$ is defined by the diagram below:

$$\require{AMScd} \begin{CD} S^1 @>{i}>> D^2 @>{q}>> S^2\\ @VV{\psi}V @VV{\hat{\alpha}}V @VV{\alpha}V\\ F @>{i}>> E @>{p}>> B \end{CD} $$

where the map $i \colon S^1 \to D^2$ is the inclusion of the boundary, the map $q \colon D^2 \to S^2$ is the quotient map taking $\partial D^2$ to a point, $p\hat{\alpha} = \alpha q$, and $\hat{\alpha} i = i \psi$. The map $\psi$ is well-defined up to homotopy.

Let $[\beta] \in \pi_1(F)$ have representative $\beta \colon S^1 \to F$ and consider $\beta.\psi.\bar{\beta}$ where $.$ denotes path concatenation and $\bar{\beta}$ is the reverse of the path $\beta$. Then $i(\beta.\psi.\bar{\beta}) = (i \beta).(i\psi).(i\bar{\beta}) = (i \beta).(\hat{\alpha}i).(i\bar{\beta})$.

From here, I think you can use a trivializing neighborhood for the basepoint in $B$ to argue that the terms in the latter expression commute (up to homotopy) and have that $i(\beta.\psi.\bar{\beta}) = \hat{\alpha}i$ (up to homotopy), but these details are eluding me at the moment. Hence $[\beta.\psi.\bar{\beta}] = [\psi] \in \pi_1(F)$.

$\endgroup$
  • $\begingroup$ @wckronholm: Why does $i(\beta.\psi.\beta)=\hat{\alpha}i$ imply that $[\beta.\psi.\beta]=[\psi]$? $\endgroup$ – user161954 Jul 14 '14 at 0:04
  • $\begingroup$ @user161954 because then both $\beta.\psi.\hat{\beta}$ and $\psi$ satisfy the same defining property given by the diagram and so are homotopy equivalent. $\endgroup$ – wckronholm Jul 14 '14 at 0:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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