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The following is the result of some "idle musings". I feel like it should be classical, but I do not even know what to look for.

Recall that in spaces (say, pointed topological spaces) there is a notion of a fibration sequence

$ F_f \xrightarrow{e} X \xrightarrow{f} Y.$

I do not want to recall the definition, but let me recall the following facts:

  1. $F_f$ is determined in an appropriate sense by $f$ (i.e. given any map $f$ there is an essentially unique way of completing to a fibration sequence).
  2. There is a canonical equivalence $F_e \simeq \Omega Y$, where $F_e \to F_f \xrightarrow{e} X$ is another fibration sequence.
  3. The induced sequence $\pi_0(F_f) \to \pi_0(X) \to \pi_0(Y)$ is exact. (Of course, exactness of a sequence of sets does not make incredibly much sense, but take $X$ and $Y$ to be loop spaces and $f$ a loop map if you want.)

From this it follows in particular that we get a long exact sequence relating the homotopy groups of $X, Y$ and $F_f$.

Now recall that given a space $X$, we can define the fundamental groupoid $\pi_{0,1}(X)$. This is a category with objects the points and morphisms the homotopy classes of paths. If $X$ is a loop space this can be made into a monoidal category in which every object is invertible, and if $X$ is a double loop space even a symmetric monoidal category. Now there is a notion of exact sequence for symmetric monoidal groupoids in which every object is invertible (see e.g. [1]).

My idle question is now the following: Given a morphism of spaces $f: X \to Y$, can we "complete" this to a sequence $\tilde{F}_f \xrightarrow{e} X \to Y$ satisfying the following properties:

  1. $\tilde{F}_f$ is determined by $f$ in an appropriate sense
  2. $\tilde{F}_e \simeq \Omega^2 Y$
  3. If $X, Y$ are double loop spaces and $f$ is a double loop map, then $\tilde{F}_f$ is also a double loop space, $e$ is a double loop map, and the sequence of symmetric monoidal groupoids in which every object is invertible $\pi_{0,1}(\tilde{F}_f) \to \pi_{0,1} X \to \pi_{0,1} Y$ is exact.

In particular then, the sequence $\tilde{F}_f \to X \to Y$ would induce a long exact sequence of "homotopy groupoids"

$\dots \to \Pi_n(\tilde{F}_f) \to \Pi_n(X) \to \Pi_n(Y) \to \Pi_{n-1}(\tilde{F}_f) \to \dots $

Here I put $\Pi_n(X) := \pi_{0,1}(\Omega^{2n} X)$. At the risk of rambling a bit, recall that any symmetric monoidal groupoid $E$ has two associated abelian groups, $\pi_0(E) = Pic(E)$ and $\pi_1(E) = Aut(1)$. Now for a loop space $X$ we get $\pi_i(\pi_{0,1}(X)) = \pi_i(X)$ for $i \in \{0,1\}$, so $\pi_0(\Pi_n(X)) = \pi_{2n}(X)$ and $\pi_1(\Pi_n(X)) = \pi_{2n+1}(X)$. At least for the definition of "exact sequence of symmetric monoidal groupoids in which every object is invertible" of [1] it is the case that this induces exact sequences on both $\pi_0$ and $\pi_1$, so the above long exact sequence automatically yields

$\dots \to \pi_{2n+2}(X) \to \pi_{2n+2}(Y) \to \pi_{2n}(\tilde{F}_f) \to \pi_{2n}(X) \to \pi_{2n}(Y) \to \dots$

$\dots \to \pi_{2n+3}(X) \to \pi_{2n+3}(Y) \to \pi_{2n+1}(\tilde{F}_f) \to \pi_{2n+1}(X) \to \pi_{2n+1}(Y) \to \dots$

[1] Enrico M. Vitale: A Picard-Brauer exact sequence of categorical groups

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  • 2
    $\begingroup$ The usual fiber satisfies your condition 3, and in fact the map $\pi_{0,1}F_f\to ker (\pi_{0,1}f)$ is an iso on $\pi_0$ and an epi on $\pi_1$. I think it follows from this that the canonical map $\tilde F_f \to F_f$ is an effective epimorphism, and it would follow from condition 2 that there is always an effective epimorphism $\Omega^2Y\to\Omega Y$, a contradiction... $\endgroup$ – Marc Hoyois May 5 '17 at 17:43

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