Transition maps of a principal bundle are smooth A smooth principal fiber bundle is a smooth fiber bundle $\pi: E \to M$ together with a Lie group $G$ and a fiber preserving right action $E \times G \to E$ which restricts to each fiber freely and transitively.
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How prove that the map $g_{\alpha \beta}: U_{\alpha}\cap U_{\beta}\to G$ is smooth, using the implicit function theorem?
 A: Here's an answer which doesn't use the implicit function theorem, but at least proves $g_{\alpha\beta}$ is smooth.  I'm going to use the notation $U_{\alpha\beta}$ to denote $U_\alpha \cap U_\beta$, just to save on typing.
First, I'm assuming the $U_{\alpha}$ are a trivializing cover of $E$.  So, in particular, $\pi^{-1}(U_{\alpha\beta})$ is diffeomorphic to $U_{\alpha\beta} \times G$ via a projection preserving $G$-equivariant diffeomorphism.  Choosing such a diffeomorphism $\psi$ once and for all, we may assume wlog that $s_\alpha, s_\beta: U_{\alpha\beta}\rightarrow U_{\alpha\beta}\times G$ and that the action of $G$ on $U_{\alpha\beta}\times G$ is given by right multiplication on the $G$ factor.
Since $s_\alpha$ and $s_\beta$ are sections, we have $s_\alpha(x) = (x, f_\alpha(x))$ and $s_\beta(x) = (x,f_\beta(x))$.  Both $f_\alpha$ and $f_\beta$ are smooth maps, since, for example, using $\pi_2:U_{\alpha\beta}\times G\rightarrow G$ for the projection, $f_\alpha = \pi_2\circ s_\alpha$ is a composition of smooth maps.
Finally, set $g_{\alpha\beta}(x) = (f_\beta(x))^{-1}\cdot f_\alpha(x)$ (where the inverse is the group inversion, and $\cdot$ refers to group multiplication.
(Technically, one can prove that if $M$ is a manifold with smooth multiplication $\mu:M\times M\rightarrow M$ satisfying the usual group axioms, then the inversion map $i:M\rightarrow M$ is automatically smooth, so $M$ is a Lie group.  The proof uses the implicit funtion theorem.)
Since group inversion and multiplication are smooth, it follows that $g_{\alpha\beta}$ is smooth.
It remains to show that the thing I called $g_{\alpha\beta}$ is the same as your $g_{\alpha\beta}$.  That is, we must show that $s_\alpha(x) = s_\beta(x) g_{\alpha\beta}$.  But we have \begin{align*} s_\beta(x) g_{\alpha\beta}(x) &= (x,f_\beta(x))\ast (f_{\beta}(x))^{-1} f_\alpha(x)\\ &= (x,f_\beta(x) (f_{\beta}(x))^{-1} f_\alpha(x))\\ &= (x,f_\alpha(x))\\ &= s_\alpha(x).\end{align*}
