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Chapter 2: Problem 5

Show that a morphism of sheaves is an isomorphism if and only if it is both injective and surjective.

Short Answer

Expert verified
A morphism of sheaves is an isomorphism if and only if it is both injective and surjective. This is due to the fact that being an isomorphism requires there be an inverse morphism, which is guaranteed by the properties of injectivity (prevents two different elements from mapping to the same location) and surjectivity (ensures all points in the target space are mapped).

Step by step solution

01

Show that an isomorphism implies both injectivity and surjectivity

Assume that \( f: \mathcal{F} \rightarrow \mathcal{G} \) is an isomorphism. Then, there exists an inverse morphism \( g: \mathcal{G} \rightarrow \mathcal{F} \). Thus, for every open set \( U \), \( g \circ f \) is the identity map on \( \mathcal{F}(U) \), i.e., \( g(f(s)) = s \) for all \( s \in \mathcal{F}(U) \). This shows that \( f \) is injective. Similarly, \( f \circ g \) is the identity map on \( \mathcal{G}(U) \), i.e., \( f(g(t)) = t \) for all \( t \in \mathcal{G}(U) \). This shows that \( f \) is surjective.
02

Show that injectivity and surjectivity imply an isomorphism

Conversely, assume that \( f: \mathcal{F} \rightarrow \mathcal{G} \) is both injective and surjective. To show that it's an isomorphism, we need to demonstrate the existence of an inverse morphism. Define \( g: \mathcal{G} \rightarrow \mathcal{F} \) as follows. For any \( t \in \mathcal{G}(U) \), because \( f \) is surjective, there exists \( s \in \mathcal{F}(U) \) such that \( f(s) = t \). Define \( g(t) = s \). Since \( f \) is injective, \( g \) is well-defined. Thus, \( f \circ g = 1_{\mathcal{G}} \) and \( g \circ f = 1_{\mathcal{F}} \), i.e., \( f \) is an isomorphism.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isomorphism
In the context of sheaves, an isomorphism signifies a very tight relationship between two sheaves. Specifically, a morphism of sheaves \( f: \mathcal{F} \rightarrow \mathcal{G}\) is termed an isomorphism if there exists another morphism \( g: \mathcal{G} \rightarrow \mathcal{F} \) such that composing them in either order gives you the identity map again.
This means that the process of applying \( f \) followed by \( g \) gets you back to where you started, for each section of the sheaves.
  • For every section \( s \) in \( \mathcal{F}(U) \), \( g(f(s)) = s \), proving that \( f \) takes \( s \) to a unique place in \( \mathcal{G}(U) \) and \( g \) brings it back.
  • Similarly, for any section \( t \) in \( \mathcal{G}(U) \), \( f(g(t)) = t \).
This defining property of isomorphisms ensures that \( \mathcal{F} \) and \( \mathcal{G} \) are, in a sense, the same sheaf, merely with labels changing.
This equivalence underpins much of the value of studying isomorphisms, as it indicates structural sameness at the fundamental level.
Injectivity
An injective morphism is analogous to a one-to-one mapping. This means every element of the first set maps to a unique element of the second set. In terms of a morphism of sheaves \( f: \mathcal{F} \rightarrow \mathcal{G} \), \( f \) is injective if and only if, whenever two sections \( s_1 \) and \( s_2 \) in \( \mathcal{F}(U) \) have the same image under \( f \) (i.e., \( f(s_1) = f(s_2) \)), then \( s_1 = s_2 \).
  • This implies there's no overlapping of images, assuring each section in \( \mathcal{F} \) gets its own unique image in \( \mathcal{G} \).
  • Injectiveness prevents merging of distinct sections in \( \mathcal{F} \), thus preserving information.
Injectivity is vital when determining if a morphism is an isomorphism because being injective ensures you can match every section back to a unique original, which is key for constructing an inverse morphism.
Surjectivity
Surjectivity in the context of morphisms of sheaves is the property that ensures every element in the target is achieved by the mapping. For \( f: \mathcal{F} \rightarrow \mathcal{G} \) to be surjective, every section \( t \) in \( \mathcal{G}(U) \) must be obtainable from some section \( s \) in \( \mathcal{F}(U) \).
This means that for every possible result \( t \), there is a preimage \( s \) such that \( f(s) = t \).
  • Surjectivity guarantees full coverage, ensuring nothing in the target set \( \mathcal{G} \) is left out.
  • It translates to the idea that \( \mathcal{G} \) can be completely "filled up" by the images of \( \mathcal{F} \).
Surjectivity is essential because it allows the creation of an inverse mapping needed for forming an isomorphism, as it ensures the morphism covers every aspect of the target sheaf, leaving no gaps.

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Most popular questions from this chapter

Let \(X\) be a noetherian scheme, and let \(\mathscr{F}\) be a coherent sheaf. (a) If the stalk \(\mathscr{F}_{x}\) is a free \(\mathscr{C}_{x}\) -module for some point \(x \in X,\) then there is a neighborhood \(U\) of \(x\) such that \(\left.\mathscr{F}\right|_{v}\) is free. (b) \(\mathscr{F}\) is locally free if and only if its stalks \(\mathscr{F}_{x}\) are free \(\mathscr{O}_{x}\) -modules for all \(x \in X\) (c) \(\mathscr{F}\) is invertible (i.e., locally free of rank 1 ) if and only if there is a coherent sheaf \(\mathscr{G}\) such that \(\mathscr{F} \otimes \mathscr{G} \cong \mathscr{O}_{X} .\) (This justifies the terminology invertible: it means that \(\mathscr{F}\) is an invertible element of the monoid of coherent sheaves under the operation \(\otimes .\)

Let \(X\) be an integral scheme. Show that the local ring \(\mathscr{O}_{\xi}\) of the generic point of \(X\) is a field. It is called the function field of \(X,\) and is denoted by \(K(X) .\) Show also that if \(U=\operatorname{Spec} A\) is any open affine subset of \(X,\) then \(K(X)\) is isomorphic to the quotient field of \(A\)

Support. Let \(\mathscr{F}\) be a sheaf on \(X\), and let \(s \in \mathscr{F}(U)\) be a section over an open set \(U\) The support of \(s\), denoted Supp s, is defined to be \(\left\\{P \in U | s_{P} \neq 0\right\\},\) where \(s_{P}\) denotes the germ of s in the stalk \(\overline{\mathscr{F}}_{p}\). Show that Supp s is a closed subset of \(U\). We define the support of \(\overline{\mathscr{F}}, \operatorname{Supp}, \overline{\mathscr{F}},\) to be \(\left\\{P \in X | \mathscr{F}_{P} \neq 0\right\\},\) It need not be a closed subset.

A morphism \(f: X \rightarrow Y\) is quasi-finite if for every point \(y \in Y, f^{-1}(y)\) is a finite set. (a) Show that a finite morphism is quasi-finite. (b) Show that a finite morphism is closed, i.e., the image of any closed subset is closed. (c) Show by example that a surjective, finite-type, quasi-finite morphism need not be finite.

Let \(A\) be a ring, let \(S=A\left[x_{0}, \ldots, x_{r}\right]\) and let \(X=\) Proj \(S\). We have seen that a homogeneous ideal \(I\) in \(S\) defines a closed subscheme of \(X\) (Ex. 3.12 ), and that conversely every closed subscheme of \(X\) arises in this way (5.16) (a) For any homogeneous ideal \(I \subseteq S\), we define the saturation \(I\) of \(I\) to be \(\left\\{s \in S | \text { for each } i=0, \ldots, r, \text { there is an } n \text { such that } x_{i}^{n} s \in I\right\\} .\) We say that \(I\) is saturated if \(I=I .\) Show that \(T\) is a homogeneous ideal of \(S\). (b) Two homogeneous ideals \(I_{1}\) and \(I_{2}\) of \(S\) define the same closed subscheme of \(X\) if and only if they have the same saturation. (c) If \(Y\) is any closed subscheme of \(X\), then the ideal \(\Gamma_{*}\left(\mathscr{I}_{Y}\right)\) is saturated. Hence it is the largest homogeneous ideal defining the subscheme \(Y\) (d) There is a \(1-1\) correspondence between saturated ideals of \(S\) and closed subschemes of \(X\).
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