Let $\pi: P \rightarrow M$ be a principal $G$ bundle. Then the gauge group is defined as the automorphisms of the bundle lifting the identity on $M$.
Suppose that $M$ is contractible, then the bundle is trivial and also the associated bundles created via a representation of $G$. In particular the bundle $Ad P = P \times_{Ad} G$ is trivial, where $Ad$ denotes the adjoint action of $G$ on itself. The gauge trasformations are also the smooth sections of $Ad P$, and therefore in the case of $M$ contractible just smooth maps from $M$ to $G$.
Given a smooth map $f':M \rightarrow G$, I am trying to reconstruct the bundle automorphism $F: P \rightarrow P$.
The obvious candidate $F(p) := p f'(\pi(p))$ only works in the case of $G$ abelian, as otherwise it is not equivariant.
So I thought the following: from a $G$-equivariant map $\hat{f}:P \rightarrow G$ one can construct $F$ by, $F(p):=p \hat{f}(p)$ so we just need to construct $\hat{f}$. For this, let $e : M \rightarrow P$ be a global section of $P$. Define $f: M \rightarrow Ad P$ by $f(m)=[e(m),f'(m)]$ and from this section obtain the $G$-equivariant map $\hat{f}:P \rightarrow G$ defined by $\hat{f}(p) = g^{-1} f'(m) g$ where $g \in G$ is such that $e(\pi(p)) g = p$.
When $M$ is contractible as above, is there any other simpler way to construct the bundle automorphism corresponding to the smooth map $M \rightarrow G$ (or the bijection of the gauge group with these maps)?
Let $P\rightarrow M$ be a principal $G$-bundle, let $(U_i,u_{ij})$ a trivialization. The restriction of $P$ to $U_i$ is isomorphic to $U_i\times G$. Let $h$ be a gauge transformation and $h_i$ its restriction to $U_i$, $h$ commutes with the action of $G$. By definition $G$ acts on the RIGHT on $P$, if $h_i(x,e)=(x,f_i(x))$, then $h_i(x,g)=h_i(x,e)g=(x,f_i(x)g)$.
Let $x\in U_i\cap U_j$, $u_{ij}h_i(x,g)=h_j(x,u_{ij}(x)g)$ implies that $(x,u_{ij}(x)f_i(x)g)=(x,f_j(x)u_{ij}(x)g)$ this implies that $u_{ij}(x)f_i(x)=f_j(x)u_{ij}(x)$ and $f_j(x)=u_{ij}(x)f_i(x)u_{ij}^{-1}(x)$, it is for this reason that $h$ is a global section of the associated bundle.
When the bundle is trivial, i.e $P=M\times G$ every map $f:M\rightarrow G$ commutes with the action of $G$ since it acts on the right, thus $h(x,g)=(x,f(x)g)$ and $h(x,g).g'=(x,f(x)g)).g'=(x,f(x)gg')=h((x,g).g')$.