Gauge group of a smooth principal bundle

Let $\pi :P\to M$ be a smooth right principal $G$-bundle as in Smooth principal $G$-bundle. A smooth gauge transformation of $P$ is a smooth diffeomorphism $\mathrm{\Phi }:P\to P$ satisfying $\pi \circ \mathrm{\Phi }=\pi ,\mathrm{\Phi }(pg)=\mathrm{\Phi }(p)g(p\in P,g\in G)\mathrm{.}$ Here diffeomorphism has the meaning of Diffeomorphisms and local diffeomorphisms of manifolds. The gauge group $\mathcal{G}(P)$ is the set of these transformations under composition. The identity is a smooth diffeomorphism and composites of smooth maps are smooth by Identity maps and composites of smooth maps are smooth; composites of diffeomorphisms are diffeomorphisms because their inverses are composites of the smooth inverses. The identity covers $M$ and commutes with the right action. If $\mathrm{\Phi },\mathrm{\Psi }\in \mathcal{G}(P)$, then $\mathrm{\Phi }\circ \mathrm{\Psi }$ covers $M$ and $(\mathrm{\Phi }\circ \mathrm{\Psi })(pg)=\mathrm{\Phi }(\mathrm{\Psi }(p)g)=\mathrm{\Phi }(\mathrm{\Psi }(p))g$, so the conditions are closed under composition. The inverse of such a $\mathrm{\Phi }$ also covers $M$: if $q=\mathrm{\Phi }(p)$ then $\pi ({\mathrm{\Phi }}^{-1}(q))=\pi (p)=\pi (q)$. Equivariance gives ${\mathrm{\Phi }}^{-1}(qg)={\mathrm{\Phi }}^{-1}(\mathrm{\Phi }(p)g)={\mathrm{\Phi }}^{-1}(\mathrm{\Phi }(pg))=pg\mathrm{.}$ Its smoothness is part of the diffeomorphism condition. Hence these maps form a group under the composition law of Group and abelian group.

The structure group $G$ acts on each fibre of $P$, while $\mathcal{G}(P)$ acts on the total bundle by automorphisms over the identity map of $M$. They are different groups: a gauge transformation may vary from fibre to fibre, and this definition assigns no topology or finite-dimensional Lie-group structure to $\mathcal{G}(P)$.