Let $\lambda = (P,\pi,M,G)$ and $\lambda' = (P',\pi',M',G')$ be two principal fiber bundles, $\lambda$ being equiped with a principal connexion whose connexion form is $\omega$. If $(f,k,\rho)$ is a morphism of principal bundles from $\lambda$ to $\lambda'$, with $f : P \to P'$ and $\rho : G \to G'$ a Lie group morphism, and such as $k : M \to M'$ is a diffeomorphism, Kobayashi & Nomizu shows (p79) that the morphism $(f,k,\rho)$ induces a unique principal connexion on $\lambda'$. But the proof assumes that $f$ and $\rho$ are surjective.
So my real question is : does the fact that $k : M \to M'$ is a diffeomorphism implies that $f$ and $\rho$ must be surjective ? It should not be difficult, but I can't prove it... I can prove that if $f$ is surjective, $\rho$ must be too, and conversly, but that's all... And I can't imagine Kobayashi & Nomizu being wrong.
No, it does not implies that $f$ and $\rho$ are surjective. Let $M$ be any manifold (for example a point) and $G$ a Lie group (for example) $Gl(n,R)$, you have trivial bundles $\pi:M\times Gl(n,R)\rightarrow M$ and $\pi':M\times Gl(n,R)\times Gl(n,R)$. The diagonal homomorphism $h:Gl(n,R)\rightarrow Gl(n,R)\times Gl(n,R)$ induces a morphism of bundle $f:M\times Gl(n,R)\rightarrow M\times Gl(n,R)\times Gl(n,R)$ by $f(x,g)=(x,g,g)$. $f$ is a morphism over the identity, but is not surjective.
But if $G=G'$ then the fact that $k$ is a diffeomorphism implies that $f$ is surjective. To See this, let $y\in P'$, there exists $x\in M$ such that $k(x)=\pi'(y)$. Let $z$ be an element of the fiber of $x$, $f(z)$ is in the fiber of $\pi(y)$, this implies that there exists $g\in G$ such that $f(z)g=y$, you have $f(zg)=f(z)g=y$.