Locally trivial fibration over a suspension

Question

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 $x_0$ ?

Answer 1

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.