Fiber-bundle : continuity of transition maps and inverse in general

Question

Let $(E,\pi,B)$ be a locally trivial fibration, with fiber a topological space $F$, $\Phi_i$ and $\Phi_j$ two trivializations over $U_i$ and $U_j$. The transition map from $i$ to $j$ is the homeomorphism $$\Phi_{ji} \colon U_i \cap U_j \times F \to U_i \cap U_j \times F, \quad (x,y) \mapsto (x,\phi_{ji,x}(y)), \quad \phi_{ji,x} \in Homeo(F).$$ Since $pr_2 \circ \Phi_{ji}$ is continuous, from the exponential law $Top(U_i \times F, F) \to Top(U_i \to F^F)$, with the compact-open topology on $F^F$ and its trace on $Homeo(F) \subset F^F$, there is a continuous map $$\phi_{ji} \colon U_i \cap U_j \to Homeo(F), \quad x \mapsto \phi_{ji,x}.$$ In general $Homeo(F)$ is not a topological group because in general the composition is not continuous (it is if $F$ is locally compact) and the inverse map is not continuous (it is if $F$ is compact, or non-compact but locally compact and locally connexe), and we cannot assume that the map $$\psi_{ji} \colon U_i \cap U_j \to Homeo(F), \quad x \mapsto \psi_{ji}(x) = (\phi_{ji,x})^{-1}$$ is continuous. But since $\Phi_{ji,x}$ is a homeomorphism, its inverse is the continuous map $$\Psi_{ij} \colon U_i \cap U_j \times F \to U_i \cap U_j \times F,\quad (x,y) \mapsto (x,\psi_{ij,x}(y)),\quad \psi_{ij,x} \in Homeo(F),$$ From the exponential law again, we get again a continuous map $$\psi_{ij} \colon U_i \cap U_j \to Homeo(F), \quad x \mapsto \psi_{ij,x},$$ and since it is obvious that $\psi_{ij,x} = (\phi_{ji,x})^{-1}$, we get that the map $\psi_{ij} \colon x \mapsto \psi_{ij,x} = (\phi_{ji,x})^{-1}$ is continuous.

So can we or can't we say that the map $x \mapsto (\phi_{ji,x})^{-1}$ is continuous in general ?

A similar reasonning can be made for composition : $x \in U_i \cap U_j \cap U_k \mapsto \phi_{ki,x} = \phi_{kj,x} \circ \phi_{ji,x}$ is continuous although the composition in $Homeo(F)$ is continuous only under conditions s.t. $F$ locally compact.

Is there something wrong hidden in what I say ? Or does that mean that the subset of $Homeo(F)$ image of the exponential (i.e. obtained through a continuous map like $\Phi_{ji}$) is in fact a topological group ?

Answer 1

I was thinking along the same lines, but I don't think it works. We do have a maps (using your notation) $$\phi_{ji} \colon U_i \cap U_j \to Homeo(F), \quad x \mapsto \phi_{ji,x}$$ and $$\psi_{ji} \colon U_i \cap U_j \to Homeo(F), \quad x \mapsto \psi_{ji}(x) = (\phi_{ji,x})^{-1}$$

But it doesn't necessarily mean that the map $$inv \colon Homeo(F) \to Homeo(F), \quad g \mapsto g^{-1}$$ is continuous, even if we replace $Homeo(F)$ by the subset/subgroup generated by all such maps $\phi_{ji,x}$.

We may also try to compose $\psi_{ji}^{-1} \circ \phi_{ji}$ to get a continuous map $Homeo(F) \to Homeo(F)$, but that doesn't work either, since $\phi_{ji}$ isn't necessarily injective.

The larger question we're working towards is whether the subset of $Homeo(F)$ generated by the $\phi$ maps is a topological group, even if $Homeo(F)$ isn't. Unfortunately, I don't think this is the case.

Let $F$ be a topological space and $Homeo(F)$ be its homeomorphism group with the compact-open topology, which may not be a topological group, i.e. whose composition or inversion maps may not be continuous.

Then consider the trivial bundle $Homeo(F) \times F \to F$. This has the obvious global trivialization \begin{align*} \Phi_{Id}: Homeo(F) \times F &\to Homeo(F) \times F \\ (p, x) &\mapsto (p, x) \end{align*} but for every $f \in Homeo(F)$, we have another global trivialization \begin{align*} \Phi_f: Homeo(F) \times F &\to Homeo(F) \times F \\ (p, x) &\mapsto (p, f(x)) \end{align*} with inverse $\Phi_{f^{-1}}$; note that the continuity of the inverse map doesn't come into play here, since $f$ isn't a function argument we're inverting.

Then the transition map of $\Phi_f \circ \Phi_{Id}^{-1}$ is simply $f$ itself, so we see that the subset of $Homeo(F)$ generated by the transition maps is simply $Homeo(F)$ itself. Therefore, if $Homeo(F)$ isn't a topological group, then this bundle shows that there exists a bundle which generates a subgroup of $Homeo(F)$ that also isn't a topological group.