I will first define the notions of exact,surjective for group schemes and then ask my question.
Let $B$ and $C$ be fppf group schemes over S. The book I am reading defines a homomorphism $f:B\to C$ over $S$ to be surjective if for every $S$-scheme $T$ and element $c \in C(T)$, there is an fppf morphism $T'\to T$ such that the image of $c$ in $C(T')$ is the image of some $b\in B(T')$.
Call a sequence of fppf group schemes $A\to B\to C$ exact at $B$ if the composition is the trivial homomorphism and the induced homomorphism $A\to \ker(g)$ is surjective. If $A$ is the kernel of $B\to C$, then the book defines the quotient $B/A$ to be $C$.
Question:
Is this well defined? That is, if $f:B\to C$ and $g:B\to D$ are two surjective homomorphisms such that $\ker(f) \cong \ker(g)$ over $B$, does this imply that $C\cong D$ so that the relevant square commutes?
My attempt:
It suffices to show that for all affine schemes $T$, there is a functorial bijection $C(T) \to D(T)$.
If I define $\hat T$ to be the inverse limit of fppf maps $T'\to T$, then since for locally finite presentation schemes, it's functor of points commutes with taking direct limits of rings, I can show that there is a bijection $C(\hat T) \to D(\hat T)$.
I don't think this is enough to show that $C\cong D$ however...