Why isn't the associated sheaf transformation onto?

What is an example of a presheaf $F$ such that the usual morphism $F\to \tilde F$ to the associated sheaf is not onto on all sections, i.e. there exists an $X$ and $F(X)\to \tilde F(X)$ is not onto?

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3 Answers

Let $X$ be a topological space with enough open sets and points (say, a Hausdorff space with infinitely many points). Take any non-trivial sheaf $\mathscr{F}$ you like on $X$, and define $\mathscr{P}(U) = \mathscr{F}(U)$ if $U \ne X$ and $\mathscr{P}(X) = 0$. It is easy to check that this defines a presheaf $\mathscr{P}$, and that $\mathscr{P}$ has the same stalks as $\mathscr{F}$, so the sheafification $\tilde{\mathscr{P}}$ of $\mathscr{P}$ is isomorphic to $\mathscr{F}$. But obviously the canonical map $\mathscr{P} \to \tilde{\mathscr{P}}$ is not surjective on sections.

Exercise. Modify the above example so that $\mathscr{P} \to \tilde{\mathscr{P}}$ is not injective on sections either.

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Let $X$ be a topological space on which there exists an unbounded continuous function$f:X\to \mathbb R$.
Take for $\mathcal F(U)$ the group $\mathcal C_b(U)$ of bounded real continuous functions on the open subset $U$.
Then $\tilde {\mathcal F}=\mathcal C$ is the sheaf of all continuous functions on $X$ and $f$ is not in the image of $\mathcal C_b (X) \to \mathcal C(X)$ .

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A nice natural example! And the strange thing is that continuous functions $f : X \to \mathbb{R}$ are automatically locally bounded because the codomain is locally compact... –  Zhen Lin Mar 20 '12 at 19:45
Ah yes, I hadn't thought of the role of local compactness of the codomain. Thanks for this insightful remark, @Zhen. –  Georges Elencwajg Mar 20 '12 at 20:50

Your question comes up naturally when one tries to intuitively define the structure sheaf of an affine scheme. Basically, if you try to define the structure sheaf of $X=\DeclareMathOperator{\Spec}{Spec}\Spec R$, your thought process could go something like this:

Any open subset $U\subset\Spec R$ is the complement of the vanishing set of an ideal $I$ of elements of $R$, which means that the set $S=\{f\in R\colon$ f does not vanish anywhere on $U\}$ could potentially include more than just the units of $R$ (it contains $\DeclareMathOperator{\rad}{rad}\rad f$, for example, if $I=\langle f\rangle$). Therefore I will try to set $\mathscr F_X(U)=S^{-1}R$, the localization of $R$ at $S$ (note that $S$ is multiplicative), which will guarantee the only ''function'' vanishing on an open set $U$ is the zero function.

More formally, let $R_f=S^{-1}R$ where $S=\{1,f,f^2,\dots\}$. Then we have defined $\mathscr F_X(U)=\displaystyle\varinjlim_{V(f)\subset X\setminus U} R_f$.

Is the above, however, a sheaf? Let's check! Let the structure sheaf be the sheaffification $\mathscr O_X$ of $\mathscr F_X$. What is true is that the distinguished open sets $X_f=X\setminus V(\left<f\right>)$ with associated rings $S^{-1}R=R_f$ as in above form a $\mathscr B$-sheaf, that is, they form a topological basis for $X$, and satisfy the sheaf axioms for coverings by distinguished open sets, i.e. after sheaffification we would have $\mathscr O_X(X_f)=\mathscr F_X(X_f)=R_f$.

The sheaffifcation then gives that $\mathscr O_X(U)=\varprojlim_{X_g\subset U}\mathscr O_X(X_g)=\displaystyle\varprojlim_{X\setminus U\subset V(g)}R_g$.

Since $R_f$ injects into $R_g$ whenever $V(f)\subset V(g)$ (i.e. $\rad g\subset\rad f$ i.e. $g^n=fh$ for some $h\in R$, so both $g$ a unit implies $f$ a unit), we have that there exists a unique a injection $\displaystyle\varinjlim_{V(f)\subset X\setminus U} R_f\to\displaystyle\varprojlim_{X\setminus U\subset V(g)}R_g$, i.e. a unique injection $\mathscr F_X(U)\to\mathscr O_X(U)$.

Here then comes your question: is this sheaffifiction surjective? This it turns out is not an easy question to answer, especially when one is beginning to study this. I had the same question some months back, and Georges Elencwajg gave a very nice, instructive example of a ring $R$ and a Zariski open subset $U$ of $\Spec R$ where this sheaffification is NOT surjective.

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