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edited Apr 13, 2017 at 12:58

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Apparently, there is an abstract nonsense argument abstract nonsense argument that shows $\mathbf{Sch}$ is concretisable. Here is a hands-on proof.

We define $U_0 : \mathbf{Sch} \to \mathbf{Set}$ to be the functor that sends a scheme to the set of points of the underlying topological space and we define $U_1 : \mathbf{Sch} \to \mathbf{Set}^\mathrm{op}$ to be the functor that sends a scheme to the disjoint union of the stalks of its structure sheaf. (This makes sense because the stalk of $f^{-1} \mathscr{O}_Y$ at $x$ is the stalk of $\mathscr{O}_Y$ at $f (x)$.) Clearly, $(U_0, U_1) : \mathbf{Sch} \to \mathbf{Set} \times \mathbf{Set}^\mathrm{op}$ is faithful, and the contravariant powerset functor $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ is also faithful, so the functor $X \mapsto U_0 X \amalg \mathscr{P} (U_1 X)$ is a faithful functor $\mathbf{Sch} \to \mathbf{Set}$.

Apparently, there is an abstract nonsense argument that shows $\mathbf{Sch}$ is concretisable. Here is a hands-on proof.

We define $U_0 : \mathbf{Sch} \to \mathbf{Set}$ to be the functor that sends a scheme to the set of points of the underlying topological space and we define $U_1 : \mathbf{Sch} \to \mathbf{Set}^\mathrm{op}$ to be the functor that sends a scheme to the disjoint union of the stalks of its structure sheaf. (This makes sense because the stalk of $f^{-1} \mathscr{O}_Y$ at $x$ is the stalk of $\mathscr{O}_Y$ at $f (x)$.) Clearly, $(U_0, U_1) : \mathbf{Sch} \to \mathbf{Set} \times \mathbf{Set}^\mathrm{op}$ is faithful, and the contravariant powerset functor $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ is also faithful, so the functor $X \mapsto U_0 X \amalg \mathscr{P} (U_1 X)$ is a faithful functor $\mathbf{Sch} \to \mathbf{Set}$.

Apparently, there is an abstract nonsense argument that shows $\mathbf{Sch}$ is concretisable. Here is a hands-on proof.

We define $U_0 : \mathbf{Sch} \to \mathbf{Set}$ to be the functor that sends a scheme to the set of points of the underlying topological space and we define $U_1 : \mathbf{Sch} \to \mathbf{Set}^\mathrm{op}$ to be the functor that sends a scheme to the disjoint union of the stalks of its structure sheaf. (This makes sense because the stalk of $f^{-1} \mathscr{O}_Y$ at $x$ is the stalk of $\mathscr{O}_Y$ at $f (x)$.) Clearly, $(U_0, U_1) : \mathbf{Sch} \to \mathbf{Set} \times \mathbf{Set}^\mathrm{op}$ is faithful, and the contravariant powerset functor $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ is also faithful, so the functor $X \mapsto U_0 X \amalg \mathscr{P} (U_1 X)$ is a faithful functor $\mathbf{Sch} \to \mathbf{Set}$.

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created Mar 18, 2014 at 19:37

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Apparently, there is an abstract nonsense argument that shows $\mathbf{Sch}$ is concretisable. Here is a hands-on proof.

We define $U_0 : \mathbf{Sch} \to \mathbf{Set}$ to be the functor that sends a scheme to the set of points of the underlying topological space and we define $U_1 : \mathbf{Sch} \to \mathbf{Set}^\mathrm{op}$ to be the functor that sends a scheme to the disjoint union of the stalks of its structure sheaf. (This makes sense because the stalk of $f^{-1} \mathscr{O}_Y$ at $x$ is the stalk of $\mathscr{O}_Y$ at $f (x)$.) Clearly, $(U_0, U_1) : \mathbf{Sch} \to \mathbf{Set} \times \mathbf{Set}^\mathrm{op}$ is faithful, and the contravariant powerset functor $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ is also faithful, so the functor $X \mapsto U_0 X \amalg \mathscr{P} (U_1 X)$ is a faithful functor $\mathbf{Sch} \to \mathbf{Set}$.