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Let $X$ be a projective variety, and $E$ be a coherent sheaf on $X$. Grothendieck has proven that there is a scheme $\mathrm{Quot}_X(E)$ parametrizing arbitrary quotient sheaves of $E$. It is probably a well-known question, but I found nothing in Google:

$$\text{Why not a scheme $\mathrm{Sub}_X(E)$ parametrizing arbitrary subsheaves of $E$?}$$

Does such a scheme exist? If yes, is it purely a question of style, like Grothendieck's definition of the projectivization $\mathbb P(V)$ as the set of one-dimensional quotients of a vector space $V$? If not, is it because cokernels of sheaves require sheafification, while kernels coincide with kernels for pre-sheaves?

Moreover, if one considers a reflexive sheaf $E$ (that is the canonical morphism $E \to E^{**}$ is an isomorphism) and its reflexive subsheaves, then there is a Sub-scheme via dualizing and taking a Quot-scheme. Is there a less restrictive constraint?

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    $\begingroup$ I think that this is a good question for mathoverflow. By the way, quot schemes parametrize locally free quotients. I guess that the answer is then that there are more locally free quotients than locally free subobjects. $\endgroup$
    – HeinrichD
    Nov 7, 2016 at 10:37
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    $\begingroup$ @HeinrichD, I believe that Quot-schemes parametrize all quotients: if $E=\mathcal O_X$ and one considers a component with Hilbert polynomial of degree zero, it is isomorphic to Hilbert scheme of points. In particular, all such sheaves have zero-dimensional support. $\endgroup$
    – evgeny
    Nov 7, 2016 at 11:58
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    $\begingroup$ For standard universal properties, you need the scheme to behave well under base change, which in these cases would mean tensor products. Tensor product is right exact, so a quotient remain a quotient, not left exact, so a sub may not remain a sub. $\endgroup$
    – Mohan
    Nov 7, 2016 at 13:57
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    $\begingroup$ I'd say it is like the definition of $P^n$, but it is not purely a `matter of style'. This essentially rephrases Mohan's comment, but here it is: On the level of points, there is of course no difference between subs and quotients: we can just think of short exact sequences $0\to A\to E\to B\to 0$. Now as you consider families (say, over $S$), you really want sequences $$0\to A\to E\boxtimes O_S\to B\to 0$$ with $A$ and $B$ flat over $S$ (so you have a family of short exact sequences). But flatness of $B$ implies flatness of $A$, so we can forget $A$ and think of $B$ only. $\endgroup$
    – t3suji
    Nov 7, 2016 at 15:58
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    $\begingroup$ A fundamental sense in which Grothendieck's definition of $\mathbf{P}(V)$ is not a "convention" is that it gives a meaningful notion of $\mathbf{P}(\mathscr{E}) := {\rm{Proj}}({\rm{Sym}}(\mathscr{E}))$ in the relative situation over a scheme $S$ when the quasi-coherent $O_S$-module $\mathscr{E}$ is not locally free of finite rank (so passing to its dual is not a real option, and so "only one" option is available, namely Grothendieck's). Also, if you read the proof of existence of Quot-schemes you'll see exactly how quotients are better-suited than subsheaves (related to Mohan's comment). $\endgroup$
    – nfdc23
    Nov 7, 2016 at 16:22

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For standard universal properties, you need the scheme to behave well under base change, which in these cases would mean tensor products. Tensor product is right exact, so a quotient remain a quotient, not left exact, so a sub may not remain a sub.

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