MathOverflow is a question and answer site for professional mathematicians. 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

Let $f: X \to Y$ be a fibration of pointed Kan complexes, and let $F$ be the fiber.

Question: How do you prove that the following diagram of homotopy groups commutes?:

$\pi_n(Y) \to \pi_{n-1}(\Omega Y)$

$\ \ \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow$

$\pi_{n-1}(F) \to \pi_{n-2}(\Omega F)$

Admittedly, I don't know for certain that it commutes, but it looks like it should.

All the arrows are boundary maps ($\delta$) from long exact sequences of homotopy groups for a fibration. The horizontal maps are isomorphisms.

The definition of $\delta$ that I'm using is that for for a fibration $X \to Y$, $\alpha \in \pi_n(Y)$, $\delta ([\alpha]) = [\beta d^0]$, where $\beta:\Delta^n \to X$ fits into the following diagram:

$ \Lambda^n_0 \to^* X$

$\ \downarrow \ \ \ \ \ \downarrow$

$\Delta^n \to^\alpha Y$

So far, I've attempted to chase elements around this diagram and use prismatic arguments, but I haven't found one that works.

$\Omega F \to \Omega X \to \Omega Y$

$\ \ \downarrow\ \ \ \ \ \ \downarrow\ \ \ \ \ \ \ \downarrow$

$PF \to PX \to PY$

$\ \ \downarrow\ \ \ \ \ \ \downarrow\ \ \ \ \ \ \ \downarrow$

$\ \ F \to \ \ X \to \ \ Y$

It feels like a useful fact that $PX \to PY\times_Y X$ is a fibration, which is true since pointed simplicial sets form a simplicial model category.

Edit: The path and loop spaces I'm using are defined by the following: For pointed simplicial sets X and Y, define the simplicial set $hom_*(X,Y)$ to have n- simplices $hom_{sSet_*}(X \wedge \Delta^n_+,Y)$, where $\Delta^n_+$ is the standard n-simplex with a disjoint basepoint.

Then I'm using $PX=hom_*(\Delta^1 , X)$

and $\Omega X=hom_*(\Delta^1/\partial \Delta^1,X)$

share|cite|improve this question
You haven't said what model for the loop space you're using... – Dan Ramras Sep 14 '10 at 23:55
Thanks, I've added that now. – user9109 Sep 15 '10 at 1:55
up vote 1 down vote accepted

The boundary map of homotopy groups is induced by a map $\partial : \Omega Y \to F$ of spaces; then the commutativity follows from the naturality of the isomorphism $\pi_n \circ \Omega \cong \pi_{n+1}$.

share|cite|improve this answer
Thanks, I found maps of spaces $\Omega Y \to PB\times_B E ^{\simeq\atop \leftarrow} F$, which cut my diagram in half and makes both parts commute. – user9109 Sep 15 '10 at 14:03

Your Answer


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.