Parallel displacement on principal bundles Let $\pi : P \to M$ be a principal bundle with structure group $G$ and connection $\Gamma$. For a fixed $x \in M$, denote by $\Omega(x)$ the space of piecewise differentiable loops based at $x$. Every $\alpha \in \Omega(x)$ determines a map $\tilde{\alpha} : P_x \to P_x$ by means of parallel displacement along $\alpha$: for $u \in P_x$, $\tilde{\alpha}(u)$ is the final point in the (unique) horizontal lift of $\alpha$ starting at $u$. The set of all maps $\tilde{\alpha}$ with $\alpha$ ranging in $\Omega(x)$ is denoted $\Phi(x)$. (cf. Kobayashi, Nomizu).
Fix $u \in P_x$ and define $i_u : \Phi(x) \to G$ as follows: for $\alpha \in \Omega(x)$, the point $\tilde{\alpha}(u)$ lies on $P_x$, so there exists an element $a = i_u(\tilde{\alpha})$ such that $\tilde{\alpha}(u) = u \cdot a$.
I already proved the map $i_u$ is an homomorphism of groups. I'm now trying to figure out if $i_u$ is injective. The kernel of $i_u$ is 
$\ker(i_u) = \{ \tilde{\alpha} \in \Phi(x) : \tilde{\alpha}(u) = u \}$.
If $\tilde{\alpha} \in \ker(i_u)$, is it true that $\tilde{\alpha}$ is the identity diffeomorphism of $P_x$?
Thanks in advance!
 A: $\alpha$ is not always a constant look if $\alpha$ is in the kernel of $i_u$. For example, if the connection is flat, all contractible loops are in $ker(i_u)$. A flat connection is defined by a representation of $h:\pi_1(M)\rightarrow G$. If $\hat M$ is the universal cover of $M$, $P$ is the quotient of $\hat M\times G$ by the diagonal action. The image of distribution $\hat M\times\{g\}, g\in G$ by the projection map $\hat M\times G\rightarrow P$ defines a flat connection on $P$.
A: Remember (or check!) that parallel transport is $G$-equivariant; that is for all $g \in G,v \in P_x$ we have $\tilde \alpha(v)\cdot g = \tilde \alpha (v\cdot g)$. 
So if $\tilde \alpha (u) = u$ and $v \in P_x$ is arbitrary, let $g \in G$ be the group element such that $u \cdot g = v$. Then we have $$\tilde \alpha (v) = \tilde \alpha( u \cdot g ) = \alpha(u) \cdot g= u \cdot g = v,$$
so $\tilde \alpha$ is the identity. Thus $i_u : \Phi(x) \to G$ is injective. 
Note that this is very different to the corresponding map $\Omega(x) \to G$ being injective - that is certainly not true. There are (at least in an intuitive sense) a lot more loops than elements of $G$.
