Let $\pi:\mathcal{X} \to S$ be a flat, family of projective varieties (here $\mathcal{X}$ and $S$ are noetherian). Let $E$ and $F$ be two locally free sheaves on $\mathcal{X}$ such that for all $s \in S$, $E\_s \cong F\_s$, where $E\_s$ and $F\_s$ are the restriction of $E$ and $F$ respectively to the fiber $\mathcal{X}\_s$ over $s$ of $\pi$. Then, does there exists an invertible sheaf $L$ on $S$ such that $E \cong F \otimes \pi^\*L$? 



Isomorphism of sheaves in families of projective varieties
As *@nfdc23* points out, even in the simplest case of $\pi=\mathrm{id}\_X$ your suggestion would amount to saying that for any two locally free sheaves of the same rank (at every point) there would be a line bundle that twists one to the other. Obviously this fails.


On the other hand if one approaches the problem a little differently, then one can get a reasonable statement which generalizes the one that motivated you to ask. 


The essential case in the statement you quote is the following:



> 
> 
> > 
> > Let $\pi:X \to S$ be a flat family of projective varieties and $\mathscr L$ a line bundle on $X$ such that for all $s \in S$, $\mathscr L\_s \simeq \mathscr O\_{X\_s}$. Then there exists an invertible sheaf $\mathscr N$ on $S$ such that $\mathscr L \simeq \pi^\*\mathscr N$.
> > 
> > 
> > 
> 
> 
> 


Then the statement you are quoting is a simple application of this for $\mathscr L\otimes \mathscr M^{-1}$.


Now, if you think about it, the above statement is actually true for arbitrary locally free sheaves:



> 
> 
> > 
> > Let $\pi:X \to S$ be a flat family of projective varieties and $\mathscr E$ a locally free sheaf of constant rank $q$ on $X$ such that for all $s \in S$, $\mathscr E\_s \simeq \mathscr O\_{X\_s}^{\oplus q}$. Then there exists a rank $q$ locally free sheaf $\mathscr G$ on $S$ such that $\mathscr E \simeq \pi^\*\mathscr G$.
> > 
> > 
> > 
> 
> 
> 


**Proof:** I claim that the same proof works. Indeed by the assumption, $h^0(X\_s, \mathscr E\_s)$ is constant, in fact equal to $q$, so $\mathscr G:=\pi\_\*\mathscr E$ is locally free of rank $q$ on $S$. Next we observe that by the assumption, the natural morphism
$$
\alpha: \pi^\*\pi\_\*\mathscr E\longrightarrow \mathscr E
$$
is surjective. Since these are locally free sheaves of the same constant rank, this can only happen if $\alpha$ is an isomorphism. $\square$




---


**Edit:** Apparently, what I originally claimed as a conclusion was a little too much to hope for (thanks to Piotr to point that out!). I guess then the above is the best statement, but it still reduces to the original in case $r=1$, so it can still be considered a direct generalization. 


