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Bredon defines bundle projection in the following way:

$\bf13.1.$ Definition. Let $X,B$ and $F$ be Hausdorff spaces and $p:X\to B$ a map. Then $p$ is called a bundle projection with fiber $F$, if each point of $B$ has neighborhood $U$ such that there is a homeomorphism $\phi:U\times F\to p^{-1}(U)$ such that $p(\phi\langle b,y\rangle)=b$ for all $b\in U$ and $y\in F$. That is, on $p^{-1}(U)$, $p$ corresponds to projection $U\times F\to U$. Such a map $\phi$ is called a trivialization of the bundle over $U$.

Then he defines Fibre Bundle

$\bf13.2.$ Definition. Let $K$ be a topological group acting effectively on the Hausdorff space $F$ as a group of homeomorphisms. Let $X$ and $B$ be Hausdorff spaces. By a fiber bundle over the base space $B$ with total space $X$, fiber $F$ and structure group $K$, we mean a bundle projection $p:X\to B$ together with a collection $\Phi$ of trivializations $\phi:U\times F\to p^{-1}(U)$, of $p$ over $U$, called charts over $U$ such that:

  1. each point of $B$ has a neighborhood over which there is a chart in $\Phi$;
  2. if $\phi:U\times F\to p^{-1}(U)$ is in $\Phi$ and $V\subset U$ then the restriction of $\phi$ to $V\times F$ is in $\Phi$;
  3. if $\phi$, $\psi\in\Phi$ are charts over $U$ then there is a map $\theta:U\to K$ such that $\psi\langle u,y\rangle=\phi\langle u,\theta(u)(y)\rangle;$ and
  4. the set $\Phi$ is maximal among collections satisfying 1, 2, and 3.

The he remarks about the condition 3. He says the map $\theta :U \rightarrow K $ exists. The only important thing we demand is the continuity of $\theta$.

And the justification is

$\quad$ Let us investigate the meaning of condition $(3)$ of Definition 13.2. Given charts $\phi$ and $\psi$ over $U$, $\phi^{-1}\psi:U\times F\to U\times F$ is a homeomorphism commuting with the projections to $U$. Thus we can write $$\phi^{-1}\psi\langle u,y\rangle=\langle u,\mu\langle u,y\rangle\rangle,$$ where $$\mu:U\times F\to F$$ is the composition $p_F\circ\phi^{-1}\psi$ with the projection $p_F:U\times F\to F$, and hence is continuous. Then $\theta:U\to K$ is given by $$\theta(u)(y)=\mu\langle u,y\rangle.$$

My confusion is the map $\theta : U \rightarrow K$ is not very clear to me. He defines how $\theta(u)$ will act on an element of $y\in Y$ $\theta(u)(y)=\mu \langle u,y\rangle$. From that why should we get a map from U to K .

I think the effectiveness of the action of $K$ on $F$ is vital. ( I guess one should use the fact $K$ acts on $F$ as same as there is an group homomorphism from $K$ to $\operatorname{Aut}(F)$ and the map is injective because the action is effective.)

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1 Answer 1

Given $u \in U$ we have defined $\theta(u)$ as a map that takes $y \in F$ to $\mu \langle u,y \rangle \in F$. Thus $\theta(u): F \to F$ certainly belongs to $\operatorname{Maps}(F,F)$, and $\theta : U \to \operatorname{Maps}(F,F)$. What remains to be seen is that the image $\theta(u)$ is in fact a homeomorphism living in $K \subset \operatorname{Maps}(F,F)$, but this is clear since the inverse map is supplied by reversing the roles of $\psi$ and $\phi$.

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Sorry, but it is not clear to me yet. As I understand $\theta (u)$ will be an element of $Aut(F)$ by reversing the role of $\phi$ and $\psi$ . Now, $K\hookrightarrow Aut(F)$ as the action of $K$ on $F$ is effective. But why Should the $\theta :U \rightarrow Aut(F)$ factors through K –  Susobhan Jul 15 at 14:37
    
I think the statement is misleading. Actually the definition of Fiber Bundle demands the transition map from $U\rightarrow Aut(F)$ to land inside K. (Remember k sits inside Aut(F), as the action of K on F is effective). One can refer the definition of fiber bundle from Steenrod's "Topology of Fibre Bundles" or from the Lecture notes of James F. Davis and Paul Kirk (Page 78-79). –  Susobhan Aug 6 at 6:35

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