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.
Best Answer
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.