Lemma 2

Let H be a Lie group, $f_T: T\rightarrow H$ and $\widetilde{f}: \widetilde{D}(G)\rightarrow H$ morphisms of Lie groups such that $f_T(p(t)) = \widetilde{f}(t)$ for all $t\in \widetilde{T}$. There exists a unique morphism of Lie groups $f: G\rightarrow H$ such that $f\circ p= \widetilde{f}$ and such that the restriction of $f$ to T is $f_T$.

Put $Z = C(G)_0$; by §1, no. 4, Cor. 1 of Prop. 4, the morphism of Lie groups $g: Z\times \widetilde{D}(G)\rightarrow G$ such that $g(z, x) =z^{-1}p(x)$ is a covering; its kernel consists of the pairs $(z, x)$ such that $p(x) =z$, for which $x\in p^{-1}(Z)\subset \widetilde{T}$. Since the morphism $(z, x) \rightarrow f_T(z^{-1}) \widetilde{f}(x)$ from $Z\times \widetilde{D}(G)$ to H maps Ker $g$ to $\{e\}$, there exists a morphism $f$ from G to H such that $f\circ p= \widetilde{f}$ and $f(z) =f_T(z)$ for $z\in Z$. But we also have $f(t) =f_T(t)$ for $t\in p( \widetilde{T})$; since $T = Z.p( \widetilde{T})$, the restriction of $f$ to T is indeed $f_T$.