I study Nitin Nitsures paper on the Construction of Hilbert and Quot Schemes and not understand the propetry (F) completely:
In previous chapter (Embedding Quot into Grassmanian) it was proved that there exist a morphism of functors
$\frak{Quot}$$^{\Phi, L}_{E/X/S} \to \frak{Grass}$$(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r))$
for appropriate positive integer $r$. All used notation is explained rigorously in the linked paper on pages 7 and 24 but let me make some short remarks:
Here we deal with special case where $V$ and $W$ are vector bundles on scheme $S$, $\pi: X := \mathbb{P}(V) \to S$ and $E = \pi^*(W)$ and $L =O_X(1)$. Futhermore $\Phi$ is the Hilbert-polynomial.
Now let come to my actually question: The author intends to show that given any locally noetherian $S$-scheme $T$ and a surjective homomorphism $f: W_T \otimes_{O_T} \operatorname{Sym}^r V_T \to \mathcal{J} $ where $\mathcal{J}$ is a locally free $O_T$-module of rank $\Phi(r)$, there exists a locally closed subscheme $T' \subset T$ with universal property (F).
Question: I not understand the universal property (F): It says: Given any locally noetherian $S$-scheme $Y$ and an $S$-morphism $\phi : Y \to T$, let $f_Y$ be the pull-back of $f$, and let $\mathcal{K}_Y = \operatorname{ker}(f_Y) = \phi^* \operatorname{ker}(f)$. [...]
Why is $\operatorname{ker}(f_Y) = \phi^* \operatorname{ker}(f)$? Indeed $\phi$ wasn't assumed to be flat, therefore there is no reason with pullback functor $\phi^*$ should be left exact.
We have exact sequence
$$0 \to \mathcal{K}:= \operatorname{ker}(f) \to W_T \otimes_{O_T} \operatorname{Sym}^r V_T \xrightarrow{f} \mathcal{J} \to 0$$
applying $\phi^*$ we obtain only
$$\phi^*\operatorname{ker}(f) \to \phi^*(W_T \otimes_{O_T} \operatorname{Sym}^r V_T) \xrightarrow{f_Y} \phi^* \mathcal{J} \to 0$$
$\phi^*$ is right exact but in general not left exact (this is only true with additional assumption ie when $\phi$ is flat).
Why here nevertheless $\operatorname{ker}(f_Y) = \phi^* \operatorname{ker}(f)$ is true?
