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The definition of fiber bundle can be found from here: Definition of Fiber Bundle

Then Bredon defines Principal bundle in the exercise as follows:

enter image description here

I am not able to show how K acts naturally on X and the rest of the exercise.

My try was-

We need to get a map $X \times K \rightarrow X$. Let $(k,x) \in K\times X$.

Then $p(x)\in B$ and by the trivial fibration condition there exists $U \ni p(x)$ such that there is homeomorphism $\varphi :U\times F \rightarrow p^{-1}(U)$ .

Send $(k,x)$ to $\varphi(p(x),k)$.

Note: By definition of Principal Bundle F and K are same.

Am I correct?

How to prove the next part of the exercise?

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I dont think you have defined an action yet. If we take $g \in K$ then we use the map $\varphi$ to define the action where if $(u,h) \in U\times K$ then $$g(u,h)=(u,gh)$$ now you need to use the fact that $K$ is the structural group to show this this is well defined.

The orbit space is $B$ since $K$ acts transitively.

If you have a section, set the values equal to the group identity element and use this to define a homeomorphism with the trivial bundle.

Both these last statement need to have the details supplied.

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Given an element $(x,k)\in X\times K$ we have to define an element of $X$. How does another element $g\in K$ come into picture? – Babai Jul 16 '14 at 15:59
I didnt understand your notation, you are still not defining an action. How should $K$ act n a fibre ? It should act by group multiplication. In your definition $k$ send the entire fibre to the element $k$, and this is not invariant under change of coordinates. – Rene Schipperus Jul 16 '14 at 16:13
K acts on F by right translation, yes that is correct. But then what is the action of K on X ? And K and F are same by definition. – Babai Jul 16 '14 at 16:20
I changed my notation to avoid confusion. – Rene Schipperus Jul 16 '14 at 16:20
I guess you want to say this- for $(x,g)\in X\times K$ consider the fibration $\varphi$ over $U$ such that $p(x) \in U$ So $x\in p^{-1}(U)$. Let $(u,h)=\varphi ^{-1}(x)$ So,send $(x,g)$ to $\varphi (x,gh)$. Am I correct? – Babai Jul 16 '14 at 16:31

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