About this editorial companion

This is the English translation of the original editorial companion. Its independent post-translation source review is complete. Build and publication remain separate gates; this text does not by itself claim a completed public English release or an English-edition DOI.

This book is an original editorial companion for moving from varieties and coordinate rings to sheaves, schemes, and cohomology. Its bridge explanations, new solutions, integrative problems, and capstone were independently written by OpenAI Codex gpt-5.6-sol, Ultra. This is not an additional lecture course by Holger Brenner or a retranslation of the source course.

The referenced lectures, problem statements, theorems, and public solutions are by Holger Brenner, chiefly from Bündel, Garben und Kohomologie (Osnabrück 2019-2020). Links to the complete BGK reader are retained for the separate HTML reader with all its source credits in the integrated edition. Their assigned targets have passed independent source review; build and publication status is recorded separately. These links identify specific sections, not answers that must be sought elsewhere. Source solutions remain source solutions; new editorial answers are not assigned the source author’s name or endorsement.

Organisation and study route

Begin with the prerequisite bridge. It distinguishes closed points from all prime points, functions from sections, stalks from fibres, and local rings from residue fields. Then follow the mastery bank in the order of BGK Units 2–15, followed by 23–27. These remain the source course numbers; the jump from 15 to 23 neither renumbers the course nor claims that Units 16–22 do not exist.

Each unit on this route has exactly three mastery items. There are 44 new editorial solutions + 13 public source solutions = 57 items across 19 route units. Seventeen units require new solution files; Units 8 and 11 already have three public solutions each counted here, so no new bank was created for either unit.

Attempt each problem before opening its solution. When checking your answer, write down the domain and codomain of every map, the base ring, and the hypotheses that allow a denominator to be inverted. The pitfall and check notes in each item help identify steps that still need proof; they are not additional problems in the count of 57 items.

Twelve integrative problems and one capstone

After the mastery bank, work through the following twelve integrative problems, each with a complete solution. They are independently written synthesis problems, not additions to Brenner’s problem list.

1. From varieties and coordinate rings to affine schemes.
2. Affine gluing and compatibility conditions.
3. The projective line from two affine charts.
4. The affine line with doubled origin.
5. Morphisms of locally ringed spaces.
6. Krull dimension and geometric examples.
7. Quasicoherent sheaves and module localisation.
8. A Čech calculation with explicit signs and quotient.
9. A computable example of projective cohomology.
10. Euler characteristic and an exact sequence.
11. A classical projective curve and its scheme.
12. Tracing sources and reconstructing hypotheses.

Finish with one integrated capstone on permanent Stacks tags and proving that a neighbourhood is affine. Its seven stages are parts of one exercise, not seven capstones. The complete answers and oral-proof rubric are for self-study; the rubric does not require a human assessor’s presence or approval. The Stacks Project Authors are credited as the authors of the downstream reference; Stacks is not a translated source course.

Thus the 57 mastery items, 12 integrative problems, one capstone, and prerequisite bridge are distinct companion layers. No new course units or official PDF pages are added: the two source courses still total 60 units and 602 official PDF pages, as frozen in the source edition. This companion does not change source problem, theorem, or unit numbers.

Terminology, sources, and usage rights

Use the BGK terminology guide alongside local definitions. Sheaf, presheaf, stalk, fibre, section, restriction, ring, field, and ringed space denote different types of objects; matching words do not replace hypotheses. Each new item points to the source problem or theorem it uses, together with the relevant revision identity or frozen witness. Source notes correcting formulas must still be read alongside the statements.

This original editorial material is licensed under CC BY-SA 4.0. The problem statements used, course translations, source editorial contributions, and media components in the BGK reader retain their respective credits and rights as recorded there; the companion’s licence does not remove or replace source-component rights. The stated identities and revision links allow readers to trace the sources without treating editorial answers as the source author’s public answers.

The provenance of production and AI-assisted review of this editorial layer is OpenAI Codex gpt-5.6-sol, Ultra. No claim is made that Holger Brenner, The Stacks Project Authors, Wikiversity, the Wikimedia Foundation, or a source institution endorses this companion, wrote its new answers, or has performed human review of it.

From varieties to schemes: what changes?

This bridge connects the starting point of a reader of classical algebraic geometry with the language of sheaves and schemes. It is an independently written bridge explanation, not a translation of an additional lecture by Holger Brenner. The algebraic prerequisites are prime and maximal ideals, quotient rings, and localisation. All rings here are commutative with identity; ring homomorphisms preserve the identity.

1. Coordinate rings are not discarded

For the classical comparison in this section only, take an algebraically closed field $k$ and a nonempty affine algebraic set $V\subseteq k^n$. Its coordinate ring is

$$A=k[V]=k[x_1,\ldots,x_n]/I(V),$$

where $I(V)$ is the ideal of all polynomials vanishing on $V$. Then $A$ is reduced: if $f^m$ vanishes on $V$, every value $f(a)$ in the field $k$ is also zero, so $f\in I(V)$. If the term variety in the source requires irreducibility, impose that condition too; in that case $A$ is an integral domain. Reduced and integral domain are not synonyms.

The prerequisite theorem connecting algebra with classical points is the Nullstellensatz: for an ideal $J\subseteq k[x_1,\ldots,x_n]$, $I(V(J))=\sqrt J$, and maximal ideals have the form $(x_1-a_1,\ldots,x_n-a_n)$ with $a_i\in k$. Consequently,

$$a\in V\quad\longleftrightarrow\quad \mathfrak m_a=\{f\in A:f(a)=0\}\in\operatorname{MaxSpec}(A).$$

The scheme construction keeps the same $A$, but its space is

$$X=\operatorname{Spec}(A) =\{\mathfrak p\subset A:\mathfrak p\text{ is a prime ideal}\}.$$

On this space, $V(J)=\{\mathfrak p:J\subseteq\mathfrak p\}$ is closed, while the sets $D(f)=\{\mathfrak p:f\notin\mathfrak p\}$ form an open basis. On a spectrum, $V(J)$ denotes a set of prime ideals; do not immediately read it as a set of tuples in $k^n$.

2. Closed points and generic points

The closure of a point $\mathfrak p$ is $V(\mathfrak p)$: a closed set $V(J)$ contains $\mathfrak p$ exactly when $J\subseteq\mathfrak p$. Thus closed points are precisely maximal ideals. The classical comparison above identifies $V$ with the subspace of closed points of $X$, not with all of $X$.

For example, for $A=k[t]$ with $k$ algebraically closed, the points are $(t-a)$, $a\in k$, and one extra point $\eta=(0)$. The closure of $\eta$ is the entire spectrum. This point records the whole line as one irreducible closed subspace; it is not a new number to be added to $k$. More generally, $\mathfrak p$ is the generic point of $V(\mathfrak p)$.

If $k$ is not algebraically closed, closed points need not be valued in $k$. For example, $(t^2+1)$ is a maximal ideal of $\mathbb R[t]$ with residue field $\mathbb R[t]/(t^2+1)\cong\mathbb C$. Always specify the base field before identifying closed points with $k$-rational points. The topological foundation is found in Brenner, Proposition 8.5.

3. From functions to sheaf sections

A presheaf $\mathcal F$ assigns an object $\mathcal F(U)$ to every open set and restriction maps to smaller open sets. Identity restrictions must be identities, and successive restriction must equal direct restriction. A sheaf adds two conditions: sections that are locally equal are equal; local sections agreeing on every overlap have exactly one gluing. This is the content needed from Brenner, Definition 4.1.

For $X=\operatorname{Spec}(A)$, the structure sheaf $\mathcal O_X$ is determined on the open basis by

$$\Gamma(D(f),\mathcal O_X)=A_f, \qquad \Gamma(X,\mathcal O_X)=A.$$

The denominator $f$ may be inverted precisely where it does not belong to the prime ideal of the point. On a general open set, a section is represented by compatible local fractions; do not require one denominator to work on the entire open set. The basis formula and its restrictions hold for any commutative ring, including rings with zero divisors; see Brenner, Lemma 9.12 and Stacks, tag 01HV.

The term function remains useful, but section emphasises that the object belongs to a sheaf on an open set. On a general scheme, a section is not merely a list of values in one fixed field.

4. Stalks, local rings, residue fields, and fibres

The stalk $\mathcal F_{\mathfrak p}$ records germs of sections around $\mathfrak p$: two representatives agree if their restrictions agree on some smaller neighbourhood. For the structure sheaf,

$$\mathcal O_{X,\mathfrak p}=A_{\mathfrak p}, \qquad \mathfrak m_{\mathfrak p}=\mathfrak pA_{\mathfrak p}, \qquad \kappa(\mathfrak p) =A_{\mathfrak p}/\mathfrak pA_{\mathfrak p} \cong\operatorname{Frac}(A/\mathfrak p).$$

The local ring records all germs, whereas the residue field records only values at the point. Evaluation is the composite $\mathcal O_X(U)\to\mathcal O_{X,\mathfrak p}\to\kappa(\mathfrak p)$ for $\mathfrak p\in U$. The affine line illustrates the difference:

At $a$, the germ $t-a$ is nonzero in the local ring, but its value in the residue field is zero. For a sheaf of modules $\mathcal F$, the fibre at the point is $\mathcal F_{\mathfrak p}\otimes_{\mathcal O_{X,\mathfrak p}} \kappa(\mathfrak p)$, not the stalk itself. For example, the stalk of $\mathcal O_X$ at $a$ is $k[t]_{(t-a)}$, while its fibre is $k$. This sheaf fibre must also be distinguished from the fibre of a scheme morphism. See Brenner, Definition 3.22, Definition 7.13, and Lemma 9.10.

5. Why nilpotents must not be silently removed

Take any field $k$ and $B=k[\varepsilon]/(\varepsilon^2)$. Every prime ideal contains $\varepsilon$, so $\operatorname{Spec}(B)$ has only the point $(\varepsilon)$. Its topological space is the same as the one-point space $\operatorname{Spec}(k)$, but their global section rings differ: they are $B$ and $k$, respectively. The section $\varepsilon$ is nonzero, although its value in the only residue field is zero. Thus even all point values need not determine a section.

A ring is reduced if it has no nonzero nilpotents; a scheme is reduced if all its local rings are reduced. The scheme $\operatorname{Spec}(B)$ above is nonreduced because $B$ itself is local and $\varepsilon\ne0$ in it. Replacing a ring $A$ by $A_{\mathrm{red}}=A/\sqrt{(0)}$ preserves the topological space of its spectrum: all prime ideals contain $\sqrt{(0)}$, and the prime-ideal correspondence for the quotient preserves closed sets. Its structure sheaf can nevertheless change, as the example $B$ shows. This is the information lost if geometry is remembered only as the zero set of equations.

6. Gluing objects and pulling back sections

A ringed space is a pair $(X,\mathcal O_X)$; it is locally ringed if every stalk of its structure sheaf is a local ring. A scheme is a space locally isomorphic to a spectrum with its structure sheaf. Here isomorphic concerns both space and sheaf, not merely a homeomorphism.

To construct a scheme from pieces $X_i$, specify open subsets $U_{ij}\subseteq X_i$ and isomorphisms $\varphi_{ij}:U_{ij}\to U_{ji}$. Besides identities and inverses, on every triple overlap the domains must agree and $\varphi_{jk}\circ\varphi_{ij}=\varphi_{ik}$ must hold. Gluing points alone is insufficient: the structure sheaves must also be glued by the same isomorphisms. Every point then still has an affine neighbourhood. This local condition does not ensure that the entire glued space is affine. The source definition is Brenner, Definition 10.1; the capstone that follows examines a global example.

A scheme morphism $f:X\to Y$ consists of a continuous map and a sheaf map

$$f^\#: \mathcal O_Y\longrightarrow f_*\mathcal O_X, \qquad (f_*\mathcal O_X)(V)=\mathcal O_X(f^{-1}(V)),$$

inducing at every $x$ a local homomorphism $\mathcal O_{Y,f(x)}\to\mathcal O_{X,x}$. Local means that the inverse image of the target maximal ideal is the source maximal ideal. This is not merely an extra condition on the point map; it connects that map with pullback of sections. Compare Brenner, Definition 7.15.

In the affine case, $\alpha:A\to B$ gives a map in the opposite direction, $f:\operatorname{Spec}(B)\to\operatorname{Spec}(A)$, with $f(\mathfrak q)=\alpha^{-1}(\mathfrak q)$. Indeed, $f^{-1}(D(a))=D(\alpha(a))$, and the section map is $A_a\to B_{\alpha(a)}$. On stalks, the map $A_{\alpha^{-1}(\mathfrak q)}\to B_{\mathfrak q}$ is local because a numerator belongs to the source prime ideal exactly when its image belongs to $\mathfrak q$. Existence and uniqueness of this morphism are Brenner, Corollary 10.10. For geometry over $k$, use $k$-algebra homomorphisms, not ring homomorphisms that forget the base structure.

A short orientation check with answers

Is $(0)$ on the affine line closed? No: its closure is the whole line. Is $t$ zero as a germ at $(t)$? No; only its evaluation is zero. Must two schemes with one-point topological spaces be isomorphic? No: $\operatorname{Spec}(k)$ and $\operatorname{Spec}(k[\varepsilon]/(\varepsilon^2))$ have different section rings. Is it correct to replace the direction $A\to B$ with $\operatorname{Spec}(A)\to\operatorname{Spec}(B)$? No; the spectrum direction is reversed. These four answers test four main transitions; they do not add counted exercises to the mastery bank.

Provenance, rights, and source route

This explanation was produced by OpenAI Codex gpt-5.6-sol, Ultra. and is licensed under CC BY-SA 4.0 as an original editorial layer. Definitions and results referenced from Bündel, Garben und Kohomologie remain credited to Holger Brenner; contributor and translation credits and source-component rights are unchanged. The frozen-version references used are Lecture 3, revision 793623, Lecture 4, revision 1003714, Lecture 7, revision 1003731, Lecture 8, revision 793632, Lecture 9, revision 793634, and Lecture 10, revision 1003733. Stacks is a downstream reference by The Stacks Project Authors, not a lecture source translated here; its rights remain separately in force. No endorsement, authorship, or human review by source authors or institutions is implied.

BGK Unit 2 mastery bank

The problem statements below come from Holger Brenner’s course. The worked solutions and checking notes were prepared independently by OpenAI Codex gpt-5.6-sol, Ultra.; they are not translations of Brenner’s public solutions. The author and source-contributor credits remain applicable, including Bocardodarapti as the contributor to the frozen worksheet revision. This editorial material is licensed under CC BY-SA 4.0 and implies neither endorsement by the author or source institutions nor human authorship or review.

This unit counts three mastery items: two new editorial solutions to Exercises 2.1 and 2.2, and the already translated public source solution to Exercise 2.4. Solution 2.4 continues to count as a source solution and is neither copied nor labelled as new editorial work here. Neither selected exercise has a public solution in the frozen source map.

The prerequisites are the definitions of a real vector bundle and local trivialisation, and Definition 2.1 on continuous sections. The base space is not assumed to be Hausdorff.

New item 1 - Nowhere-zero sections and triviality of line bundles

Source: Exercise 2.1. Statement identifier: Reelles Geradenbündel/Trivial/Nullstellenfreier Schnitt/Aufgabe. Fixed witness: revision 1048817, source page 111598. The exercise number is determined by frozen worksheet revision 602852; the revision of the transcluded statement is recorded separately above.

Source statement

Show that a real line bundle $L\to X$ over a topological space $X$ is trivial if and only if it has a continuous section that is nowhere zero.

Complete editorial solution

Write the bundle projection as $p:L\to X$. Each fibre $L_x=p^{-1}(x)$ is a one-dimensional real vector space. The nowhere-zero condition means $s(x)\ne0_x$ in each fibre, not avoidance of a single zero point common to the whole total space.

First direction. If the bundle is trivial, there is a bundle isomorphism

$$\tau:L\longrightarrow X\times\mathbb R$$

that preserves the base point and is linear on each fibre. Set $s(x):=\tau^{-1}(x,1)$. This map is continuous as the composite of $x\mapsto(x,1)$ and $\tau^{-1}$. Since $\tau$ is over $X$, we have $p(s(x))=x$, so $s$ is a section. The linear isomorphism $\tau_x:L_x\to\mathbb R$ sends $0_x$ to $0$, whereas $\tau_x(s(x))=1$. Thus the section is nowhere zero.

Conversely. Let $s$ be a nowhere-zero continuous section. Define

$$\Phi:X\times\mathbb R\longrightarrow L, \qquad \Phi(x,t)=t\,s(x).$$

Since $s(x)$ is a basis of the one-dimensional space $L_x$, the map $t\mapsto t\,s(x)$ is a linear isomorphism $\mathbb R\to L_x$. Thus $\Phi$ is bijective and linear on each fibre. We must still prove that $\Phi$ and its inverse are continuous; a continuous bijection alone is not enough to give a bundle isomorphism.

Take any local trivialisation

$$\tau_U:p^{-1}(U)\longrightarrow U\times\mathbb R$$

with $U\subseteq X$ open. In these coordinates, the section $s$ has the form

$$\tau_U(s(x))=(x,a(x)),\qquad x\in U,$$

for a continuous function $a:U\to\mathbb R$. Its continuity follows from that of $\tau_U\circ s|_U$ and the second-coordinate projection. Since $s$ is nowhere zero, $a(x)\ne0$ throughout $U$. In these coordinates, $\Phi$ and its inverse are given by

$$(x,t)\longmapsto(x,t\,a(x)), \qquad (x,b)\longmapsto\left(x,\frac{b}{a(x)}\right).$$

Both formulas are continuous: multiplication in $\mathbb R$ is continuous, and $x\mapsto1/a(x)$ is continuous because $a$ is nowhere zero. The sets $U\times\mathbb R$ form an open cover of the domain of $\Phi$, while the sets $p^{-1}(U)$ form an open cover of the domain of its inverse. Continuity on an open cover gives global continuity in both directions.

Hence $\Phi$ is a bundle isomorphism $X\times\mathbb R\cong L$, and $L$ is trivial. Both directions have been proved.

Pitfalls and checks

Choosing a nonzero vector separately in every fibre does not yet produce a continuous section. It is continuity that ensures the coordinate function $a$ and the inverse trivialisation are continuous. Check the inverse formula in one chart using

$$\frac{t\,a(x)}{a(x)}=t, \qquad \frac{b}{a(x)}\,a(x)=b.$$

Rank one is used precisely when a single nonzero vector is declared to be a basis. In higher rank, a single nowhere-zero section does not by itself trivialise the whole bundle.

New item 2 - The image of a section is a closed subspace

Source: Exercise 2.2. Statement identifier: Reelles Vektorbündel/Schnitt/Abgeschlossene Teilmenge/Aufgabe. Fixed witness: revision 1048838, source page 111631. The exercise number comes from frozen worksheet revision 602852.

Source statement

Let $s:X\to V$ be a continuous section of a real vector bundle $p:V\to X$ over a topological space $X$. Show that the image $s(X)\subseteq V$ is a closed subset homeomorphic to $X$.

Complete editorial solution

Homeomorphism with the base. If $s(x)=s(y)$, apply $p$ and use $p\circ s=\operatorname{Id}_X$ to obtain $x=y$. Thus the map

$$\bar s:X\longrightarrow s(X),\qquad x\longmapsto s(x),$$

is bijective. It is continuous for the subspace topology on $s(X)$: if $O$ is open in $V$, then $\bar s^{-1}(O\cap s(X))=s^{-1}(O)$ is open in $X$. Its inverse is the continuous restriction

$$p|_{s(X)}:s(X)\longrightarrow X.$$

Indeed, $p(s(x))=x$, and if $v=s(x)\in s(X)$, then $s(p(v))=s(x)=v$. Hence $\bar s$ is a homeomorphism.

Closedness in the total space. Take a local trivialisation

$$\tau_U:p^{-1}(U)\longrightarrow U\times\mathbb R^r.$$

There is a continuous map $f:U\to\mathbb R^r$ with $\tau_U(s(x))=(x,f(x))$. Moreover,

$$s(X)\cap p^{-1}(U)=s(U),$$

because $s(x)\in p^{-1}(U)$ exactly when $x=p(s(x))\in U$. Therefore $\tau_U(s(U))$ is the graph

$$\operatorname{Graph}(f) =\{(x,v)\in U\times\mathbb R^r\mid v=f(x)\}.$$

The map

$$h:U\times\mathbb R^r\longrightarrow\mathbb R^r, \qquad h(x,v)=v-f(x)$$

is continuous. Since $\{0\}$ is closed in $\mathbb R^r$, this graph is the closed subset

$$\operatorname{Graph}(f)=h^{-1}(\{0\})$$

of $U\times\mathbb R^r$. Via the homeomorphism $\tau_U$, it follows that $s(X)\cap p^{-1}(U)$ is closed in $p^{-1}(U)$.

The sets $p^{-1}(U)$ from all the trivialisations form an open cover of $V$. On each such set,

$$(V\setminus s(X))\cap p^{-1}(U)$$

is open in $p^{-1}(U)$, hence also open in $V$. Their union is $V\setminus s(X)$. Thus the complement of $s(X)$ is open, and $s(X)$ is closed in $V$.

Pitfalls and checks

Do not infer closedness from $p\circ s=\operatorname{Id}_X$ alone. For general continuous maps, this identity gives a topological embedding, but does not by itself give a closed image. Here the proof uses the local model $U\times\mathbb R^r$ and the closedness of $\{0\}$ in a real vector space. The base space $X$ need not be Hausdorff.

As a check, if $s$ is the zero section, then $f=0$ and $h(x,v)=v$. The local statement becomes the closedness of $U\times\{0\}$, exactly as expected.

BGK Unit 3 mastery bank

The problem statements come from Holger Brenner’s course. The following worked solutions and checking notes were prepared independently by OpenAI Codex gpt-5.6-sol, Ultra.; they are not translations of public source solutions. The source credits remain applicable, including Bocardodarapti as the contributor to the frozen worksheet revision. This editorial material is licensed under CC BY-SA 4.0 and implies neither endorsement by the author or source institutions nor human authorship or review.

The three mastery items for Unit 3 comprise the two new editorial solutions here to Exercises 3.3 and 3.16, together with the already translated public source solution to Exercise 3.1. Solution 3.1 continues to count as a source solution and is not repeated here. The frozen source map states that Exercises 3.3 and 3.16 have no public solutions.

The prerequisites used are the gluing data for the Möbius strip from Unit 2, the definition of the tensor product of bundles, and the definitions of a presheaf and a stalk in terms of germ classes. Neither the sheaf property nor cohomology is required.

New item 1 - The tensor square of the Möbius strip

Source: Exercise 3.3. Statement identifier: Möbiusband/Tensorprodukt/Trivial/Aufgabe. Fixed witness: revision 846097, source page 111727. The exercise number is determined by frozen worksheet revision 619301.

Source statement

Show that the tensor product of the Möbius strip with itself is a trivial line bundle.

Complete editorial solution

Write $L\to S^1$ for the real line bundle of the Möbius strip, with

$$S^1=\{(x,y)\in\mathbb R^2\mid x^2+y^2=1\}.$$

Use the cover and gluing data from Example 2.11:

$$U=S^1\setminus\{(0,1)\}, \qquad V=S^1\setminus\{(0,-1)\}.$$

On $U\cap V$ we have $x\ne0$. If $t_U$ and $t_V$ are the fibre coordinates in the two trivialisations, the transition convention is

$$t_V=\varepsilon(x,y)t_U, \qquad \varepsilon(x,y)= \begin{cases} 1,&x>0,\\ -1,&x<0. \end{cases}$$

The function $\varepsilon$ is continuous on the intersection, since its two components are open and the formula is constant on each component.

By Definition 3.2, the transition map of $L\otimes L$ is obtained by tensoring the two transition maps. On a fibre, the map on pure tensors is

$$a\otimes b\longmapsto (\varepsilon a)\otimes(\varepsilon b) =\varepsilon^2(a\otimes b)=a\otimes b.$$

Pure tensors span the tensor product, so this map is the identity on all of $\mathbb R\otimes_{\mathbb R}\mathbb R$. Under the linear identification $a\otimes b\mapsto ab$, the fibre is $\mathbb R$, and the tensor-bundle transition is always $1$, on both components of the intersection.

Here is the resulting global trivialisation. Let $e_U(p)$ and $e_V(p)$ be the local basis vectors of $L_p$ with coordinate $1$ in their respective charts. Since coordinates change according to the formula above, on the intersection we have

$$e_U(p)=\varepsilon(p)e_V(p).$$

Hence

$$e_U(p)\otimes e_U(p) =\varepsilon(p)^2 e_V(p)\otimes e_V(p) =e_V(p)\otimes e_V(p).$$

Thus the following two rules define a single well-defined map:

$$\begin{aligned} \Phi:L\otimes L&\longrightarrow S^1\times\mathbb R,\\ a\,e_U(p)\otimes e_U(p)&\longmapsto(p,a) \quad(p\in U),\\ a\,e_V(p)\otimes e_V(p)&\longmapsto(p,a) \quad(p\in V). \end{aligned}$$

In each chart, $\Phi$ is a trivialisation, so it is continuous and a linear bijection on every fibre. Its inverse is given by the two local formulas

$$(p,a)\longmapsto a\,e_U(p)\otimes e_U(p), \qquad (p,a)\longmapsto a\,e_V(p)\otimes e_V(p),$$

which are continuous and agree on the intersection. Local continuity on an open cover gives global continuity in both directions. Thus $\Phi$ is an isomorphism of line bundles, and

$$L\otimes L\cong S^1\times\mathbb R.$$

Pitfalls and checks

The tensor product is not the direct sum. The rank of $L\otimes L$ is $1\cdot1=1$, whereas the rank of $L\oplus L$ is $1+1=2$. The calculation $(-1)(-1)=1$ here is a calculation of tensor-product transitions, not a new gluing instruction for a picture of the strip.

As a check, the local sections $e_U\otimes e_U$ and $e_V\otimes e_V$ agree on the intersection and are nonzero in every fibre. Together they give a nowhere-zero global continuous section. New item 1 for Unit 2 provides a second check that this line bundle is trivial.

New item 2 - Stalks of the product of two presheaves

Source: Exercise 3.16. Statement identifier: Prägarbe/Produkt/Halm/Aufgabe. Fixed witness: revision 1083990, source page 111822. The exercise number comes from frozen worksheet revision 619301.

Source statement

Let $\mathcal F$ and $\mathcal G$ be presheaves on a topological space $X$, and let $\mathcal F\times\mathcal G$ be their product presheaf. Show that for every point $P\in X$,

$$(\mathcal F\times\mathcal G)_P =\mathcal F_P\times\mathcal G_P.$$

Complete editorial solution

We prove the source equality by constructing a canonical bijection, meaning a bijection independent of the choice of neighbourhoods or germ representatives. There is no hypothesis that $\mathcal F$ or $\mathcal G$ is a sheaf; it is enough that both are presheaves of sets.

On an open set $U$, the value of the product presheaf is $\mathcal F(U)\times\mathcal G(U)$, and restriction is componentwise. For $V\subseteq U$, the formula is

$$(s,t)|_V=(s|_V,t|_V).$$

The identity and composition properties of restrictions follow from those of the two presheaves. Thus the product object is indeed a presheaf.

By Definition 3.21 and Definition 3.22, a germ $s_P\in\mathcal F_P$ is represented by a section $s\in\mathcal F(U)$ on an open neighbourhood $U$ of $P$. Representatives $s\in\mathcal F(U)$ and $s'\in\mathcal F(U')$ give the same germ exactly when there is an open neighbourhood $W$ with $P\in W\subseteq U\cap U'$ and $s|_W=s'|_W$.

The canonical map. Define

$$\Theta:(\mathcal F\times\mathcal G)_P \longrightarrow\mathcal F_P\times\mathcal G_P, \qquad (s,t)_P\longmapsto(s_P,t_P).$$

If the pairs $(s,t)$ and $(s',t')$ have the same germ, they agree after restriction to some common neighbourhood. Their components also agree there, so $s_P=s'_P$ and $t_P=t'_P$. Hence $\Theta$ is well-defined.

Surjectivity. Take any $(a,b)\in\mathcal F_P\times\mathcal G_P$. Choose a representative $s\in\mathcal F(U)$ of $a$ and $t\in\mathcal G(V)$ of $b$, with $P\in U$ and $P\in V$. They may not yet have the same domain. Since $U\cap V$ is still an open neighbourhood of $P$, we have a pair

$$(s|_{U\cap V},t|_{U\cap V}) \in(\mathcal F\times\mathcal G)(U\cap V).$$

The germ of this pair is sent by $\Theta$ to $(a,b)$: restriction to a smaller neighbourhood does not change the germ. Thus $\Theta$ is surjective.

Injectivity. Suppose $(s,t)\in(\mathcal F\times\mathcal G)(U)$ and $(s',t')\in(\mathcal F\times\mathcal G)(U')$ have the same image under $\Theta$. This means

$$s_P=s'_P,\qquad t_P=t'_P.$$

The first equality gives an open neighbourhood $P\in W_{\mathcal F}\subseteq U\cap U'$ with $s|_{W_{\mathcal F}}=s'|_{W_{\mathcal F}}$. The second equality gives an open neighbourhood $P\in W_{\mathcal G}\subseteq U\cap U'$ with $t|_{W_{\mathcal G}}=t'|_{W_{\mathcal G}}$. Set

$$W:=W_{\mathcal F}\cap W_{\mathcal G}.$$

The set $W$ is still an open neighbourhood of $P$. Restricting once more and using composition of restrictions, both equalities hold simultaneously on $W$. Hence

$$(s,t)|_W=(s|_W,t|_W) =(s'|_W,t'|_W)=(s',t')|_W.$$

Thus $(s,t)_P=(s',t')_P$, and $\Theta$ is injective. Together with surjectivity, this proves the required canonical bijection.

Explicitly, its inverse sends the pair of germs $(s_P,t_P)$ to the germ $(s|_{U\cap V},t|_{U\cap V})_P$. The injectivity just proved ensures that the result is independent of the chosen representatives. Thus all choices in the construction of the inverse have been checked.

Pitfalls and checks

Do not pair two representatives before making their domains agree. A section on $U$ and a section on $V$ yield a section of the product only after both have been restricted to $U\cap V$.

The decisive step is taking a finite intersection of neighbourhoods. The intersection of two open neighbourhoods is still open and contains $P$. The same proof does not automatically apply to infinite products, because an infinite intersection of open neighbourhoods need not be open. To check the inverse, starting with $(s,t)_P$ on one neighbourhood $U$ gives back $(s|_U,t|_U)_P=(s,t)_P$; starting with $(s_P,t_P)$ gives back the two original germs.

BGK 4 mastery bank: gluing and stalks

The following problem statements come from Holger Brenner’s course. The complete solutions, explanations, and brief checks are new editorial material prepared by OpenAI Codex gpt-5.6-sol, Ultra., not translated source solutions. In the Unit 4 freeze, there are no public source solutions for these three selected exercises. Credits to the author and source contributors remain applicable; this text does not claim human authorship or review. The problem statements are unchanged. This material is licensed under CC BY-SA 4.0.

The main prerequisite is Definition 4.1 on sheaves: sections that agree locally are equal, and a family of sections agreeing on intersections has a unique gluing. For a sheaf of sets, $\mathcal F(\varnothing)$ has exactly one element; for a sheaf of groups, this value is the trivial group. Both include the empty cover in the sheaf axiom.

New solution 1 - The product of two sheaves, Exercise 4.1

Source: Exercise 4.1 in the reader, identifier Garbe/Produkt/Aufgabe, page 111820, fixed revision 1082920. The exercise number follows Worksheet 4, revision 1003857; the identity of the transcluded exercise page is recorded separately in the Unit 4 manifest.

Brenner’s problem statement. Let $\mathcal F$ and $\mathcal G$ be sheaves on a topological space $X$. Prove that the assignment

$$\mathcal H(U):=\mathcal F(U)\times\mathcal G(U)$$

with the natural product maps as restrictions defines a sheaf on $X$.

Independent editorial solution. For open sets $V\subseteq U$, define

$$\rho^{\mathcal H}_{U,V}(s,t) :=\bigl(\rho^{\mathcal F}_{U,V}(s), \rho^{\mathcal G}_{U,V}(t)\bigr).$$

Restriction from $U$ to itself is the identity. If $W\subseteq V\subseteq U$, then for every pair $(s,t)$,

$$\rho^{\mathcal H}_{V,W}\rho^{\mathcal H}_{U,V}(s,t) =\bigl(s|_W,t|_W\bigr) =\rho^{\mathcal H}_{U,W}(s,t).$$

Thus $\mathcal H$ is, to begin with, a presheaf.

Take an open cover $U=\bigcup_{i\in I}U_i$. If two pairs $(s,t),(s',t')\in\mathcal H(U)$ have equal restrictions to every $U_i$, equality of pairs means

$$s|_{U_i}=s'|_{U_i},\qquad t|_{U_i}=t'|_{U_i}$$

for every $i$. The local equality axiom for $\mathcal F$ gives $s=s'$, and the same axiom for $\mathcal G$ gives $t=t'$. Hence the pairs are equal.

Now take a compatible family $(s_i,t_i)\in\mathcal H(U_i)$. Compatibility on $U_i\cap U_j$ means precisely the two systems of equalities

$$s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j},\qquad t_i|_{U_i\cap U_j}=t_j|_{U_i\cap U_j}.$$

The sheaf property of $\mathcal F$ gives a unique $s\in\mathcal F(U)$ restricting to all the $s_i$. Likewise, there is a unique $t\in\mathcal G(U)$ restricting to all the $t_i$. The pair $(s,t)\in\mathcal H(U)$ glues the original family and is unique by the local equality property just proved.

Finally, $\mathcal H(\varnothing)$ is a product of two singleton sets and is therefore a singleton. Thus the empty cover also satisfies the axiom. Both sheaf conditions hold for every cover, and $\mathcal H$ is a sheaf.

Pitfall and check. Here the product is taken on each open set, with componentwise restrictions. Specifying the sets $\mathcal F(U)\times\mathcal G(U)$ alone does not specify a presheaf. As a check, the two projections $(s,t)\mapsto s$ and $(s,t)\mapsto t$ commute with restrictions precisely because of the definition above.

New solution 2 - Sections on two disjoint pieces, Exercise 4.2

Source: Exercise 4.2 in the reader, identifier Garbe/Unzusammenhängender Raum/Produkt/Aufgabe, page 111900, fixed revision 1082921. The exercise number follows Worksheet 4, revision 1003857.

Brenner’s problem statement. Let $\mathcal G$ be a sheaf on a disconnected topological space decomposed as

$$X=U\mathbin{\uplus}V,$$

where $U,V$ are open, nonempty, and disjoint. Show that

$$\mathcal G(X)=\mathcal G(U)\times\mathcal G(V).$$

Independent editorial solution. The equality in the statement is a canonical identification through restriction. The map to be proved bijective is

$$R:\mathcal G(X)\longrightarrow\mathcal G(U)\times\mathcal G(V), \qquad s\longmapsto(s|_U,s|_V).$$

If $R(s)=R(t)$, then $s|_U=t|_U$ and $s|_V=t|_V$. Since $U,V$ cover $X$, the local equality axiom gives $s=t$. Thus $R$ is injective.

For surjectivity, choose any pair $(a,b)\in\mathcal G(U)\times\mathcal G(V)$. The only intersection on which a comparison is needed is $U\cap V=\varnothing$. The two restrictions

$$a|_{\varnothing},\ b|_{\varnothing}\in\mathcal G(\varnothing)$$

are automatically equal, since $\mathcal G(\varnothing)$ is a singleton. Thus $(a,b)$ is a compatible family on the cover $\{U,V\}$. The gluing axiom gives a section $s\in\mathcal G(X)$ with $s|_U=a$ and $s|_V=b$. Therefore $R(s)=(a,b)$, so $R$ is surjective.

The inverse of $R$ sends $(a,b)$ to its unique gluing. No additional choice enters this construction; that is why the identification is canonical. If $\mathcal G$ is a sheaf of groups or rings, restrictions are homomorphisms, so $R$ is also an isomorphism for that algebraic structure, not merely a bijection of sets.

Pitfall and check. For a cover whose members are not disjoint, not every pair of sections can be glued. The permissible pairs must satisfy $a|_{U\cap V}=b|_{U\cap V}$. In this exercise the condition is automatic because the intersection is empty, not because gluing ignores compatibility. Nor does the statement literally identify sections with pairs before the map $R$ has been specified.

New solution 3 - The skyscraper sheaf, Exercise 4.9

Source: Exercise 4.9 in the reader, identifier Wolkenkratzergarbe/Gruppe/Garbeneigenschaft/Aufgabe, page 139888, fixed revision 1081969. The exercise number follows Worksheet 4, revision 1003857. The German typo in the source’s final instruction does not change the mathematical statement.

Brenner’s problem statement. Let $X$ be a topological space, $P\in X$, and $G$ a commutative group. For each open set $U\subseteq X$, set

$$\mathcal G(U)= \begin{cases} G,&P\in U,\\ 0,&P\notin U. \end{cases}$$

With the natural restrictions, prove that $\mathcal G$ is a sheaf of commutative groups, determine $\mathcal G_P$, and, if $P$ is closed, determine $\mathcal G_Q$ for every $Q\ne P$.

Independent editorial solution. For $V\subseteq U$, the restriction $\rho_{U,V}$ is the identity $G\to G$ if $P\in V$; the unique homomorphism $G\to0$ if $P\in U$ but $P\notin V$; and the identity of the trivial group $0\to0$ if $P\notin U$. The case $P\notin U$ but $P\in V$ cannot occur. These three cases immediately give the identity restriction and composition laws: once a restriction is zero, every subsequent restriction remains zero. Thus $\mathcal G$ is a presheaf of commutative groups.

Take an open cover $U=\bigcup_{i\in I}U_i$. If $P\notin U$, all the groups $\mathcal G(U)$, $\mathcal G(U_i)$, and those on intersections are trivial. There is only one possible compatible family, and its gluing is unique. This also includes $U=\varnothing$.

If $P\in U$, there is at least one index $i_0$ with $P\in U_{i_0}$. Take a compatible family $g_i\in\mathcal G(U_i)$. For every $i$ with $P\in U_i$, the point $P$ also belongs to $U_i\cap U_{i_0}$. Both restrictions to this intersection are identities on $G$, so compatibility gives

$$g_i=g_{i_0}\quad\text{in }G.$$

For indices with $P\notin U_i$, the section $g_i$ must be zero. Thus $g:=g_{i_0}\in G=\mathcal G(U)$ restricts to every $g_i$: via the identity on pieces containing $P$, and via $G\to0$ on the others. This gluing is unique, because restriction to $U_{i_0}$ is the identity. The same uniqueness argument shows that two sections over $U$ agreeing locally are equal. Hence $\mathcal G$ is a sheaf of commutative groups.

For the stalk at $P$, every open neighbourhood $W$ of $P$ has $\mathcal G(W)=G$, and all restrictions between such neighbourhoods are identities. Concretely, a germ represented by $g\in G$ on a neighbourhood is determined only by $g$: two representatives give the same germ exactly when those elements of $G$ agree. Hence

$$\mathcal G_P\cong G.$$

Finally, suppose $P$ is closed and $Q\ne P$. The set $X\setminus\{P\}$ is open and contains $Q$. Every germ at $Q$ can be represented by a section on an open neighbourhood $W$ of $Q$. After restriction to the smaller neighbourhood

$$W\cap\bigl(X\setminus\{P\}\bigr),$$

the section lies in the zero group. Thus every germ at $Q$ is zero, and

$$\mathcal G_Q=0\qquad(Q\ne P,\ P\text{ closed}).$$

Pitfall and check. The hypothesis that $P$ is closed is needed in the last step, not in constructing the sheaf. If every neighbourhood of $Q$ instead contains $P$, all the groups used to form the stalk at $Q$ are $G$ with identity restrictions, so that stalk is also isomorphic to $G$. Do not assume that every point of a topological space is closed. As another check, when $G=0$, the entire sheaf and all its stalks are indeed zero.

BGK 5 mastery bank: sheafification and quotient stalks

The following problem statements come from Holger Brenner’s course. The complete solutions, explanations, and brief checks are new editorial material prepared by OpenAI Codex gpt-5.6-sol, Ultra., not translated source solutions. Credits to the author and source contributors remain applicable; this text does not claim human authorship or review. No endorsement by the author or source institutions is implied. This material is licensed under CC BY-SA 4.0.

Unit 5 already has a public source solution to Exercise 5.5. That solution continues to count as a source solution and is not recreated here. The two new solutions below, to Exercises 5.2 and 5.11, complete the three mastery items for Unit 5. Both problem statements are preserved.

The notation $s_P$ denotes the germ of a section $s$ at $P$, while $\mathcal F_P$ is the stalk of the presheaf or sheaf $\mathcal F$. Equality of two germs means that their representatives agree after restriction to a sufficiently small open neighbourhood; they need not already agree on their original neighbourhoods.

New solution 1 - The universal property of sheafification, Exercise 5.2

Source: Exercise 5.2 in the reader, identifier Prägarbe/Vergarbung/Universelle Eigenschaft/Aufgabe, page 111906, fixed revision 1083991. The exercise number follows Worksheet 5, revision 619386; the revision of the transcluded exercise page is recorded separately in the Unit 5 manifest.

Brenner’s problem statement. Let $\mathcal F$ be a presheaf on a topological space $X$, and let $\widetilde{\mathcal F}$ be its sheafification. For every presheaf morphism $\psi:\mathcal F\to\mathcal G$ to a sheaf $\mathcal G$, prove that there is exactly one morphism

$$\widetilde\psi:\widetilde{\mathcal F}\longrightarrow\mathcal G$$

factoring $\psi$ through the canonical morphism $\eta:\mathcal F\to\widetilde{\mathcal F}$. That is, $\widetilde\psi\circ\eta=\psi$.

Independent editorial solution. We construct the component $\widetilde\psi_U$ for each open set $U$, then prove that all components commute with restrictions. Use Definition 5.1: a section $\sigma\in\widetilde{\mathcal F}(U)$ is a family of germs $(\sigma_P)_{P\in U}$ locally arising from sections of $\mathcal F$. Thus there is an open cover $U=\bigcup_iU_i$ and $s_i\in\mathcal F(U_i)$ with

$$\sigma|_{U_i}=\eta_{U_i}(s_i), \qquad (s_i)_Q=\sigma_Q\quad(Q\in U_i).$$

We want to glue the sections

$$t_i:=\psi_{U_i}(s_i)\in\mathcal G(U_i).$$

To check compatibility, take $Q\in U_i\cap U_j$. The germs $(s_i)_Q$ and $(s_j)_Q$ agree. Since $\psi$ commutes with restrictions, it induces a stalk map $\psi_Q$, so

$$(t_i)_Q=\psi_Q((s_i)_Q) =\psi_Q((s_j)_Q)=(t_j)_Q.$$

By Lemma 4.4, the stalkwise test for equality of sections, applied to the sheaf $\mathcal G$ over $U_i\cap U_j$, we obtain

$$t_i|_{U_i\cap U_j}=t_j|_{U_i\cap U_j}.$$

Since $\mathcal G$ is a sheaf, the family $(t_i)$ has a unique gluing $t\in\mathcal G(U)$. Define $\widetilde\psi_U(\sigma):=t$.

This construction is independent of the cover or local representatives. Indeed, suppose $(V_j,r_j)$ is another choice of representatives for $\sigma$. At each point $Q\in U_i\cap V_j$, we have $(s_i)_Q=(r_j)_Q=\sigma_Q$. The same stalk argument gives

$$\psi_{U_i}(s_i)|_{U_i\cap V_j} =\psi_{V_j}(r_j)|_{U_i\cap V_j}.$$

The two glued sections have equal restrictions on the refinement cover $\{U_i\cap V_j\}_{i,j}$ of $U$. Local uniqueness in $\mathcal G$ shows that the results are equal. Thus $\widetilde\psi_U$ is well-defined.

Now take an open set $W\subseteq U$. The restriction $\sigma|_W$ is represented by $s_i|_{U_i\cap W}$ on the cover $\{U_i\cap W\}_i$. Since $\psi$ is a presheaf morphism,

$$\psi_{U_i\cap W}(s_i|_{U_i\cap W}) =\psi_{U_i}(s_i)|_{U_i\cap W}.$$

Thus $\widetilde\psi_W(\sigma|_W)$ and $\widetilde\psi_U(\sigma)|_W$ have equal local restrictions. Uniqueness of gluing gives

$$\widetilde\psi_W(\sigma|_W) =\widetilde\psi_U(\sigma)|_W.$$

Hence this family of components is a sheaf morphism.

To check the factorisation, take $s\in\mathcal F(U)$. The section $\eta_U(s)$ has representative $s$ on all of $U$, so the construction with the one-member cover gives

$$\widetilde\psi_U(\eta_U(s))=\psi_U(s).$$

Thus $\widetilde\psi\circ\eta=\psi$.

For uniqueness, suppose $\theta:\widetilde{\mathcal F}\to\mathcal G$ is another morphism with $\theta\circ\eta=\psi$. For a section $\sigma$ and the local representatives $s_i$ above, naturality of $\theta$ gives

$$\theta_U(\sigma)|_{U_i} =\theta_{U_i}(\eta_{U_i}(s_i)) =\psi_{U_i}(s_i) =\widetilde\psi_U(\sigma)|_{U_i}.$$

The sections are equal by the sheaf property of $\mathcal G$. This holds for all $U$ and $\sigma$, so $\theta=\widetilde\psi$. For $U=\varnothing$, both sheaves take singleton values, so the component and all its identities are also unique. The complete proof does not assume that $\mathcal F$ is already a sheaf.

Pitfall and check. Do not define $\widetilde\psi_U$ only on sections coming from $\eta_U(\mathcal F(U))$: $\eta_U$ need not be surjective. Representatives are available locally, and it is the sheaf property of the target $\mathcal G$ that permits gluing. If $\mathcal F$ is already a sheaf, $\eta$ is an isomorphism by Lemma 5.2(4); the formula above then gives $\widetilde\psi=\psi\circ\eta^{-1}$, as expected.

New solution 2 - Stalks of a quotient sheaf, Exercise 5.11

Source: Exercise 5.11 in the reader, identifier Garben von Gruppen/Untergarbe/Quotientengarbe/Halm/Aufgabe, page 112024, fixed revision 1082924. The exercise number follows Worksheet 5, revision 619386.

Brenner’s problem statement. Let $\mathcal G$ be a sheaf of commutative groups, $\mathcal F\subseteq\mathcal G$ a subsheaf of groups, and $\mathcal G/\mathcal F$ its quotient sheaf. Prove that, for every point $P\in X$,

$$(\mathcal G/\mathcal F)_P=\mathcal G_P/\mathcal F_P.$$

Independent editorial solution. This equality means a canonical group isomorphism. We must not replace the quotient sheaf by the quotient of sections on each open set. Following Definition 5.8, first form the presheaf of groups

$$\mathcal Q(U):=\mathcal G(U)/\mathcal F(U), \qquad \rho^{\mathcal Q}_{U,V}([g]):=[g|_V].$$

These restrictions are well-defined: if $g-h\in\mathcal F(U)$, then $g|_V-h|_V\in\mathcal F(V)$ because $\mathcal F$ is a subsheaf. The quotient sheaf is $\widetilde{\mathcal Q}$. By Lemma 5.2(2), which applies to every presheaf, the canonical map induces an isomorphism

$$\mathcal Q_P\xrightarrow{\ \sim\ } \widetilde{\mathcal Q}_P =(\mathcal G/\mathcal F)_P.$$

It therefore suffices to determine $\mathcal Q_P$.

The quotient homomorphisms on each open set form a presheaf morphism $q:\mathcal G\to\mathcal Q$. On stalks, it gives

$$q_P:\mathcal G_P\longrightarrow\mathcal Q_P, \qquad g_P\longmapsto[g]_P.$$

If two sections $g\in\mathcal G(U)$ and $h\in\mathcal G(V)$ represent the same germ at $P$, they agree on some open neighbourhood $P\in W\subseteq U\cap V$. Their quotient classes then also agree in $\mathcal Q(W)$, so the formula for $q_P$ is independent of the representative. Addition of germs is computed after shrinking to a common neighbourhood; since every $q_U$ is a homomorphism, so is $q_P$.

The map $q_P$ is surjective. Every element $\xi\in\mathcal Q_P$ has a representative $a\in\mathcal Q(U)$ on some open neighbourhood $U$ of $P$. By the definition of a quotient group, there is $g\in\mathcal G(U)$ with $a=[g]$. Thus $\xi=[g]_P=q_P(g_P)$.

Next, the inclusion $\mathcal F\subseteq\mathcal G$ induces a stalk inclusion $\mathcal F_P\subseteq\mathcal G_P$, by Lemma 4.5. We show that this subgroup is exactly the kernel of $q_P$.

If $g_P\in\ker q_P$, then $[g]_P=0$ in $\mathcal Q_P$. By the definition of equality of germs, there is an open neighbourhood $W$ of $P$, contained in the domain of the representative $g$, such that

$$[g|_W]=0\quad\text{in }\mathcal G(W)/\mathcal F(W).$$

This means precisely that $g|_W\in\mathcal F(W)$, so the original germ $g_P$ belongs to $\mathcal F_P$. Conversely, every element of $\mathcal F_P$ has a representative $f\in\mathcal F(W)$. The image of $f$ in $\mathcal Q(W)$ is zero, so $q_P(f_P)=0$. Thus

$$\ker q_P=\mathcal F_P.$$

The first isomorphism theorem for commutative groups now gives the explicit isomorphism

$$\mathcal G_P/\mathcal F_P\xrightarrow{\ \sim\ }\mathcal Q_P, \qquad g_P+\mathcal F_P\longmapsto[g]_P.$$

Composing it with the sheafification isomorphism yields

$$\mathcal G_P/\mathcal F_P \xrightarrow{\ \sim\ }(\mathcal G/\mathcal F)_P,$$

giving the required canonical identification. The construction uses a representative on a sufficiently small neighbourhood; it makes no claim that a global representative is always available.

Pitfall and check. The germ equality $[g]_P=0$ means that there is a neighbourhood $W$ with $g|_W\in\mathcal F(W)$; it does not immediately mean that $g\in\mathcal F(U)$ on its original domain. Sheafification can change global sections but not stalks. As boundary checks, if $\mathcal F=0$, the formula gives the stalk $\mathcal G_P$; if $\mathcal F=\mathcal G$, both sides are the zero group.

BGK 6 mastery exercises

The following three problem statements come from Holger Brenner’s course on Wikiversity. The complete solutions and learning checks are new editorial material prepared independently by OpenAI Codex gpt-5.6-sol, Ultra. These are not translated public solutions by Brenner: the frozen source map records the absence of public solutions for all three exercises. Their mathematical statements and hypotheses are unchanged.

This text is licensed under CC BY-SA 4.0, with source credits preserved. No claim of human authorship, endorsement, or review is made for these editorial solutions. The learning sequence moves from sections on open sets to stalks and then to sheaf morphisms.

New item 1 - Brenner Exercise 6.4: a split sequence

Source: Exercise 6.4 in the BGK reader. Frozen entity: Topologische Gruppen/Spaltende Sequenz/Garbenversion/Aufgabe, pageid 112025, revision 1050308. The exercise number follows Worksheet 6, revision 900086.

Source problem statement

Let $F$ and $H$ be commutative topological groups, $G=F\times H$ with the product topology, and

$$0\longrightarrow F\longrightarrow G\longrightarrow H\longrightarrow0$$

the associated product short exact sequence. Prove that, for every topological space $X$, there is a short exact sequence of sheaves

$$0\longrightarrow C^0(-,F)\longrightarrow C^0(-,G) \longrightarrow C^0(-,H)\longrightarrow0,$$

and that the rightmost map remains surjective after global evaluation on $X$.

Complete independent solution

Use additive notation for all three groups. The inclusion and projection in question are $j(a)=(a,0_H)$ and $p(a,b)=b$. For each open $U\subseteq X$, define

$$\begin{aligned} j_U:C^0(U,F)&\longrightarrow C^0(U,G),& j_U(a)(x)&=(a(x),0_H),\\ p_U:C^0(U,G)&\longrightarrow C^0(U,H),& p_U(b)(x)&=p(b(x)). \end{aligned}$$

Both maps are group homomorphisms. They also produce continuous functions, since the inclusion $j$ and projection $p$ are continuous. Composition with a fixed function commutes with restriction to an open set, so the families $j_U,p_U$ are presheaf morphisms.

The presheaf $C^0(-,F)$ is a sheaf: local functions agreeing on intersections determine exactly one function on their union, and that function is continuous because continuity can be tested on an open cover. The same argument applies to $G$ and $H$. Thus these are indeed sheaf morphisms.

Now check exactness for every $U$. If $j_U(a)=0$, then $(a(x),0_H)=(0_F,0_H)$ for every $x\in U$, so $a=0$; hence $j_U$ is injective. The composite $p_Uj_U$ is zero. Conversely, if $p_U(b)=0$, each $b(x)$ has the form $(a(x),0_H)$. The function $a=\operatorname{pr}_F\circ b$ is continuous and $j_U(a)=b$. Thus

$$\ker p_U=\operatorname{im}j_U.$$

For surjectivity, take any $h\in C^0(U,H)$ and set

$$\sigma_U(h)(x):=(0_F,h(x)).$$

This function is continuous, and $p_U\sigma_U(h)=h$. The family $\sigma_U$ itself is compatible with restrictions. Thus $\sigma$ is a sheaf morphism with $p\sigma=\operatorname{Id}$: the sheaf sequence even splits. Every section in the kernel comes from a section on the left, and every section on the right has a preimage on the same open set; this proves sheaf exactness, not just a formal statement about a complex.

In particular, for $U=X$, the formula $h\mapsto(x\mapsto(0_F,h(x)))$ gives a right inverse on global sections. Hence

$$C^0(X,G)\longrightarrow C^0(X,H)$$

is surjective, as required. The entire argument also applies to $U=\varnothing$, when the group of maps has just one element.

Pitfall and check

In general, surjectivity of a sheaf morphism guarantees only local preimages, not global ones. Here the decisive extra step is the existence of a single global right inverse $H\to F\times H$, $h\mapsto(0_F,h)$. Check directly that $\sigma_U(h)|_V=\sigma_V(h|_V)$ for $V\subseteq U$; this is why the splittings on all open sets constitute a sheaf splitting.

New item 2 - Brenner Exercise 6.9: pushforward from a point

Source: Exercise 6.9 in the BGK reader. Frozen entity: Topologischer Raum/Punkt/Vorschub/Wolkenkratzergarbe/Aufgabe, pageid 112028, revision 1084500. The exercise number follows Worksheet 6, revision 900086.

Source problem statement

Let $X$ be a topological space, $P\in X$, and $i:\{P\}\to X$ the inclusion. For a sheaf of commutative groups $\mathcal F$ on $\{P\}$, describe $i_*\mathcal F$ on the open sets of $X$. Determine its stalks when $P$ is a closed point.

Complete independent solution

Write $A:=\mathcal F(\{P\})$. Since $\mathcal F$ is a sheaf of groups, $\mathcal F(\varnothing)=0$: the gluing condition for the empty cover gives exactly one section on the empty set. These two open sets give all the sheaf data on a one-point space.

By Definition 6.9, for open $U\subseteq X$,

$$(i_*\mathcal F)(U)=\mathcal F(i^{-1}(U)) =\begin{cases} A,&P\in U,\\ 0,&P\notin U. \end{cases}$$

If $V\subseteq U$, there are three possibilities. If $P\in V$, both section groups are $A$ and the restriction is the identity $A\to A$. If $P\in U$ but $P\notin V$, the restriction is the unique homomorphism $A\to0$. If $P\notin U$, the restriction is $0\to0$. The possibility $P\notin U$ but $P\in V$ cannot occur. Thus the entire restriction structure is determined. This pushforward is a sheaf by Lemma 6.10, which applies to every continuous map and every sheaf on its domain.

At $P$, every open neighbourhood contains $P$. All the groups forming the stalk are $A$, and all restriction homomorphisms are identities. The map $A\to(i_*\mathcal F)_P$ sends $a$ to the germ of the section $a$ on $X$. It is surjective because every germ representative comes from a copy of $A$, and injective because restrictions never identify two distinct elements. Thus

$$(i_*\mathcal F)_P\cong A.$$

Now suppose $\{P\}$ is closed and take $Q\ne P$. The set $X\setminus\{P\}$ is open and contains $Q$. Every open neighbourhood $U$ of $Q$ can be shrunk to $U\cap(X\setminus\{P\})$. On that smaller neighbourhood, the pushforward section group is zero. Consequently every germ representative at $Q$ becomes zero after restriction, so