Well-definedness of the action of the structure group of a principal bundle on the total space.

Question

Find the definition of a fiber bundle here- Definition of Fiber Bundle

I am having difficulty in proving that the natural action of $K$ on $X$ is well-defined:

Let us recall how does K acts on X $(x,g)\in X \times K$ consider the fibration $\varphi_{U}$ over $U$ such that $p(x)\in U$ So $x\in p^{-1}(U)$. Let $(p(x),h)=\varphi_{U}^{-1}(x)$ So,send $(x,g)$ to $\varphi_{U}(p(x),gh)$.

Let $x\in V$ also. Let $(p(x),h_1)=\varphi_{V}^{-1}(x)$.

How do we show that- $\varphi_{U}(p(x),gh)=\varphi_{V}(p(x),gh_1)$?

Also can someone give any hints how to show the orbit space is isomorphic to B.

Answer 1

By definition of fiber bundle, there exists $\theta_{U,V}:U\cap V\to K$ such that $\phi_V(b,k)=\phi_U(b,k\cdot\theta_{U,V}(b))$ on $p^{-1}(U\cap V)$. Evaluating this equality on $(p(x),h_1)$, we get: $$\phi_U(p(x),h)=x=\phi_V(p(x),h_1)=\phi_U(p(x),h_1\cdot\theta(p(x)))$$

This means that $\theta_{U,V}(p(x))=h_1^{-1}\cdot h$. In particular,

$$\phi_V(p(x),gh_1)=\phi_U(p(x),gh_1\theta_{U,V}(p(x)))=\phi_U(p(x),gh_1h_1^{-1}h)=\phi_U(p(x),gh)$$