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I am trying to prove the representability of the Quotient functor.

I have the following problem.

Let $\phi \colon T \to S$ be a morphism of noetherian schemes and let $F$ be a coherent sheaf on $\mathbb{P}_S$, which is flat over $S$. Let $\Phi\colon \mathbb{P}^n_T\to \mathbb{P}^n_S$ be the corresponding morphism of projective spaces. Denote by $\pi_T\colon \mathbb{P}^n_T\to T$ and by $\pi_S\colon \mathbb{P}^n_S\to S$ the projection morphisms. Suppose we have an integer $n$ such that the global base change morphism $$ \phi^*(\pi_S)_* F(n)\to (\pi_T)_* \Phi^* F(n)$$ is an isomorphism, $F(n)$ is generated by its global sections and $(\pi_S)_*F(n)$ is locally free. The same holds for $\Phi^*F$.

Now I am asking if the following diagram is commutative $$\require{AMScd} \begin{CD} \Phi^*(\pi_S)^*(\pi_S)_*F(n)@> >> \Phi^* F(n)\\ @VVV @VVV \\ (\pi_T)^* (\pi_T)_* \Phi^*F(n)@>{}>> \Phi^* F(n); \end{CD} $$ where the top horizontal map is given by the inverse image of the canonical map $(\pi_S)^*(\pi_S)_*F(n)\to F(n)$ under $\Phi$ and the bottom horizontal map is just the usual canonical morphism. The left vertical map is an isomorphism given by the functoriality of the inverse image functor and the base change isomorphism and the right vertical map is just the identity.

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    $\begingroup$ This certainly looks commutative to me. Chasing a global section of $F(n)$ (over a chart of $S$) around seems ok (is there some subtlety I'm missing?). Also, in the first displayed equation, shouldn't that be $\Phi^* F(n)$ not $\phi^* F(n)$? $\endgroup$ Aug 20, 2014 at 22:56

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