Let $S$ be a scheme and let $v : \mathcal E \longrightarrow \mathcal F$ be a morphism of quasi-coherent $\mathcal O_{S}$-modules.
Let $\mathcal F$ be finite locally free. Then the locus $v=0$ is closed, that is the functor $$F : (Sch/S)^{opp} \longrightarrow (Sets)$$ $$ F(T) = \{ f \in Hom_S(T,S)| f^*(v) = 0\}$$ is represented by a closed subscheme of S.
The proof is given in the book. But at the last line I got stuck.
Gist of the proof: By using $\mathcal F$ finite locally free we can say that $$\mathcal Hom (\mathcal E , \mathcal F) = (\mathcal E \otimes \mathcal F^{\vee})^{\vee}$$
and also if $f:T\longrightarrow S$ is a morphism of schemes, then $$ f^*(v)= 0\;\;iff\;\; f^*(v') = 0$$
Where $v'$ is the corresponder of $v$ in $(\mathcal E \otimes \mathcal F^{\vee})^{\vee}$.
Now in the proof they say that $f^*(v') = 0$ iff f factors through the closed subscheme $V(\mathscr I)$. Where $\mathscr I$ is the image of $v$ and a quasi coherent ideal of $\mathscr O_S$. In this "factors through" part I am completely stuck.
My attempt: The goal is to give a bijection between $$Hom (T,{V(\mathscr I)}) \longrightarrow F(T)$$ Let $f \in Hom (T,{V(\mathscr I)}) $ need to show that $f^*(v)$ is 0.
I was trying to show that at the stalk level $$f^*(v)_{t} :(\mathcal E \otimes \mathcal F^{\vee})_{(f(t))} \otimes {_\mathcal {O_{S,f(t)}}} \mathcal O_{T,t} \longrightarrow \mathcal O_{S,f(t)}$$ it is 0 for every $t \in T$. But I am stuck. PLease help me!!
Edit : Definition of $V(\mathscr I)$ $$V(\mathscr I) = \{s \in S : \mathscr I_{s} \neq \mathcal O_{S,s}\}$$ or $$V(\mathscr I) = \{s \in S : \mathscr I_{s} = 0\}$$