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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

Shyscraper Sheares. Let \(X\) be a topological space. let \(P\) be a point, and let \(A\) be an abelian group. Define a sheaf \(i_{P}(A)\) on \(X\) as follows: \(i_{p}(A)(\mathcal{L})=A\) if \(P \in L, 0\) otherwise. Verify that the stalk of \(i_{P}(A)\) is \(A\) at every point \(Q \in\left\\{P \text { ; }^{-} \text {, and } 0\right.\) elsewhere, where \(\\{P \text { ; - denotes the closure of the set consisting of the point } P\) Hence the name "skyscraper sheaf." Show that this sheaf could also be described \(\operatorname{as} i_{*}(A),\) where \(A\) denotes the constant sheaf \(A\) on the closed subspace \(\\{P\\}^{-},\) and \(i:\\{P\\}^{-} \rightarrow X\) is the inclusion.

If \(\widetilde{\psi}\) is a coherent sheaf on a noetherian formal scheme \(\vec{x},\) which can be generated by global sections, show in fact that it can be generated by a finite number of its global sections.

The Grothendieck Group of a Nonsingular Curce. Let \(X\) be a nonsingular curve over an algebraically closed field \(k .\) We will show that \(K(X) \cong \operatorname{Pic} X \oplus \mathbf{Z},\) in several steps. (a) For any divisor \(D=\sum n_{i} P_{i}\) on \(X\), let \(\psi(D)=\sum n_{i \hat{i}}\) ' \(\left(k\left(P_{i}\right)\right) \in K(X),\) where \(k\left(P_{i}\right)\) is the skyscraper sheaf \(k\) at \(P_{t}\) and 0 elsewhere. If \(D\) is an effective divisor, let \(\mathrm{C}_{D}\) be the structure sheaf of the associated subscheme of codimension \(1,\) and show that \(\psi(D)=\dot{\gamma}\left(C_{D}\right) .\) Then use (6.18) to show that for any \(D, \psi(D)\) depends only on the linear equivalence class of \(D,\) so \(\psi\) defines a homomorphism \(\psi: \mathrm{Cl} X \rightarrow K(X)\) (b) For any coherent sheaf \(\mathscr{F}\) on \(X\), show that there exist locally free sheaves \(\delta_{0}\) and \(\mathscr{E}_{1}\) and an exact sequence \(0 \rightarrow \mathscr{E}_{1} \rightarrow \mathscr{B}_{0} \rightarrow \mathscr{H} \rightarrow 0 .\) Let \(r_{0}=\operatorname{rank} \delta_{0}\) \(r_{1}=\operatorname{rank} \delta_{1},\) and define det \(\tilde{\mathscr{H}}=\left(\bigwedge^{r_{0}} \mathscr{E}_{0}\right) \otimes\left(\bigwedge^{r_{1}} \delta_{1}\right)^{-1} \in \operatorname{Pic} X .\) Here \(\wedge \mathrm{de}\) notes the exterior power (Ex. 5.16). Show that det \(\mathscr{H}\) is independent of the resolution chosen, and that it gives a homomorphism det: \(K(X) \rightarrow\) Pic \(X\) Finally show that if \(D\) is a divisor, then \(\operatorname{det}(\psi(D))=\mathscr{L}(D)\) (c) If \(\mathscr{F}\) is any coherent sheaf of rank \(r,\) show that there is a divisor \(D\) on \(X\) and an exact sequence \(0 \rightarrow \mathscr{P}(D)^{\oplus r} \rightarrow \mathscr{H} \rightarrow \mathscr{I} \rightarrow 0\), where \(\mathscr{J}\) is a torsion sheaf. Con- clude that if \(\mathscr{F}\) is a sheaf of rank \(r,\) then \(\gamma(\mathscr{F})-r \gamma\left(\mathcal{O}_{X}\right) \in \operatorname{Im} \psi\) (d) Using the maps \(\psi,\) det, rank,and \(1 \mapsto \gamma\left(C_{X}\right)\) from \(\mathbf{Z} \rightarrow K(X),\) show that \(K(X) \cong\) Pic \(X \oplus \mathbf{Z}\)

Let \(X\) be a scheme. For any \(x \in X\). let \(C_{\lambda}\) be the local ring at \(x\), and \(m_{\lambda}\) its maximal ideal. We define the residue field of \(x\) on \(X\) to be the field \(k(x)=C_{x} m_{x} .\) Now let \(K\) be any field. Show that to give a morphism of Spec \(K\) to \(X\) it is equivalent to give a point \(x \in X\) and an inclusion \(\operatorname{map} k(x) \rightarrow K\).

Support. Recall the notions of support of a section of a sheaf, support of a sheaf, and subsheaf with supports from (Ex. 1.14 ) and (Ex. 1.20 ). (a) Let \(A\) be a ring, let \(M\) be an \(A\) -module, let \(X=\operatorname{Spec} A,\) and let \(\mathscr{F}=\tilde{M}\) For any \(m \in M=\Gamma(X, \overline{\mathscr{F}}),\) show that Supp \(m=V(\text { Ann } m),\) where Ann \(m\) is the annihilator of \(m=\\{a \in A | a m=0\\}\) (b) Now suppose that \(A\) is noetherian, and \(M\) finitely generated. Show that \(\operatorname{Supp} \mathscr{F}=V(\operatorname{Ann} M)\) (c) The support of a coherent sheaf on a noetherian scheme is closed. (d) For any ideal a \(\subseteq A,\) we define a submodule \(\Gamma_{\mathrm{a}}(M)\) of \(M\) by \(\Gamma_{\mathrm{a}}(M)=\) \(\left\\{m \in M | a^{n} m=0 \text { for some } n>0\right\\} .\) Assume that \(A\) is noetherian, and \(M\) any \(A\) -module. Show that \(\Gamma_{\mathrm{a}}(M)^{\sim} \cong \mathscr{H}_{Z}^{0}(\mathscr{F}),\) where \(Z=V(\mathrm{a})\) and \(\mathscr{F}=\tilde{M}\) \([\text {Hint}: \text { Use (Ex. } 1.20)\) and (5.8) to show a priori that \(\mathscr{H}_{Z}^{0}(\mathscr{F})\) is quasi-coherent. Then show that \(\left.\Gamma_{\mathrm{a}}(M) \cong \Gamma_{\mathrm{z}}(\mathscr{F}) .\right]\) (e) Let \(X\) be a noetherian scheme, and let \(Z\) be a closed subset. If \(\mathscr{F}\) is a quasicoherent (respectively, coherent) \(O_{X}\) -module, then \(\mathscr{H}_{Z}^{0}(\mathscr{F})\) is also quasicoherent (respectively, coherent).
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