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 $\Phi_{ji}$ 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 ?

  • $\begingroup$ It seems you just showed that while continuity of the inverse doesn't follow from the structure of Homeo(F) alone, it does follow from the specific type of maps we get from fiber bundles. $\endgroup$ – user2520938 Jan 22 at 10:51

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