8.7: Solutions for Chapter 7
- Page ID
- 55031
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In the commutative diagram below, suppose the (B, C, B ′, C′) square is a pullback:
We need to show that the (A, B, A ′, B ′) square is a pullback iff the (A, C, A ′, C′) rectangle is a pullback.
Suppose first that (A, B, A ′, B ′) is a pullback, and take any (X, p, q) as in the following diagram:
where q ; f ; g ' = p ; h\(_{3}\). Then by the universal property of the (B, C, B ', C' ) pullback, we get a unique dotted arrow r making the left-hand diagram below commute:
In other words r ; h\(_{2}\) = g ; f ′ andr ; g = p. Then by the universal property of the (A, B, A ′, B ′) pullback, we get a unique dotted arrow r ' : X → A making the right-hand diagram commute, i.e. r ' ; f = r and r ' ; h\(_{1}\) = q. This gives the existence of an r with the required property, r ' ; f = r and r ' ; f = r ; g = p. To see uniqueness, suppose given another morphisms r\(_{0}\) such that r\(_{0}\) ; f ; g = p and r\(_{0}\) ; h\(_{1}\) = q:
Then by the uniqueness of r, we must have r\(_{0}\) ; f = r, and then by the uniqueness of r ′, we must have r\(_{0}\) = r ′. This proves the first result.
The second is similar. Suppose that (A, C, A ′, C′) and (B, C, B ′, C ′) are pullbacks and suppose given a commutative diagram of the following form:
i.e. where r ; h\(_{2}\) = q ; f ′. Then letting p := r ; g, we have
p ; h\(_{3}\) = r ; g ; h\(_{3}\) = r ; h\(_{2}\) ; g ' = q ; f ' ; g '
so by the universal property of the (A, C, A ′, C ′) pullback, there is a unique morphism r′ : X → A such that r ' ; f ; g = p and r\(_{0}\) ; h\(_{1}\) = q, as shown:
But now let r\(_{0}\) := r ' ; f . It satisfies r\(_{0}\) ; g = p and r\(_{0}\) ; h\(_{2}\) = q ; f ', and r satisfies the same equations: r ; g = p and r ; h\(_{2}\) = q ; f ′. Hence by the universal property of the (B, C, B ′,C ′) pullback r\(_{0}\) = r ′. It follows that r′ is a pullback of the (A, B, A ′, B ′) square, as desired.
A function f : A → B is injective iff for all a\(_{1}\), a\(_{2}\) \(\in\) A, if f (a\(_{1}\)) = f (a\(_{2}\)) then a\(_{1}\) = a\(_{2}\).
It is a monomorphism iff for all sets X and functions g\(_{1}\), g\(_{2}\) : X → A, if g\(_{1}\) ; f = g\(_{2}\) ; f then g\(_{1}\) = g\(_{2}\). Indeed, this comes directly from the universal property of the pullback from Definition 7.5,
because the dashed arrow is forced to equal both g\(_{1}\) and g\(_{2}\), thus forcing g\(_{1}\) = g\(_{2}\).
1. Suppose f is a monomorphism, let a\(_{1}\), a\(_{2}\) \(\in\) A be elements, and suppose f (a\(_{1}\)) = f (a\(_{2}\)).
Let X = {∗} be a one element set, and let g\(_{1}\), g\(_{2}\) : X → A be given by g\(_{1}\)(∗) := a\(_{1}\) and g\(_{2}\)(∗) := a\(_{2}\). Then g\(_{1}\) ; f = g\(_{2}\) ; f , so g\(_{1}\) = g\(_{2}\), so a\(_{1}\) = a\(_{2}\).
2. Suppose that f is an injection, let X be an y set, and let g\(_{1}\), g\(_{2}\) : X → A be such that g\(_{1}\) ; f = g\(_{2}\) ; f. We will have g\(_{1}\) = g\(_{2}\) if we can show that g\(_{1}\)(x) = g\(_{2}\)(x) for every x \(\in\) X. So take any x \(\in\) X; since f (g\(_{1}\)(x)) = f (g\(_{2}\)(x)) and f is injective, we have g\(_{1}\)(x) = g\(_{2}\)(x) as desired.
1. Suppose we have a pull back as shown, where i is an isomorphism:
Let j := i\(^{-1}\) be the inverse of i, and consider g := (f ; j): A → B ′. Then g ; i = f, so by the existence part of the universal property, there is a map j : A → A ' such that j ' ; i ' = id\(_{A}\) and j ' ; f ' = f ; j. We will be done if we can show i ' ; j ' = id\(_{A'}\). One checks that (i ' ; j ' ) ; i ' = i ' and that (i ' ; j ' ) ; f ' = i ' ; f ; j = f ' ; i ; j = f ' .
But id\(_{A'}\) also satisfies those properties : id\(_{A'}\) ; i ' = i ' and id\(_{A'}\) ; f ' = f ', so by the uniqueness part of the universal property, (i ' ; j ' ) = id\(_{A'}\).
2. We need to show that the following diagram is a pullback:
So take any object X and morphisms g : X → A and h : X → B such that g ; f = h ; id\(_{B}\).
We need to show there is a unique morphism r : X → A such that r ; id\(_{A}\) = g and r ; f = h. That’s easy: the first requirement forces r = g and the second requirement is then fulfilled.
Consider the diagram shown left, in which all three squares are pullbacks:
The front and bottom squares are the same—the assumed pullback—and the right-hand square is a pullback because f is assumed monic. We can complete it to the commutative diagram shown right, where the back square and top square are pullbacks by Exercise 7.7. Our goal is to show that the left-hand square is a pullback.
To do this, we use two applications of the pasting lemma, Exercise 7.4. Since the right-hand face is a pullback and the back face is a pullback, the diagonal rectangle (lightly drawn) is also a pullback. Since the front face is a pullback, the left-hand face is also a pullback.
The following is an epimono factorization of f :
1. If V is a quantale with the stated properties, then
• I serves as a top element : v ≤ I for all v \(\in\) V.
• v ⊗ w serves as a meet operation, i.e. it satisfies the same universal property as \(\land\), namely v ⊗ w is a greatest lower bound for v and w.
Now the operation satisfies the same universal property as exponentiation (hom-object) does, namely v ≤ (w -o x) iff v ⊗ W ≤ x. So V is a cartesian closed category, and of course it is a preorder.
2. Not every cartesian closed preorder comes from a quantale with the stated properties, because quantales have all joins and cartesian closed preorders need not. Finding a counterexample a cartesian closed preorder that is missing some joins—takes some ingenuity, but it can be done. Here’s one we came up with:
This is the product preorder \(\mathbb{N}\)\(^{op}\) × \(\mathbb{N}\)\(^{op}\): its objects are pairs (a, b) \(\in\) \(\mathbb{N}\) × \(\mathbb{N}\) with (a, b) ≤ (a ′, b ′) iff, in the usual ordering on \(\mathbb{N}\) we have a ′ ≤ a and b ′ ≤ b. But you can just look at the diagram.
It has a top element, (0, 0), and it has binary meets, (a, b) \(\land\) (a ′, b ′) = (max(a, a ′), max(b, b ′)). But it has no bottom element, so it has no empty join. Thus we will be done if we can show that for each x, y, the hom-object x -o y exists. The formula for it is x -o y = \(\bigvee\){w | w \(\land\) x ≤ y}, i.e. we need these particular joins to exist. Since y \(\land\) x ≤ y, we have y ≤ x -o y. So we can replace the formula with x -o y = \(\bigvee\){w | y ≤ w and w \(\land\) x ≤ y}. But the set of elements in \(\mathbb{N}\)\(^{op}\) × \(\mathbb{N}\)\(^{op}\) that are bigger than y is finite and nonempty.\(^{2}\) So this is a finite nonempty join, and \(\mathbb{N}\)\(^{op}\) × \(\mathbb{N}\)\(^{op}\) has all finite nonempty joins: they are given by inf.
Let m : \(\mathbb{Z}\) → \(\mathbb{B}\) be the characteristic function of the inclusion \(\mathbb{N}\) \(\subseteq\) \(\mathbb{Z}\).
1. \(\lceil{m(−5)}\rceil\) = false. 2. \(\lceil{m(0)}\rceil\) = true.
1. The characteristic function \(\lceil{id_{\mathbb{N}}\rceil\) : \(\mathbb{N}\) → \(\mathbb{B}\) sends each n \(\in\) \(\mathbb{N}\) to true.
2. Let !\(_{\mathbb{N}}\) : Ø → \(\mathbb{N}\) be the inclusion of the empty set. The characteristic function \(\lceil{!_{\mathbb{N}}\rceil\) : \(\mathbb{N}\) → \(\mathbb{B}\) sends each n \(\in\) \(\mathbb{N}\) to false.
1. The sort of thing (*?*) we’re looking for is a subobject of B, say A \(\subseteq\) B. This would have a characteristic function, and we’re trying to find the A for which the characteristic function is ¬: \(\mathbb{B}\) → \(\mathbb{B}\).
2. The question now asks “what is A?” The answer is {false} \(\subseteq\) B.
1. Here is the truth table for P = (P \(\land\) Q):
- Yes!
- The characteristic function for P ⇒ Q is the function \(\lceil{⇒}\rceil\) : \(\mathbb{B}\) × \(\mathbb{B}\) → \(\mathbb{B}\) given by the first, second, and fourth column of Eq. (A.3).
- It classifies the subset {(true, true), (false, true), (false, false) \(\subseteq\) \(\mathbb{B}\) × \(\mathbb{B}\).
Say that \(\lceil{E}\rceil\), \(\lceil{P}\rceil\), \(\lceil{T}\rceil\) : \(\mathbb{N}\) → \(\mathbb{B}\) classify respectively the subsets E := {n \(\in\) \(\mathbb{N}\) | n is even}, P := {n \(\in\) \(\mathbb{N}\) | n is prime}, and T := {n \(\in\) \(\mathbb{N}\) | n ≥ 10} of \(\mathbb{N}\).
1. \(\lceil{E}\rceil\)(17) = false because 17 is not even.
2. \(\lceil{P}\rceil\)(17) = true because 17 is prime.
3. \(\lceil{T}\rceil\)(17) = true because 17 ≥ 10.
4. The set classified by (\(\lceil{E}\rceil\) \(\land\) \(\lceil{P}\rceil\)) \(\lor\) \(\lceil{T}\rceil\) is that of all natural numbers that are either above 10 or an even prime. The smallest three elements of this set are 2, 10, 11.
1. The 1-dimensional analogue of an \(\mathcal{E}\)-ball around a point x \(\in\) \(\mathbb{R}\) is B(x, \(\mathcal{E}\)) := {x′ \(\in\) R || x − x ′ | < \(\mathcal{E}\)}, i.e. the set of all points within \(\mathcal{E}\) of x.
2. A subset U \(\subseteq\) \(\mathbb{R}\) is open if, for every x \(\in\) U there is some \(\mathcal{E}\) > 0 such that B(x, \(\mathcal{E}\)) \(\subseteq\) U.
3. Let U\(_{1}\) := {x \(\in\) \(\mathbb{R}\) | 0 < x < 2} and U\(_{2}\) := {x \(\in\) \(\mathbb{R}\) | 1 < x < 3}. Then U := U\(_{1}\) \(\cup\) U\(_{2}\) = {x \(\in\) \(\mathbb{R}\) | 0 < x < 3}.
4. Let I = {1, 2, 3, 4, ...} and for each i \(\in\) I let U := {x \(\in\) \(\mathbb{R}\) | \(\frac{1}{i}\) < x < 1}, so we have U\(_{1}\) (\subseteq\) U\(_{2}\) (\subseteq\) U\(_{3}\) (\subseteq\) · · · .
Their union is U := \(\bigcup\)\(_{i \in I}\)U\(_{i}\) = {x \(\in\) \(\mathbb{R}\) | 0 < x < 1}.
1. The coarse topology on X is the one whose only open sets are X (\subseteq\) X and Ø (\subseteq\) X. This is a topology because it contains the top and bottom subsets, it is closed under finite intersection (the intersection A ∩ B is Ø iff one or the other is Ø), and it is closed under arbitrary union (the union \(\bigcup_{i \in I} A_{i}\) is X iff A\(_{i}\) = X for some i \(\in\) I ).
2. The fine topology on X is the one where every subset A (\subseteq\) X is considered open. All the conditions on a topology say “if such-and-such then such-and-such is open,” but these are all satisfied because everything is open!
3. If (X, P(X)) is discrete, (Y, Op\(_{Y}\)) is any topological space, and f : X → Y is any function then it is continuous. Indeed, this just means that for any open set U (\subseteq\)Y the preimage f\(^{ −1}\)(U) (\subseteq\) X is open, and everything in X is open.
1. The Hasse diagram for the Sierpinsky topology is Ø → {1} → {1, 2}.
2. A set \(\left(U_{i}\right)_{i \in I}\) covers U iff either
• I = Ø andU = Ø; or
• U\(_{i}\) = U for some i \(\in\) I.
In other words, the only way that some collection of these sets could cover another set U is if that collection contains U or if U is empty and the collection is also empty.
Let (X, Op) be a topological space, suppose that Y (\subseteq\) X is a subset, and consider the subspace topology Op\(_{?∩Y}\) .
1. We want to show that Y \(\in\) Op\(_{?∩Y}\). We need to find B \(\in\) Op such that Y = B ∩ Y; this is easy, you could take B = Y or B = X, or anything in between.
2. We still need to show that Op\(_{?∩Y}\) contains Ø and is closed under finite intersection and arbitrary union. Ø = Ø ∩ Y, so according to the formula, Ø \(\in\) Op\(_{?∩Y}\) . Suppose that A\(_{1}\), A\(_{2}\) \(\in\) Op\(_{?∩Y}\) . Then there exist B\(_{1}\), B\(_{2}\) \(\in\) Op with A\(_{1}\) = B\(_{1}\) ∩ Y and A\(_{2}\) = B\(_{2}\) ∩ Y. But then A\(_{1}\) ∩ A\(_{2}\) = (B\(_{1}\) ∩ Y) ∩ (B\(_{2}\) ∩ Y) = (B\(_{1}\) ∩ B\(_{2}\)) ∩ Y, so it is in Op\(_{?∩Y}\) since B\(_{1}\) ∩ B\(_{2}\) \(\in\) Op.
The same idea works for arbitrary unions: given a set I and A\(_{i}\) for each i \(\in\) I, we have A\(_{i}\) = B\(_{i}\) ∩ Y for some B\(_{i}\) \(\in\) Op, and
\(\bigcup_{i \in I} A_{i}=\bigcup_{i \in I}\left(B_{i} \cap Y\right)=\left(\bigcup_{i \in i} B_{i}\right) \cap Y \in \mathbf{O p}_{? \cap Y}\)
Let’s imagine a V-category C, where V is the quantale corresponding to the open sets of a topological space (X, Op). Its Hasse diagram would be a set of dots and some arrows between them, each labeled by an open set U (\subseteq\) Op. It might look something like this:
Recall from Section 2.3 that the ‘distance’ between two points is computed by taking the join, over all paths between them, of the monoidal product of distances along that path. For example, C(B, C) = (U\(_{3}\) \(\land\) U\(_{1}\)) \(\lor\) (U\(_{4}\) \(\land\) U\(_{2}\)), because \(\land\) is the monoidal product in V.
In general, we can thus imagine the open set C(a, b) as a kind of ‘size restriction’ for getting from a to b, like bridges that your truck needs to pass under. The size restriction for getting from a to itself is X: no restriction. In general, to go on any given route (path) from a to b, you have to fit under every bridge in the path, so we take their meet. But we can go along any path, so we take the join over all paths.
1. The fiber of f over a is {a\(_{1}\), a\(_{2}\)}.
2. The fiber of f over c is {c\(_{1}\)}.
3. The fiber of f over d is Ø.
4. A function f ′ : X → Y for which every fiber has either one or two elements is shown below.
Refer to Eq. (A.4).
1. Here is a drawing of all six sections over V\(_{1}\) = {a, b, c}:
2. When V\(_{2}\) = {a, b, c, d}, there are no sections: Sec\(_{f}\)(V\(_{2}\)) = Ø.
3. When V\(_{3}\) = {a, b, d, e}, the set Sec\(_{f}\)(V\(_{3)\))) has 2 ∗ 3 ∗ 1 ∗ 2 = 12 elements.
Sec\(_{f}\)({a, b, c}) and Sec\(_{f}\)({a, c}) are drawn as the top row (six-element set) and bottom row (two-element set) below, and the restriction map is also shown:
1. Let g\(_{1}\) := (a\(_{1}\), b\(_{1}\)) and g\(_{2}\) := (b\(_{2}\), e\(_{1}\)); these do not agree on the overlap.
2. No, there’s no section g \(\in\) Sec\(_{f}\)(U\(_{1}\) \(\cup\) U\(_{2}\)) for which g|\(_{U_{2}\) = g\(_{1}\) and g|\(_{U_{2}\) = g\(_{2}\)
No, there is not a one-to-one correspondence between sheaves on M and vector fields on M. The relationship between sheaves on M and vector fields on M is that the set of all vector fields on M corresponds to one sheaf, namely Sec\(_{\pi}\), where \(\pi\): TM → M is the tangent bundle as described in Example 7.46. There are so many sheaves on M that they don’t even form a set (it’s just a ‘collection’); again, one member of this gigantic collection is the sheaf Sec\(_{\pi}\) of all possible vector fields on M.
1. The Hasse diagram for the Sierpinsky topology is Ø → {1} → {1, 2} .
2. A presheaf F on Op consists of any three sets and any two functions F({1, 2}) → F({1}) → F(Ø) between them.
3. Recall from Exercise 7.31 that the only non-trivial covering (a covering of U is non-trivial if it does not contain U) occurs when U = Ø in which case the empty family over U is a cover.
4. As explained in Example 7.36, F will be a sheaf iff F(Ø) \(\cong\) {1}. Thus we the category of sheaves is equivalent to that of just two sets and one function F({1, 2}) → F({1}).
The one-point space X = {1} has two open sets, Ø and {1}, and every sheaf S \(\in\) Shv(X) assigns S(Ø) = {()} by the sheaf condition (see Example 7.36). So the only data in a sheaf S \(\in\) Shv(X) is the set S({1}). This is how we get the correspondence between sets and sheaves on the one point space. According to Eq. (7.50), the subobject classifier Ω: Op(X)\(^{op}\) → Set in Shv(X) should be the functor where Ω({1}) is the set of open sets of {1}. So we’re hoping to see that there is a one-to-one correspon- dence between the set Op({1}) and the set \(\mathbb{B}\) = {true, false} of booleans. Indeed there is: there are two open sets of {1}, as we said, Ø and {1}, and these correspond to false and true respectively.
By Eqs. (7.50) and (7.51) the definition of Ω(U) is Ω(U) := {U′ \(\in\) Op | U′ \(\subseteq\) U}, and the definition of the restriction map for V (\subseteq\) U is U ′ → U ′ ∩ V.
1. It is functorial: given W \(\subseteq\) V \(\subseteq\) U and U ′ \(\subseteq\) U, we indeed have (U ′ ∩ V) ∩ W U ′ ∩ W, since W \(\subseteq\) V. For functoriality, we also need preservation of identities, and this amounts to U ′ ∩ U = U ′ for all U ′ \(\subseteq\) U.
2. Yes, a presheaf is just a functor; the above check is enough.
We need a graph homomorphism of the following form:
There is only one that classifies G ′, and here it is. Let’s write γ := \(\lceil{G′}\rceil\).
- Since D is missing from G ′, we have γ(D) = 0(vertex: missing).
- Since vertices A, B, C are present in G ′ we have γ(A) = γ(B) = γ(C) = V (vertex: present).
- The above forces γ(i) = (V, 0; 0)(arrow from present vertex to missing vertex: missing).
- Since the arrow f is in G ′, we have γ( f ) = (V, V; A) (arrow from present vertex to present vertex: present).
- Since the arrows g and h are missing in G ′, we have γ(g) = γ(h) = (V, V; 0)(arrow from present vertex to present vertex: missing).
With U = \(\mathbb{R}\) − {0} \(\subseteq\) \(\mathbb{R}\), we have:
1. The complement of U is \(\mathbb{R}\) − U{0} and ¬U is its interior, which is ¬U.
2. The complement of ¬U is \(\mathbb{R}\) − \(\mathbb{R}\), and this is open, so ¬¬U = \(\mathbb{R}\).
3. It is true that U \(\subseteq\) ¬¬U.
4. It is false that ¬¬U \(\subseteq\) \(^{?}\) U.
1. If for any V \(\in\) Op we have ⊤ \(\land\) V = V then when V = X we have ⊤ \(\land\) X := ⊤ ∩ X = X, but anything intersected with X is itself, so ⊤ = ⊤ ∩ X = X.
2. (⊤ \(\lor\) V) := (X \(\cup\) V) = X holds and (V ⇒ X) \(\bigcup_{\{R \in \mathbf{O p} \mid R \cap V \subseteq X\}} R=X\) holds because (X ∩ V) \(\subseteq\) X.
3. If for any set V \(\in\) Op we have (⊥ \(\lor\) V) = V, then when V = ∅ we have (⊥\(\lor\) Ø) = (⊥ ∪ Ø) = Ø, but anything unioned with Ø is itself, so ⊥ = ⊥ ∪ Ø = Ø.
4. (⊥ \(\land\) V) = (Ø ∩ V) = Ø holds, and (⊥ ⇒ V) = \(\bigcup_{\{R \in \mathbf{O p} \mid R \cap Ø \subseteq X\}} R=X\) holds because (X ∩ Ø) \(\subseteq\) V.
S is the sheaf of people, the set of which changes over time: a section in S over any interval of time is a person who is alive throughout that interval. A section in the subobject {S | p} over any interval of time is a person who is alive and likes the weather throughout that interval of time.
We need an example of a space X, a sheaf S \(\in\) Shv(X), and two predicates p,q: S → Ω for which \(p(s) \vdash_{s: S} q(s)\) holds. Take X to be the one-point space, take S to be the sheaf corresponding to the set S = \(\mathbb{N}\), let p(s) be the predicate “24 ≤ s ≤ 28,” and let q(s) be the predicate “s is not prime.” Then \(p(s) \vdash_{s: S} q(s)\) holds.
As an informal example, take X to be the surface of the earth, take S to be the sheaf of vector fields as in Example 7.46 thought of in terms of wind-blowing. Let p be the predicate “the wind is blowing due east at somewhere between 2 and 5 kilometers per hour” and let q be the predicate “the wind is blowing at somewhere between 1 and 5 kilometers per hour.” Then \(p(s) \vdash_{s: S} q(s)\) holds. This means that for any open set U, if the wind is blowing due east at somewhere between 2 and 5 kilometers per hour throughout U, then the wind is blowing at somewhere between 1 and 5 kilometers per hour throughout U as well.
We have the predicate p : \(\mathbb{N}\) × \(\mathbb{Z}\) → \(\mathbb{B}\) given by p(n, z) iff n ≤ |z|.
1. The predicate ∀(z: Z).p(n, z) holds for {0} \(\subseteq\) \(\mathbb{N}\).
2. The predicate ∃(z: Z).p(n, z) holds for \(\mathbb{N}\) \(\subseteq\) \(\mathbb{N}\).
3. The predicate ∀(n: N).p(n, z) holds for Ø \(\subseteq\) \(\mathbb{Z}\).
4. The predicate ∃(n: N).p(n, z) holds for \(\mathbb{Z}\) \(\subseteq\) \(\mathbb{Z}\).
Suppose s is a person alive throughout the interval U. Apply the above definition to the example p(s, t) = “person s is worried about news t” from above.
1. The formula says that ∀(t :T).p(s, t) “returns the largest open set V \(\subseteq\) U for which p(s|\(_{V}\), t) = V for all t \(\in\) T(V).” Note that T(V) is the set of items that are in the news throughout the interval V. Substituting, this becomes “the largest interval of time V \(\subseteq\) U over which person s is worried about news t for every item t that is in the news throughout V.” In other words, for V to be nonempty, the person s would have to be worried about every single item of news throughout V. My guess is that there’s a festival happening or a happy kitten somewhere that person s is not worried about, but maybe I’m assuming that person s is sufficiently mentally “normal.” There may be people who are sometimes worried about literally everything in the news; we ask you to please be kind to them.
2. Yes, it is exactly the same description.
Suppose s is a person alive throughout the interval U. Apply the above definition to the example p(s, t) = “person s is worried about news t” from above.
1. The formula says that ∃(t : T). p(s, t) “returns the union V = \(\bigcup_{i}\)V\(_{i}\) of all the open sets V\(_{i}\) for which there exists some t\(_{i}\) \(\in\) T(V\(_{i}\)) satisfying p(s|\(_{V_{i}}\) , t\(_{i}\)) = V\(_{i}\).” Substituting, this becomes “the union of all time intervals V\(_{i}\) for which there is some item ti in the news about which s is worried throughout V\(_{i}\).” In other words it is all the time that s is worried about at least one thing in the news. Perhaps when s is sleeping or concentrating on something, she is not worried about anything, in which case intervals of sleeping or concentrating would not be subsets of V. But if s said “there’s been such a string of bad news this past year, it’s like I’m always worried about something!,” she is saying that it’s like V = “this past year.”
2. This seems like a good thing for “there exists a piece of news that worries s” to mean: the news itself is allowed to change as long as the person’s worry remains. Someone might disagree and think that the predicate should mean “there is one piece of news that worries s throughout the whole interval V.” In that case, perhaps this person is working within a different topos, e.g. one where the site has fewer coverings. Indeed, it is the notion of covering that makes existential quantification work the way it does.
It is clear that if j(j(q)) = j(q) then j(j(q)) ≤ j(q) by reflexivity. On the other hand, assume the hypothesis, that p ≤ j(p) for all U \(\subseteq\) X and p \(\in\) Ω(U). If j(j(q)) ≤ j(q), then letting p := j(q) we have both j(p) ≤ p and p ≤ j(p). This means p \(\cong\) j(p), but Ω is a poset (not just a preorder) so p = j(p), i.e. j(j(q)) = j(q) as desired.
Let S be the sheaf of people and j be “assuming Bob is in San Diego...”
1. Take p(s) to be “s likes the weather.”
2. Let U be the interval 2019/01/01–2019/02/01. For an arbitrary person s \(\in\) S(U), p(s) is a subset of U, and it means the subset of U throughout which s likes the weather.
3. Similarly j(p(s)) is a subset of U, and it means the subset of U throughout which, assuming Bob is in San Diego, s liked the weather. In other words, j(p(s)) is true whenever Bob is not in San Diego, and it is true whenever s likes the weather.
4. It is true that p(s) ≤ j(p(s)), by the ‘in other words’ above.
5. It is true that j(j(p(s)) = j(p(s), because suppose given a time during which “if Bob is in SanDiego then if Bob is in San Diego then s likes the weather.” Then if Bob is in San Diego during this time then s likes the weather. But that is exactly what j(p(s)) means.
6. Take q(s) to be “s is happy.” Suppose “if Bob is in SanDiego then both s likes the weather and s is happy.” Then both “if Bob is in San Diego then s likes the weather” and “if Bob is in San Diego then s is happy” are true too. The converse is equally clear.
We have o\(_{[a, b]}\) := {[d,u] \(\in\) \(\mathbb{I}\)\(\mathbb{R}\) | a < d ≤ u < b}.
1. Since 0 ≤ 2 ≤ 6 ≤ 8, we have [2, 6] \(\in\) o\(_{[0, 8]}\) by the above formula.
2. In order to have [2, 6] \(\in\)\(^{?}\) o\(_{[0, 5]}\) ∪ o\(_{[4, 8]}\), we would need to have either [2,6] \(\in\)\(^{?}\) o\(_{[0, 5]}\) or [2,6] \(\in\)\(^{?}\) o\(_{[4, 8]}\). But the formula does not hold in either case.
A subset U \(\subseteq\) \(\mathbb{R}\) is open in the subspace topology of \(\mathbb{R}\) \(\subseteq\) \(\mathbb{I}\)\(\mathbb{R}\) iff there is an open set U′ \(\subseteq\) \(\mathbb{I}\)\(\mathbb{R}\) with U =b U′ ∩ \(\mathbb{R}\). We want to show that this is the case iff U is open in the usual topology. Suppose that U is open in the subspace topology. Then U = U′ ∩ \(\mathbb{R}\), where U′ \(\subseteq\) \(\mathbb{I}\)\(\mathbb{R}\) is the union of some basicopens, U′ = \(\bigcup)_{i \in I}\)o\(_{[a_{i}, b_{i}]}\), where o\(_{[a_{i}, b_{i}]}\) = {[d, u] \(\in\) \(\mathbb{I}\)\(\mathbb{R}\) | a\(_{i}\) < d < u < b\(_{i}\)}. Since \(\mathbb{R}\) = {[x, x] \(\in\) \(\mathbb{I}\)\(\mathbb{R}\)}, the intersection U′ ∩ \(\mathbb{R}\) will then be
\(U=\bigcup_{i \in I}\left\{x \in \mathbb{R} \mid a_{i}<x<b_{i}\right\}\)
and this is just the union of open balls B(m\(_{i}\), r\(_{i}\)) where \(m_{i}:=\frac{a_{i}+b_{i}}{2}\) is the midpoint and \(r_{i}:=\frac{b_{i}-a_{i}}{2}\) is the radius of the interval (a\(_{i}\), b\(_{i}\)). The open balls B(m\(_{i}\), r\(_{i}\)) are open in the usual topology on \(\mathbb{R}\) and the union of opens is open, so U is open in the usual topology. Suppose that U is open in the usual topology.
Then U = \(\bigcup)_{j \in J}\)B(m\(_{j}\), \(\mathcal{E}\)\(_{j}\)) for some set J. Let a\(_{j}\) := m\(_{j}\) − \(\mathcal{E}\)\(_{j}\) and b\(_{j}\) := m\(_{j}\) + \(\mathcal{E}\)\(_{j}\). Then
\(U=\bigcup_{j \in J}\left\{x \in \mathbb{R} \mid a_{j}<x<b_{j}\right\}=\bigcup_{j \in J}\left(o_{\left[a_{j}, b_{j}\right]} \cap \mathbb{R}\right)=\left(\bigcup_{j \in J} o_{\left[a_{j}, b_{j}\right]}\right) \cap \mathbb{R}\)
which is open in the subspace topology.
Fix any topological space (X,Op\(_{X}\)) and any subset R \(\subseteq\) \(\mathbb{I}\)\(\mathbb{R}\) of the interval domain. Define H\(_{X}\)(U) := { f : U ∩ R → X | f is continuous}.
1. H\(_{X}\) is a presheaf: given V \(\subseteq\) U the restriction map sends the continuous function f : U ∩ R → X to its restriction along the subset V ∩ R \(\subseteq\) U ∩ R.
2. It is a sheaf: given any family U\(_{i}\) of open sets with U = \(\bigcup_{i}\)U\(_{i}\) and a continuous function f\(_{i}\) : U\(_{i}\) ∩ R → X for each i, agreeing on overlaps, they can be glued together to give a continuous functionon all of U ∩ R, since U ∩ R = (\(\bigcup_{i}\)U\(_{i}\)) ∩ R = \(\bigcup_{i}\)U\(_{i}\)(U\(_{i}\) ∩R).