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Let $T$ be a scheme (probably integral noetherian) and $X$ a smooth projective variety. Let $K,K',A,A'$ be locally free coherent sheaves on $X\times T$. There are exact sequences:

$0\rightarrow K\rightarrow A\rightarrow E\rightarrow 0$

and

$0\rightarrow K'\rightarrow A'\rightarrow E'\rightarrow 0$

which are exact when restricted to any fiber $X\times t$. In particular, $E$ and $E'$ are flat over $T$ by a result in Chapter 8 of Matsumura textbook.

Is $\mathcal{H}om(E,E')$ flat over $T$?

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  • $\begingroup$ For completeness' sake, could you make the reference to Matsumura a bit more specific? Which book by Matsumura do you mean and what is the result you're quoting? $\endgroup$ – R. van Dobben de Bruyn Jan 13 '17 at 3:10
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No. For example, let $X = T = C$ be a smooth projective curve (but we keep the notation $X \times T$ to indicate the separate functions of the two). Let $E = \mathcal O_{\Delta_C}$, and let $E' = \mathcal O_{x \times T}$ for a point $x \in X$.

Since both the maps $\Delta_C \to X \times T \to T$ and $x \times T \to X \times T \to T$ are isomorphisms, we see that $E$ and $E'$ are flat over $T$. They also admit a presentation of the form you describe: we can take $A = A' = \mathcal O_{X \times T}$, with $K = \mathcal O_{X \times T}(-\Delta_C)$ and $K' = \mathcal O_{X \times T}(-x \times T)$. The sequences $$\begin{array}{ccccccccc}0 & \to & \mathcal O_{X \times T}(-\Delta_C) & \to & \mathcal O_{X \times T} & \to & \mathcal O_{\Delta_C} & \to & 0\\0 & \to & \mathcal O_{X \times T}(-x\times T) & \to & \mathcal O_{X \times T} & \to & \mathcal O_{x \times T} & \to & 0\\\end{array}$$ are both exact, hence remain so in each fibre $X \times t$ (e.g. use right exactness of $-\otimes \mathcal O_{X \times t}$ plus the fact that the left hand side is locally free of rank $1$, so that the first map is injective by a rank count).

But $E|_{X \times t} = \mathcal O_t$ and $E'|_{X \times t} = \mathcal O_x$ (with the identification $X \times t \cong X$). Hence, $\mathcal Hom(E,E')$ jumps at $x = t$: there are no homomorphisms when $t \neq x$, but there is one when $t = x$.

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  • $\begingroup$ @S.D.: Ah, you're absolutely right. My argument is wrong because $\mathcal Hom$ does not commute with taking fibres (and in fact a direct computation gives $\mathcal Hom(E, E') = 0$). I will try to find a different example for this question. $\endgroup$ – R. van Dobben de Bruyn Sep 16 '18 at 16:18
  • $\begingroup$ Note: the given answer is still incorrect as far as I can tell. I am no longer active on MO, so it's unlikely that I will come back later to give a correct argument. I'm leaving this up here in the hope someone finds it useful. $\endgroup$ – R. van Dobben de Bruyn Dec 11 '18 at 3:45

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