# Why do these two constructions of a sheaf associated to a module on a projective scheme agree?

Let $B$ be a graded ring an $M$ be a $\mathbb Z$-graded $B$-module. We can associate to $M$ a sheaf of modules on $\operatorname{Proj} B$ by defining the sheaf on the principal open sets $D_+(f)$ to be $(M_{(f)})^\sim$ (where this is the associated module in the affine case) and checking that these agree on overlaps.

However, there is a second construction, which appears on page $165$ of Qing Liu's Algebraic Geometry and Arithmetic Curves. Consider the canonical injection $f:\operatorname{Proj B}\rightarrow \operatorname{Spec} B$. This is continuous. Consider the sheaf $\mathcal F = M^\sim$ on $\operatorname{Spec} B$ (again, this is the associated sheaf in the affine case). We can take the subsheaf of $f^{-1}\mathcal F$ consisting of all degree 0 elements to be the sheaf associated to $M$ on $\operatorname{Proj} B$.

(As @Hanno notes in the comments, there is an issue here. There doesn't seem to be a way to turn this into a sheaf of modules, because $f$ is a merely a continuous map, not a morphism. I would also appreciate comments on this issue.)

I would like to know why these constructions agree. I know that $(f^{-1} \mathcal F)_{\mathfrak p} = M_\mathfrak p$, so taking the homogeneous elements, we see that the two sheafs have the same stalks. But I am rather uncomfortable with the inverse image construction, and do not see how to construct an isomorphism.

• Isn't a morphism of sheaves an isomorphism iff it's an isomorphism on stalks? I seem to recall being told that something like that is true (although I don't know the details). – Qiaochu Yuan Jul 5 '13 at 5:44
• @QiaochuYuan That's very true. But here we only know the stalks are isomorphic individually. We don't yet have a sheaf morphism. – Potato Jul 5 '13 at 6:01
• Could you tell in more detail how you want to turn $f^{-1}{\mathcal F}$ into a sheaf of ${\mathcal O}_{\text{Proj}(B)}$-modules? It is a priori only a module over $f^{-1}{\mathcal O}_{\text{Spec}(B)}$, and $f$ only a map of topological spaces, not equipped with an algebraic component usually taken to induce from a module over $f^{-1}{\mathcal O}_Y$ a module over ${\mathcal O}_X$. – Hanno Jul 5 '13 at 6:58
• @Hanno I am not sure about that myself! This appears on Liu, Algebraic Geometry and Arithmetic Surfaces, page 165. – Potato Jul 5 '13 at 7:54
• Dear Potato, As a side comment, have you thought geometrically about what these various constructions mean? For example, suppose that $B = \mathbb C[x,y]$ with the usual grading, so that its Proj is $\mathbb P^1$ and its Spec is $\mathbb A^2$, both over $\mathbb C$. In this case, it might help you to think about what the map $f: \mathbb P^1 \to \mathbb A^2$ actually is. (Certainly, this example makes it clear that it's not a morphism of schemes.) Regards, – Matt E Jul 8 '13 at 6:29