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3

You say at first that $(X,\mathcal{O}_X)$ is a locally ringed space, but then talk about schemes afterwards. Note that there are many locally ringed spaces which are far from being schemes. The most likely definition of quasicoherent sheaf on a locally ringed space is a sheaf $\mathcal{F}$ so that there is an open cover $\{U_i\}_{i\in I}$ of $X$ so that over ...

3

Yes that's true, the statement you want to prove is that if $U$ is any open subset of any ringed space $X$ then for $x\in U$ the natural map $\mathcal O_{U,x}\to\mathcal O_{X,x}$ is an isomorphism. It's pretty straightforward, for instance for surjectivity if you have an element $f_x\in\mathcal O_{X,x}$ then it is represented by a function $f\in\mathcal O_X(... 2$\mathcal{A}$is a sheaf of abelian groups (or vector spaces), not a sheaf of$G$-modules, and an easy way to answer your questions is that : yes sections of a sheaf can be seen as some kind of functions. Let me expand a bit on this and then take a look at the definitions you linked. If you have a very general sheaf$\mathcal{F}$on a space$X$, a section$s$... 2 Define$A_n = k[[t^{\frac{1}{n}}]]$for every$n \in \mathbb{N}$. Note that if$n$divides$m$, then$A_m$is a flat$A_n$-module. Indeed, we have $$A_{m} = \bigoplus_{j=1}^{\frac{m}{n}}A_n\cdot t^{\frac{j}{m}}$$ So it is even a free$A_n$-module. Consider a morphism$\phi:A^{\oplus s} \rightarrow A$. We denote its kernel by$K$. We have$\phi(e_i) = f_i$... 2 Note that$Hom_U(\tilde{M},\tilde{N})$identifies naturally with$Hom_A(M,N)$. Indeed, given$f : \tilde{M} \rightarrow \tilde{N}$, its action on the global sections is a$A$-linear map$M \rightarrow N$. Conversely, given$g: M \rightarrow N$, we can consider the maps$g_{D(a)}: M_a=\tilde{M}(D(a)) \rightarrow N_a=\tilde{N}(D(a))$which are the ... 2 You are almost correct: the image of$s_i$in$\mathscr{F}_x$is the equivalence class of$(X,s_i)$. 1 Why do you think the first map needs to be surjective? This is easily seen to be false by example: consider holomorphic functions on$\mathbb C$. The natural map$\mathcal O_{\mathbb C}(\mathbb C) \to \mathcal O_{\mathbb C}(\mathbb C \setminus \left\{0\right\})$is not surjective since$f(z) = 1/z$is not in its image, so pairing with the restriction map to ... 1 Yes, as long as$\operatorname{Spec} A$is an open affine set in$X$. More generally if$U$is an open subset of a ringed space$(X,\mathcal O_X)$, and$\mathcal O_U$is the sheaf of rings on$U$defined by$\mathcal O_U(V) = \mathcal O_X(V)$for all open sets$V \subset U \subset X$, then$\mathcal O_{X,x} = \mathcal O_{U,x}$for all$x \in U$. This is ... 1 For any sheaf on any topological space$X$, if$U=\coprod U_i$is a disjoint union of open sets then$\mathcal{F}(U)=\prod\mathcal{F}(U_i)$as it can easily be seen from the gluing condition of a sheaf. Thus on a discrete space$X$, we have$U=\coprod_{x\in U}\{x\}$and so$\mathcal{F}(U)=\prod_{x\in U}\mathcal{F}(\{x\})$. Moreover$\mathcal{F}(\{x\})=\...

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As requested, the two comments above are turned into an answer. You need to prove the adjunction directly. Note that a morphism of presheaves $A\to {i_p}_*B$ is the data for all U of a morphisms $A(U)\to{i_p}_*B(U)$ such that some diagrams commute. For all $U\not\ni p$, there is no data here since ${i_p}_*B(U)$ is the terminal object of your category and ...

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Let $(s_p)_{p \in U} \in \prod_{p \in U} \mathscr F_p$ be a compatible germ. By definition, there exists a cover $\{U_i\}_{i \in I}$ for $U$, and elements $f_i \in \mathscr F(U_i)$ such that for all $q \in U_i$ we have $s_q = [f_i, U_i] \in \mathscr F_q$. If $q \in U_i \cap U_j$, then $[f_i, U_i]=s_q=[f_j, U_j]$ in $\mathscr F_q$, hence [f_i|_{U_i \cap U_j}...

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