are surjective group scheme homomorphisms also surjective on global sections?

Question

Suppose $P,G$ are group schemes over $S$ (not necessarily commutative), where $G$ is finite and constant (so isomorphic to $n$ disjoint copies of $S$)

Suppose there is a surjection homomorphism $f : P\rightarrow G$ of group schemes. Must the corresponding homomorphism of groups $f(S) : P(S)\rightarrow G(S)$ be surjective as well?

This is equivalent to asking, for every section $g\in G(S)$, does there exist a section $p\in P(S)$ such that $f(S)\circ p = g$ ?

Edit: In particular I'm thinking of the situation where in addition to the above, $S$ is connected with geometric point $s$ and $P$ is an inverse limit of finite etale group schemes over $S$, so we can think of $P$ as a profinite group $P_s$ with $\pi_1(S,s)$ acting on it via group automorphisms, and $G$ as a finite group $G_s$ with the trivial action of $\pi_1(S,s)$. Similarly, $f$ is then just a group homomorphism $P_s\rightarrow G_s$ respecting the $\pi_1(S,s)$-action, so the equation is, does every preimage $f^{-1}(g)$ for $g\in G_s$ contain a $\pi_1(S,s)$-invariant element?

Intuition: The preimages $f^{-1}(g)$ should be all isomorphic (I believe this statement is true for abelian schemes, and in general one would think that group schemes are "locally the same everywhere"). Since the preimage of the identity clearly contains a section, then so must every other preimage.

Answer 1

I only know about affine group schemes, so assume $G, P$ are affine and $S = \mathrm{Spec} \ k$. Then if $k$ is algebraically closed this is true. But in general one needs to take a faithfully flat extension of $k$ to find a preimage.

See section 7.7, particularly 7.7.4, of Milne's notes here.