Let $\mathscr{E}, \mathscr{F}$ be quasicoherent modules over a scheme $S$, $\mathscr{F}$ of finite type, and $v: \mathscr{E} \to \mathscr{F}$ a homomorphism. Consider the functor $F: (Sch/S)^{op} \to Sets$ defined by $F(T) = \{f \in \operatorname{Hom}_S(T, S): f^*(v) \mbox{ is surjective} \}$, for $S$-schemes $T$. I want to show that $F$ is represented by an open subscheme of $S$ (this is proposition 8.4 in Görtz-Wedhorn).
To do so, it is sufficient to find an open subschemes $U \subseteq S$ such that $f: T \to S$ factors through $U$ iff $f^*(v)$ is surjective. As $\mathscr{F}$ is of finite type, $\operatorname{Supp}(\operatorname{coker}(v))$ is closed. Let $U = S \setminus \operatorname{Supp}(\operatorname{coker}(v))$. If $f$ factors through $U$, then, for $t \in T$, $\operatorname{coker}(f^*(v))_t = f^*(\operatorname{coker}(v))_t = \operatorname{coker}(v)_{f(t)} \otimes_{\mathscr{O}_{S, f(t)}} \mathscr{O}_{T, t} = 0$ - which implies that $f^*(v)$ is surjective.
I am having trouble proving the reciprocal: if $f^*(v)$ is surjective, then $f$ factors through $U$. The main reason of my struggle is because $\operatorname{coker}(f^*(v))_t =\operatorname{coker}(v)_{f(t)} \otimes_{\mathscr{O}_{S, f(t)}} \mathscr{O}_{T, t} = 0$ doesn't seem to imply that $\operatorname{coker}(v)_{f(t)} = 0$. Can anyone help me?
You can show your final claim - that is, if $\renewcommand{\coker}{\operatorname{coker}}\coker(v)_{f(t)}\otimes_{\mathcal{O}_{S,f(t)}} \mathcal{O}_{T,t}=0$, we must have $\coker(v)_{f(t)}=0$. As $\coker(v)_{f(t)}$ is finitely generated over $\mathcal{O}_{S,f(t)}$, the tensor product $\coker(v)_{f(t)}\otimes_{\mathcal{O}_{S,f(t)}} \mathcal{O}_{T,t}$ is a finitely generated $\mathcal{O}_{T,t}$-module, so we can apply Nakayama: this module is zero iff it's quotient by the action of the maximal ideal is zero.
Let $\mathfrak{m}_t$ denote the maximal ideal of $\mathcal{O}_{T,t}$. Then $\left(\coker(v)_{f(t)}\otimes_{\mathcal{O}_{S,f(t)}} \mathcal{O}_{T,t}\right)/\mathfrak{m}_t\left(\coker(v)_{f(t)}\otimes_{\mathcal{O}_{S,f(t)}} \mathcal{O}_{T,t}\right)$ is isomorphic to $\left(\coker(v)_{f(t)}/\mathfrak{m}_{f(t)}\right)\otimes_{\mathcal{O}_{S,f(t)}/\mathfrak{m}_{f(t)}} (\mathcal{O}_{T,t}/\mathfrak{m}_t)$. As $\mathcal{O}_{T,t}/\mathfrak{m}_t$ is a nonzero $\mathcal{O}_{S,f(t)}/\mathfrak{m}_{f(t)}$-vector space, this tensor product is zero iff the LHS vanishes. But that occurs iff $f(t)\in U$, so we're done.