Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 5 down vote accepted

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.

share|improve this answer

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 $\mathcal C_b (X) \to \mathcal C(X)$ .

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.