Could anyone help me? It follows that this presheaf is locally the same as the sheaf of continuous functions. You don't even need a topology to say what a presheaf is. Is "their" just a misspelling of "they're" in this quote of Melville? rev 2020.10.9.37784, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Let me amplify on that, starting with the topological space case. My DM wouldn't let my character use Acrobatics to escape an Ankheg's grapple, even after it was asleep. (where we have made use of the Yoneda lemma) which is exact by definition of sheaf. Making statements based on opinion; back them up with references or personal experience. Because the same thing is true for open neighborhoods of $q$ which don't contain $p$, the condition on the functions in this presheaf has no effect on sufficiently small open sets. It is obvious that there is a parallel between the definition of structure sheaf of $\operatorname{Spec}(A)$ Donu, good point about presheaves. For example, since $\mathcal{U}_i \subset \emptyset$, then we have a restriction map $\mathcal{F}(\emptyset) \rightarrow \mathcal{F}(\mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. It is just the sheaf which lets you glue those pieces together - i.e. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Use MathJax to format equations. Definition of the structure sheaf of a complex analytic space, Locally constant sheaf on irreducible space is constant. This morphism is an isomorphism since it has as explicit inverse the group morphism $j:A\to \mathcal C_{A,x}$ sending $a\in A$ to $j(a)= \operatorname [{const}(a)]_x$, the germ at $x$ of the constant function $\operatorname {const}(a): X\to A$ sending any point of $X$ to $a$. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Making statements based on opinion; back them up with references or personal experience. so in (3), all maps are identity maps, so the map between stalks should be identity right? A presheaf of sets is the canonical co-completion of a category: You take a (small) category $S$ which does not allow glueing (=has not all colimits) and then $Pre(S)$ has all colimits. I think it suffices to show that $\mathscr{O}'_{\mathfrak{p}}=\varinjlim\limits_{\mathfrak{p}\in U}\mathscr{O}'(U)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}\mathscr{O}'(X_f)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}A_f=A_{\mathfrak{p}}$. There are two ways a pre-sheaf can fail to be a sheaf, both seem reasonable expectations from a function: 1) One would reasonably expect that the only function that is locally $0$ is the zero function. Since I got in touch with schemes, I think of a presheaf or a sheaf as of a space. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. So, is there a naturally occurring pre-sheaf on $\operatorname{Spec}(A)$ (which in general is not a sheaf) that exists for any ring $A$ such that its sheafification gives exactly the structure sheaf $\mathscr O$ of $\operatorname{Spec}(A)$? You can read chapter 2 in the very readable book "Sheaf theory" written by Tennison to get more details. Is "their" just a misspelling of "they're" in this quote of Melville? What is "d---d" in "I’m d—d if I don’t fill you". A classical example (which Hartshorne himself gives) is the presheaf $\mathcal{F}$ of constant $\mathbb{Z}$-valued functions. Your $ \mathfrak A^+$ is exactly the sheaf $\mathcal C_A$, where $A$ is endowed with the discrete topology. 2. If U is connected, then these locally constant functions are constant. $U=\text{Spec }(A_f)$ means the set of all prime ideals in $A$ which are correspondent with the prime ideals in $A_f$. The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. sheafification. rev 2020.10.9.37784, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, $\emptyset = \cup_i \mathcal{U}_i, \, \mathcal{U}_i=\emptyset$, $\mathcal{F}(\emptyset) \rightarrow \mathcal{F}(\mathcal{U}_i)$.
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