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EDIT: The original question has been answered, but another difficulty in the proof has appeared. See below.

Let $f,g : X \to Y$ be two morphisms of schemes such that the induced pullback functors $f^* , g^* : Qcoh(Y) \to Qcoh(X)$ are isomorphic. Can we conclude $f=g$?

If $X$ and $Y$ are affine, then this is quite easy; simply use naturality to conclude that the isomorphism $f^* M \cong g^* M$ is multiplication with a global unit in $\mathcal{O}_Y$, which is independent from $M$, and deduce $f^\# = g^\#$. Ok then the claim is also true if $X$ is not affine. But what happens when $Y$ is not affine? The problem is basically, that you cannot lift sections to global sections. Also, the reduction to the affine case works only if we already know that

There is an open affine covering $\{U_i\}$ of $Y$, such that $f^{-1}(U_i) = g^{-1}(U_i)$ and the pushforward with $U_i \to Y$ maps quasicoherent modules to quasicoherent modules.

Laurent Moret-Bailly has proven below that $f$ and $g$ coindice as topological maps. Thus, only the latter concerning the pushforwards is unclear (to me). Everything is ok when $Y$ is noetherian or quasiseparated. What about the general case?

PS: I'm also interested in questions like this one; is it possible to recover scheme theoretic properties from the categories of quasi-coherent modules? If anyone knows literature about this going beyond Gabriel's and Rosenberg's, please let me know.

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    $\begingroup$ Probably. Definitely if use q-coh sheaves on etale sites, not Zar. sites. Functor assigning to any scheme its Zar. (or etale) loc. ringed topos is fully faithful (proved in SGA4 VIII, & explained in Prop. 3.1.1 and Thm. 3.1.3 in draft of "Univ. property of non-arch. analytification" on my webpage), so suffices that functor from loc. ringed topos to tensor category of q-coh sheaves is faithful. By main result (Thm. 5.11) of Lurie's "Tannaka duality & geom. stacks", fully faithful in etale case. Surely pf works in Zar. case for schemes. It's been a while since I read it. Check for yourself. $\endgroup$
    – BCnrd
    Commented Aug 30, 2010 at 17:24
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    $\begingroup$ Martin, small addendum: in section 6 Lurie gives the argument for faithfulness (it is the much easier part), which is all you need. But his notion of "geometric stack" covers only those schemes which are quasi-compact and have affine diagonal. So I should make some caveats on it's relevance to your situation. But if you look at his arguments then perhaps you'll get some idea for what to do (and surely his arguments become even simpler in the scheme setting, especially with the Zariski topology). Good luck. $\endgroup$
    – BCnrd
    Commented Aug 30, 2010 at 17:34
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    $\begingroup$ Dear Martin: one more small technical point is that you need to clarify in what sense you're saying that the pullback functors are isomorphic. That is, as functors between mere abelian categories, or as tensor functors, etc. It seems plausible that to make a proof you should assume the isomorphism is as functors between tensor categories. (I'm not saying I see a counterexample if you weaken to just an isomorphism of functors between abelian categories, but that kind of hypergenerality does seem a bit silly, and perhaps false.) $\endgroup$
    – BCnrd
    Commented Aug 30, 2010 at 18:36
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    $\begingroup$ Martin, you almost certainly want to assume the tensor compatibility. All it means is the obvious necessary (and extremely reasonable) condition that the isomorphism of functors respects the formation of tensor products on both sides (in a manner respecting the associativity, symmetry, and identity objects for the tensor products on both sides). It sounds like you throw away too much structure if you ignore that aspect in general. Again, look at Lurie's arguments (specialized to your situation with schemes, so some of his complications will go away). $\endgroup$
    – BCnrd
    Commented Aug 30, 2010 at 22:30
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    $\begingroup$ Martin, sure, assume whatever you want if it makes a proof work. My main point was that in view of Lurie's result and its proof, probably you should keep track of tensor compatibility and nothing more (at least under some finiteness hypotheses). He uses quasi-compactness to get a smooth cover by an affine scheme (in your case even an etale cover: disjoint union of constituents of a finite open affine covering) and then needs affine diagonal for some effective descent argument (if I remember correctly). By the way, the LaTeX command \rightrightarrows is very useful. $\endgroup$
    – BCnrd
    Commented Aug 31, 2010 at 4:39

3 Answers 3

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If $Z$ is a closed subscheme of $Y$, then $f^*\mathcal{O}_Z$ and $g^*\mathcal{O}_Z$ are isomorphic $\mathcal{O}_X$-modules. So they have the same support, hence $f^{-1}(Z)=g^{-1}(Z)$. This implies that $f=g$ set-theoretically because every point of $Y$ is the intersection of the subsets containing it which are either open or closed. Then you can reduce to the affine case.

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  • $\begingroup$ Ah this is the argument I was lookin' for. You mean the pushforward of $\mathcal{O}_Z$ with $Z \to X$, right? $\endgroup$
    – Martin Brandenburg
    Commented Aug 31, 2010 at 9:24
  • $\begingroup$ I see that $f^{-1}(Z)$ is the support of $f^{-1} i_* \mathcal{O}_Z$, but $f^* i_* \mathcal{O}_Z$ could have a smaller support and this depends on $f$. $\endgroup$
    – Martin Brandenburg
    Commented Aug 31, 2010 at 9:30
  • $\begingroup$ Yes. Of course the $\mathcal{O}_X$-module structure of this thing is not enough to get the surjection $\mathcal{O}_X\to\mathcal{O}_Z$ but it does determine the support. $\endgroup$
    – Laurent Moret-Bailly
    Commented Aug 31, 2010 at 9:31
  • $\begingroup$ Ah I see: $(f^* i_* \mathcal{O}_Z)_x=0$ if and only if $\mathcal{O}_{X,x}$ is generated by $f^\#(I_{f(x)})$, if and only if $I_{f(x)} = \mathcal{O}_{Y,f(x)}$ since $f^\#$ is a local homomorphism! Thus we don't have to know $f^\#$. $\endgroup$
    – Martin Brandenburg
    Commented Aug 31, 2010 at 9:51
  • $\begingroup$ See my edit in the question; I think we need more than $f=g$ set-theoretically in order to reduce to the affine case. Or do I miss something? $\endgroup$
    – Martin Brandenburg
    Commented Aug 31, 2010 at 17:38
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One can use the following argument. Assume for a start that we have a pair of morphisms $f_1:X_1 \to Y$ and $f_2:X_2 \to Y$. Then one can recover the fiber product $X_1\times_Y X_2$ as follows. Just take the functor $f_1^*\otimes f_2^*:QCoh(Y\times Y) \to QCoh(X_1\times X_2)$ and apply it to the surjection $O_{Y\times Y} \to \Delta_*O_Y$ --- you will get the surjection $O_{X_1\times X_2} \to O_{X_1\times_Y X_2}$. Note that this reconstruction of the fiber produc depends only on the functors $f_1^*$, $f_2^*$.

Now, returning to the original question, applying the above to $X_1 = X$, $X_2 = Y$, $f_1 = f$, $f_2 = 1_Y$, we can reconstruct $X\times_Y Y$ which is nothing but the graph of $f$. Similarly, replacing $f$ by $g$ we can reconstruct the graph of $g$. We conclude that if $f^* = g^*$ then the graphs of $f$ and $g$ are the same, hence $f = g$.

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  • $\begingroup$ Sorry, I'm confused: if we only know the pullback functors $f_1^{\ast}$ and $f_2^{\ast}$ as abstract functors between tensor categories, how do we form $f_1^{\ast} \otimes f_2^{\ast}$? More precisely, in what sense do you intrinsically construct the (tensor) category of quasi-coherent sheaves on a fiber product from that on the "factors"? For example, not every q-coh sheaf on the fiber product is obtained as a tensor product of pullbacks of such from the factors... $\endgroup$
    – BCnrd
    Commented Aug 30, 2010 at 18:48
  • $\begingroup$ If schemes are sufficiently good then every sheaf is a cokernel of a morphism of tensor products of pullbacks, I think this is enough to define the functor. If the schemes are not so good, you can try to define the functor locally and then glue. Note that we don't need to define the category of the product --- it can be considered as given, we only have to define the functor. $\endgroup$
    – Sasha
    Commented Aug 30, 2010 at 18:58
  • $\begingroup$ I don't understand your proof at all. As Brian already noted, the functor is $(f_1 \times f_2)^*$ and it is not clear at all why this only depends on $f_1^*$ and $f_2^*$. Also, $\Delta_* \Omega_Y$ is not quasi-coherent in general, and the pullback with $f_1 \times f_2$ cannot be $\mathcal{O}_(X_1 \times_Y X_2)$ since this sheaf is not defined on $X_1 \times X_2$ (maybe a simple pullback is just not written down here; anyway we forget something). The surjectivity assertions are only clear to me when $Y$ is separated. $\endgroup$
    – Martin Brandenburg
    Commented Aug 30, 2010 at 23:08
  • $\begingroup$ Certainly, I assume that $Y$ is separable. After that, the only thing needed for the argument is an existence of a surjection from a tensor product of pullbacks to the sheaf of ideals of the diagonal --- indeed then we have a right-exact sequence of the form $p_1^*F \otimes p_2^*G \to O_{Y\times Y} \to \Delta^*O_Y \to 0$. Taking the pullback we get a right-exact sequence $p_1^*f_1^*F \otimes p_2^*f_2^* G \to O_{X_1\times X_2} \to O_{X_1\times_Y X_2} \to 0$, hence we can reconstruct the fiber product as a cokernel of the first map, which depends only on the functors $f_1^*$ and $f_2^*$. $\endgroup$
    – Sasha
    Commented Aug 31, 2010 at 4:15
  • $\begingroup$ And when I write $O_{X_1\times_Y X_2}$ I add implicitly the pushforward for the natural map $X_1\times_Y X_2 \to X_1\times X_2$, certainly the pushforward is a quasicoherent sheaf on $X_1\times X_2$. $\endgroup$
    – Sasha
    Commented Aug 31, 2010 at 4:16
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Made some mistake in my reply.

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