# Locally trivial fibration over a suspension

For $$X$$ a paracompact space, I am trying to classify all locally trivial fibration with base the suspension $$SX = X \times [-1,1]\, /\, (X \times \{-1\} \cup X \times \{1\})$$, and fiber-type a space $$F$$ such that $$G_F = Homeo(F)$$ with the C.O. topology is a topological group.

I duplicate the reasonning for classification of fibration with fiber-type $$F$$ over the sphere $$S^n$$. In that case, denoting the set of isomorphism classes of such fibrations by $$[FB(S^n,F)]$$, we get that $$[FB(S^n,F)] \simeq \pi_{n-1}(G_F) / \pi_{0}(G_F),$$ where $$\pi_{n-1}(G_F)$$ is the set of pointed homotopy classes $$(S^{n−1}, ∗) → (G_F , Id_F )$$, and the action of $$\pi_{0}(G_F)$$ on $$\pi_{n-1}(G_F)$$ is induced by the action of $$G_F$$ on itself by conjugaison.

In the case of base $$SX$$, we have $$SX = CX_+ \cup CX_-$$ and $$X \simeq CX_+ \cap CX_-$$, with the cones $$CX_{\pm}$$ being paracompact and contractile, so every fibration over $$CX_+$$ or $$CX_-$$ is trivial. If $$\Phi_{\pm}$$ are the corresponding trivializations, $$f_0 \in G_F$$ a fixed homeomorphism of $$F$$, and $$x_0 \in X$$ an arbitrary base-point, we can always choose $$\Phi_{\pm}$$ s.t. $$\Phi_+(x_0) = \Phi_-(x_0) = f_0$$, and so the transition function from $$\Phi_+$$ to $$\Phi_-$$ determine an application $$\phi \colon (X,x_0) \to (G_F,Id_F)$$ As in the case of $$S^n$$, the action of $$G_F$$ on itself by conjugaison induces an action on the homotopy classes $$[\phi]$$ which depends only of the path-component of the element in $$G_F$$. Lets denote the set of isomorphism classes of fibration with fiber $$F$$ by $$[FB(SX,F)]$$, and the set of equivalence classes of homotopy class under the action of $$\pi_{0}(G_F)$$ by $$[(X,x_0),(G_F,Id_F)]_* = [(X,x_0),(G_F,Id_F)]\,/\,\pi_{0}(G_F)$$ For $$[\xi] \in [FB(SX,F)]$$ and $$[\phi]_* \in [(X,x_0),(G_F,Id_F)]_*$$, we can define an application $$\theta \colon [FB(SX,F)] \to [(X,x_0),(G_F,Id_F)]_*, \quad [\xi] \mapsto [\phi]_*$$ This is well-defined : choosing another set of trivializations $$\Psi_{\pm}$$ gives a function $$\psi$$ s.t. $$[\psi]_* = [\phi]_*$$, changing the choice of the value at $$x_0$$ from $$f_0$$ to $$g_0$$ also gives applications in the same equivalence class ; it is also easy to show that two isomorphic fibration $$\xi$$ and $$\xi'$$ gives function in the same equivalence class.

Now, $$\theta$$ is a bijection : it is surjective because from an application $$\phi\colon (X,x_0) \to (G_F,Id_F)$$ we can reconstruct a fibration over $$SX$$ with $$\phi$$ as its transition function unique up to iso, an injective because if $$\xi$$ gives $$\phi$$ and $$\xi'$$ gives $$\phi'$$ and $$[\phi]_* = [\phi']_*$$, we can modify $$\phi$$ s.t. $$\phi$$ and $$\phi'$$ are homotopic by an homotopy $$H$$ and construct a fibration $$\xi_H$$ over $$SX \times [0,1]$$, and $$SX$$ being paracompact we get $$\xi \simeq \xi_{\psi}|_{SX \times {0}} \simeq \xi_{\psi}|_{SX \times {1}} \simeq \xi′$$ So we apparently get that $$[FB(SX,F)] \simeq [(X,x_0),(G_F,Id_F)]_*$$

But here $$[FB(SX,F)]$$ cannot depend of the choice of the base-point, but it seems to me that $$[(X,x_0),(G_F,Id_F)]_*$$ does depend on that choice if $$X$$ is not a homogeneous space like $$S^n$$.

So I think there must be a mistake somewhere, but where ??

Or contrary to the intuition, does $$[(X,x_0),(G_F,Id_F)]_*$$ is independant of the choice of $$x_0$$ ?

## 1 Answer

It is independent of the choice of base point.

Let $$Map((X,x_0), (G_F,id))$$ be the based mapping space (based at $$x_0$$). Let $$Map(X, G_F)$$ be the free mapping space. Then we have a split short exact sequence of topological groups:

$$Map((X,x_0), (G_F,id)) \to Map(X, G_F) \stackrel{ev_{x_0}}{\to} G_F$$

(These are groups using pointwise multiplication). The last map, which evaluates at $$x_0$$, is split by identifying $$G_F$$ with the constant maps. Note that the inclusion of the constant maps doesn't depend on the choice of basepoint.

In any case this induces a split short exact sequence on the corresponding groups of path components: $$\pi_0Map((X,x_0), (G_F,id)) \to \pi_0Map(X, G_F) \stackrel{ev_{x_0}}{\to} \pi_0G_F$$

Now the group structure on $$\pi_0Map((X,x_0), (G_F,id))$$ might depend on identifying it as the kernel of the map $$ev_{x_0}$$, which of course depends on the base point.

However we only care about it as a $$\pi_0G_F$$-set. From the above split exact sequence it follows that as sets with $$\pi_0G_F$$-action $$[(X,x_0), (G_F,id)] = \pi_0Map((X,x_0), (G_F,id))$$ is isomorphic to the $$\pi_0G_F$$-set of, say, left cosets: $$\pi_0Map((X,x_0), (G_F,id)) \cong \pi_0Map(X, G_F)/ \pi_0G_F$$ The right-hand side is independent of the choice of basepoint $$x_0$$, which tells us, perhaps surprisingly, that already $$[(X,x_0), (G_F,id)]$$ is independent of basepoint.

It further follows that the quotient of this set by $$\pi_0G_F$$, which is your $$[(X,x_0), (G_F,id)]_*$$, is also independent of basepoint.

• Nice ! thx a lot. – ychemama Mar 19 at 14:55