Does the canonical morphism commute with direct image functor? 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.
 A: yes, the diagram from your question commutes.
But commutativity does not depend on your particular choice of schemes and sheaves. Neither does it depend on the fact that your base change is an isomorphism.
Using the notation from EGA 0,4 I consider a pullback in the category of Noetherian schemes
$$\require{AMScd}
\begin{CD}
 X^´@> >g^´> X\\
@Vf^´VV @VfVV \\
 S^´@>{}>g> S
\end{CD}
 $$
In the following, map (1) denotes your top horizontal map derived from applying ${g^´}^*$ to the projection formula of $f$,
$$\require{AMScd}
\begin{CD}
{g^´}^*( \sigma_f ): {g^´}^* f^* f_* {  }\mathop \longrightarrow {g^´}^*\\
\end{CD}
$$
According to EGA 0,4.4.3.3 the map
$$\require{AMScd}
\begin{CD}
\sigma_f: f^* f_* {  }\mathop \longrightarrow id\\
\end{CD}
$$
evaluates global sections along the fibres of $f$ on the points of the fibres.
I claim: Map (1) is equal to the composition of map (2), your left vertical map derived from the base change,
$$\require{AMScd}
\begin{CD}
 {g^´}^* f^* f_* = {f^´}^* g^* f_* @> >> {f^´}^* {f^´}_* {g^´}^*\\
\end{CD}
 $$
with map (3), your bottom horizontal map derived from the projection formula of $f^´$ applied to ${g^´}^*$,
$$\require{AMScd}
\begin{CD}
\sigma _{f^´} ({g^´}^*): {f^´}^* {f^´}_* {g^´}^*@> >> {g^´}^*\\
\end{CD}
 $$
Please check my proof: Consider a coherent sheaf $\mathcal{F}$ on $X$. I may assume $S^`$ = Spec $A’$ and $S$ = Spec $A$ affine like in Hartshorne III,9.3.
The base change
$$\require{AMScd}
\begin{CD}
g^* f_* \mathcal{F}@> >> {f^´}_*{g^´}^*\mathcal{F}\\
\end{CD}
$$
reduces to the $A^´$-linear map 
$$
\begin{CD}
id \otimes \rho_{{g^´}(\mathcal F)}: A^´\otimes _A H^0(X, \mathcal{F}) \mathop{\>\>\longrightarrow} H^0(X^´,{g^´}^*\mathcal{F})
\end{CD}
$$
which derives from an $A$-linear map
$$\require{AMScd}
\begin{CD}
\rho_{g^´(\mathcal F)}: H^0(X, \mathcal{F}) \mathop \longrightarrow H^0(X^´,{g^´}^*\mathcal{F})
\end{CD}
$$
Following EGA 0,4.4.3.2 the latter map results from the adjoint of the projection formula of $g^´$
$$\require{AMScd}
\begin{CD}
\rho_{g^´}: \mathcal{id} @> >> {g^´}_*{g^´}^*
\end{CD}
$$
This map identifies sections on the basis with global sections along the fibres of $g^´$: Each section of $\mathcal{F}$ on $U \subset X$ is by definition also a section of ${g^´}^*\mathcal{F}$ on ${g^´}^{-1}(U)$, hence a section of ${g^´}_*({g^´}^*\mathcal{F})$ on $U$.
The given diagram of sheaves commutes iff the induced diagram of stalks commutes. I consider an arbitrary point ${x^´} \in {X^´}$ and suitable affine neighbourhouds Spec $B^´$ of ${x^´} \in {X^´}$ and Spec $B$ of $x:={g^´}(x^´) \in X$, such that the dual diagram in the category of k-algebras
$$\begin{array}
´B^´ & \stackrel{}{\leftarrow} & B \\
\uparrow{} & & \uparrow{} \\
A^´ & \stackrel{}{\leftarrow} & A
\end{array}
$$
commutes; notably $B^´ = B \otimes_A A^´$.
Then map (2) reduces to
$$\require{AMScd}
\begin{CD}
B^´\otimes _{A^´} ({A^´} \otimes _{A} H^0(X, \mathcal{F})) @> >> B^´ \otimes _{A^´} H^0(X^´,{g^´}^*\mathcal{F})) 
\end{CD}
$$
i.e.
$$\require{AMScd}
\begin{CD}
B \otimes _{A} (A^´ \otimes_A H^0(X, \mathcal{F})) {\longrightarrow } B\otimes _{A}H^0(X^´,{g^´}^*\mathcal{F})
\end{CD}
$$
$$\require{AMScd}
\begin{CD}
b \otimes a^´ \otimes s \mapsto b \otimes a^´ \rho_{g^´}(s)
\end{CD}
$$
Map (3) reduces to
$$\require{AMScd}
\begin{CD}
B^´\otimes _{A^´} H^0(X^´, {g^´}^*\mathcal{F}) = B \otimes_A H^0(X^´, {g^´}^*\mathcal{F})@> >> ({g^´}^*\mathcal{F})_{x^´}
\end{CD}
$$
$$\require{AMScd}
\begin{CD}
b \otimes s^´ \mapsto b * {s^´}_{x^´}
\end{CD}
$$
Hence the composition of map (3) with map (2) reduces to
$$\require{AMScd}
\begin{CD}
B\otimes _{A} (A^´ \otimes_A H^0(X, \mathcal{F})) @> >> ({g^´}^*\mathcal{F})_{x^´} = B^´ \otimes _B \mathcal{F}_x
\end{CD}
$$
$$\require{AMScd}
\begin{CD}
b\otimes a^´ \otimes s \mapsto b * a^´ * (\rho_{g^´}(s))_{x^´} = b * a^´ \otimes s_x
\end{CD}
$$
On the other hand, map (1) reduces to the same map
$$\require{AMScd}
\begin{CD}
B^´\otimes_{B} (B \otimes_A H^0(X, \mathcal{F})) = B \otimes_A (A^´ \otimes_A H^0(X, \mathcal{F})) @> >> B^´ \otimes _B \mathcal{F}_x
\end{CD}
$$
i.e.
$$\require{AMScd}
\begin{CD}
B \otimes_A (A^´ \otimes_A H^0(X, \mathcal{F}) @> >> B^´ \otimes _B \mathcal{F}_x 
\end{CD}
$$
$$\require{AMScd}
\begin{CD}
b\otimes a^´ \otimes s \mapsto b * a^´ \otimes s_x
\end{CD}
$$
q.e.d.
I assume your notation of the global base change contains a typo: Bold phi and normal phi should be exchanged.
Why your assumptions about flatness, global generation, locally free image sheaves, isomorphy of base change? In which context do you consider “representability of the Quotient functor”?
