7S Chapter 7 Solutions

solution

Solutions to the exercises of 7 Sketches, Chapter 7: 7S Chapter 7 Exercises. Index: Map of Content.

Solution 7.4

proof — Exercise 7.4

Let the diagram be $A\stackrel{f}{\to }B\stackrel{g}{\to }C$ over $A^{\prime }\stackrel{f^{\prime }}{\to }{B}^{\prime }\stackrel{g^{\prime }}{\to }{C}^{\prime }$ with verticals $h_1,h_2,h_3$, and suppose the right square is a pullback.

Left square pullback $\Rightarrow$ rectangle pullback. Given $p:X\to C$ and $q:X\to A^{\prime }$ with $q;f^{\prime };g^{\prime }=p;h_3$, the right pullback yields a unique $r:X\to B$ with $r;h_2=q;f^{\prime }$ and $r;g=p$; the left pullback then yields a unique $r^{\prime }:X\to A$ with $r^{\prime };f=r$ and $r^{\prime };h_1=q$. Hence $r^{\prime };f;g=p$ and $r^{\prime };h_1=q$. If $r_0$ also satisfies these, then $r_0;f$ satisfies the defining equations of $r$, so $r_0;f=r$, and then $r_0=r^{\prime }$ by uniqueness in the left square.

Rectangle pullback $\Rightarrow$ left square pullback. Given $r:X\to B$ and $q:X\to A^{\prime }$ with $r;h_2=q;f^{\prime }$, set $p:=r;g$. Then $p;h_3=r;h_2;g^{\prime }=q;f^{\prime };g^{\prime }$, so the rectangle yields a unique $r^{\prime }:X\to A$ with $r^{\prime };f;g=p$ and $r^{\prime };h_1=q$. Now $r^{\prime };f$ and $r$ both satisfy $(-);g=p$ and $(-);h_2=q;f^{\prime }$, so by uniqueness in the right pullback $r^{\prime };f=r$. Uniqueness of $r^{\prime }$ follows from uniqueness for the rectangle. $■$

Sources: 7 Sketches, Exercise 7.4 and Solution A.7.

Solution 7.6

proof — Exercise 7.6

The pullback definition says: for all $g_1,g_2:X\to A$, if $g_1;f=g_2;f$ then $g_1=g_2$ (the mediating arrow into the pullback $A$ must equal both $g_1$ and $g_2$).

1. If $f$ is mono and $f(a_1)=f(a_2)$, take $X=\{\ast \}$, $g_i(\ast )=a_i$. Then $g_1;f=g_2;f$, so $g_1=g_2$, so $a_1=a_2$.
2. If $f$ is injective and $g_1;f=g_2;f$, then for each $x$, $f(g_1(x))=f(g_2(x))$ gives $g_1(x)=g_2(x)$; so $g_1=g_2$.

Lean

import Mathlib
#check @CategoryTheory.mono_iff_injective

Sources: 7 Sketches, Exercise 7.6 and Solution A.7.

Solution 7.7

proof — Exercise 7.7

1. Let $j=i^{-1}$ and $g:=f;j:A\to B^{\prime }$. Since $g;i=f={\mathrm{id}}_A;f$, the universal property gives $j^{\prime }:A\to A^{\prime }$ with $j^{\prime };i^{\prime }={\mathrm{id}}_A$ and $j^{\prime };f^{\prime }=f;j$. For the other composite: $(i^{\prime };j^{\prime });i^{\prime }=i^{\prime }$ and $(i^{\prime };j^{\prime });f^{\prime }=i^{\prime };f;j=f^{\prime };i;j=f^{\prime }$; ${\mathrm{id}}_{A^{\prime }}$ satisfies the same two equations, so by uniqueness $i^{\prime };j^{\prime }={\mathrm{id}}_{A^{\prime }}$.
2. Given $g:X\to A$ and $h:X\to B$ with $g;f=h$, the unique $r:X\to A$ with $r;{\mathrm{id}}_A=g$ and $r;f=h$ is $r=g$.

Lean

import Mathlib
open CategoryTheory Limits
#check @CategoryTheory.IsPullback.of_horiz_isIso    -- a square with iso horizontals is a pullback
#check @CategoryTheory.IsPullback.of_id_fst          -- the identity square over f is a pullback

Sources: 7 Sketches, Exercise 7.7 and Solution A.7.

Solution 7.8

proof — Exercise 7.8

Build a cube: the front and bottom faces are the given pullback, the right face is the square $(A,A,A,B)$ with identities and $f$ — a pullback because $f$ is mono (Definition 7.5) — and the back and top faces are the “identity” squares of 7S Exercise 7.7 (2), which are pullbacks. By the pasting lemma, right face + back face pullbacks make the diagonal rectangle a pullback; then front face pullback + rectangle pullback make the left face $(A^{\prime },A^{\prime },A^{\prime },B^{\prime })$ with identities and $f^{\prime }$ a pullback, which says $f^{\prime }$ is mono. Hence monomorphisms are stable under pullback.

Lean

import Mathlib
open CategoryTheory Limits
#check @CategoryTheory.Limits.pullback.fst_of_mono    -- Mono g → Mono (pullback.fst f g)
#check @CategoryTheory.Limits.pullback.snd_of_mono    -- Mono f → Mono (pullback.snd f g)

Sources: 7 Sketches, Exercise 7.8 and Solution A.7.

Solution 7.9

example program — Exercise 7.9

$\underline{3}\twoheadrightarrow \underline{2}\hookrightarrow \underline{3}$: first the surjection $1,2\mapsto 1$, $3\mapsto 2$ onto the image $\{1,2\}$, then the inclusion of the image into $\underline{3}$.

Julia

using Catlab
f = FinFunction([1, 1, 2], 3)
e, m = epi_mono(f)
collect(e), collect(m)          # ([1, 1, 2], [1, 2])
compose(e, m) == f              # true

Sources: 7 Sketches, Exercise 7.9 and Solution A.7.

Solution 7.11

proof — Exercise 7.11

1. $I$ is a top element and $v\otimes w$ satisfies the universal property of the Meet $v\wedge w$, so the monoidal structure is cartesian. The quantale’s $⊸$ satisfies $v\le (w⊸x)⟺v\otimes w\le x$, which is the universal property of the Exponential Object $x^w$. So $\mathcal{V}$ is a cartesian closed preorder — a complete Heyting Algebra.
2. No: quantales have all joins, but a cartesian closed preorder need not. Example: ${\mathbb{N}}^{\mathrm{op}}\times {\mathbb{N}}^{\mathrm{op}}$ (the Product Preorder, $(a,b)\le (a^{\prime },b^{\prime })$ iff $a^{\prime }\le a$ and $b^{\prime }\le b$). It has top $(0,0)$, meets $(a,b)\wedge (a^{\prime },b^{\prime })=(\max (a,a^{\prime }),\max (b,b^{\prime }))$, but no bottom element, hence no empty join. Yet $x⊸y=\bigvee \{w\mid w\wedge x\le y\}=\bigvee \{w\mid y\le w,w\wedge x\le y\}$ exists because only finitely many $w$ lie above $y=(a,b)$ (exactly $(a+1)(b+1)$ of them), and finite nonempty joins are given by componentwise $\min$.

Sources: 7 Sketches, Exercise 7.11 and Solution A.7.

Solution 7.16

example — Exercise 7.16

1. $\mathsf{false}$: $-5\notin \mathbb{N}$. 2. $\mathsf{true}$: $0\in \mathbb{N}$.

Haskell

charN :: Integer -> Bool
charN = (>= 0)          -- charN (-5) == False, charN 0 == True

Sources: 7 Sketches, Exercise 7.16 and Solution A.7.

Solution 7.17

example — Exercise 7.17

1. Every $n$ is in the image, so $┌\mathrm{id}┐(n)=\mathsf{true}$ for all $n$ — the predicate $\mathsf{true}$, classifying the top Subobject.
2. Nothing is in the image, so $┌!┐(n)=\mathsf{false}$ for all $n$ — the predicate $\mathsf{false}$, classifying the bottom subobject.

Sources: 7 Sketches, Exercise 7.17 and Solution A.7.

Solution 7.19

example — Exercise 7.19

1. A Subobject of $\mathbb{B}$ — i.e. a subset $A\subseteq \mathbb{B}$ (a mono into $\mathbb{B}$); characteristic maps classify subobjects of their domain (Subobject Classifier).
2. $A=\{\mathsf{false}\}$: $\neg (b)=\mathsf{true}$ iff $b\in \{\mathsf{false}\}$. See Internal Logic of a Topos.

Sources: 7 Sketches, Exercise 7.19 and Solution A.7.

Solution 7.20

example — Exercise 7.20

$P$	$Q$	$P\wedge Q$	$P=(P\wedge Q)$
t	t	t	t
t	f	f	f
f	t	f	t
f	f	f	t

2. Yes — this is material implication.
3. $\Rightarrow :\mathbb{B}\times \mathbb{B}\to \mathbb{B}$ given by the last column.
4. It classifies $\{(\mathsf{t},\mathsf{t}),(\mathsf{f},\mathsf{t}),(\mathsf{f},\mathsf{f})\}\subseteq \mathbb{B}\times \mathbb{B}$ — the Equalizer of $\wedge$ and ${\pi }_1$ (Internal Logic of a Topos).

Sources: 7 Sketches, Exercise 7.20 and Solution A.7.

Solution 7.21

example program — Exercise 7.21

1. $\mathsf{false}$. 2. $\mathsf{true}$. 3. $\mathsf{true}$. 4. The set is “even primes or $\ge 10$” $=\{2\}\cup \{10,11,12,\dots \}$; smallest three: $2,10,11$.

JuliaHaskell

using Primes
E(n) = iseven(n); P(n) = isprime(n); T(n) = n >= 10
[n for n in 0:20 if (E(n) && P(n)) || T(n)][1:3]      # [2, 10, 11]

isPrime n = n > 1 && all (\d -> n `mod` d /= 0) [2 .. n - 1]
classified = [ n | n <- [0 ..], (even n && isPrime n) || n >= 10 ]   -- take 3 → [2,10,11]

Sources: 7 Sketches, Exercise 7.21 and Solution A.7.

Solution 7.27

example — Exercise 7.27

1. $B(x,ϵ)=\{x^{\prime }\in \mathbb{R}\mid |x-x^{\prime }|<ϵ\}=(x-ϵ,x+ϵ)$.
2. $U$ is open iff for every $x\in U$ there is $ϵ>0$ with $B(x,ϵ)\subseteq U$ (Topological Space).
3. $U_1=(0,2)$, $U_2=(1,3)$ cover $U=(0,3)$.
4. $U_i=(\frac{1}{i},1)$ for $i\in \{1,2,3,\dots \}$ cover $U=(0,1)$.

Sources: 7 Sketches, Exercise 7.27 and Solution A.7.

Solution 7.29

proof — Exercise 7.29

1. It contains $X$ and $\varnothing$; $A\cap B=\varnothing$ unless both are $X$; a union is $X$ iff some member is $X$. All three axioms hold.
2. Every subset is open, so every “such-and-such is open” conclusion holds trivially.
3. Continuity requires $f^{-1}(U)$ open in $X$ for each open $U\subseteq Y$; everything in $X$ is open.

Lean

import Mathlib
#check @continuous_of_discreteTopology     -- every map out of a discrete space is continuous
#check @DiscreteTopology
#check @continuous_bot                      -- ⊥ is the discrete topology in Mathlib's order

Sources: 7 Sketches, Exercise 7.29 and Solution A.7.

Solution 7.31

example — Exercise 7.31

1. $\varnothing \to \{1\}\to \{1,2\}$, a three-element chain.
2. $(U_i{)}_{i\in I}$ covers $U$ iff either $I=\varnothing$ and $U=\varnothing$, or $U_i=U$ for some $i$: since the opens are totally ordered, a union equals its largest member. So the only cover not containing $U$ itself is the empty cover of $\varnothing$.

Sources: 7 Sketches, Exercise 7.31 and Solution A.7.

Solution 7.32

proof — Exercise 7.32

1. $Y=X\cap Y$ with $X\in \mathrm{Op}$.
2. $\varnothing =\varnothing \cap Y$. If $A_i=B_i\cap Y$ then $A_1\cap A_2=(B_1\cap B_2)\cap Y$ and ${\bigcup }_i{A}_i=({\bigcup }_i{B}_i)\cap Y$, and $B_1\cap B_2$, ${\bigcup }_i{B}_i$ are open in $X$.
3. The preimage of $B\in \mathrm{Op}$ under the inclusion is $B\cap Y$, which is open by definition.

Lean

import Mathlib
#check @instTopologicalSpaceSubtype
#check @isOpen_induced_iff              -- IsOpen s ↔ ∃ t, IsOpen t ∧ f ⁻¹' t = s
#check @continuous_subtype_val

Sources: 7 Sketches, Exercise 7.32 and Solution A.7.

Solution 7.34

annotation — Exercise 7.34

A $\mathcal{V}$-category has objects and, for each pair $a,b$, an open set $\mathcal{C}(a,b)\subseteq X$, with $X\subseteq \mathcal{C}(a,a)$ and $\mathcal{C}(a,b)\cap \mathcal{C}(b,c)\subseteq \mathcal{C}(a,c)$. Think of $\mathcal{C}(a,b)$ as a size restriction for getting from $a$ to $b$ — bridges your truck must fit under. Going from $a$ to itself has no restriction ($X$). Along a path you must fit under every bridge (meet $=\cap$), and you may take any path (join $=\cup$): as in Matrix Multiplication in a Quantale, $\mathcal{C}(B,C)=(U_3\cap U_1)\cup (U_4\cap U_2)$ for the two paths of the book’s example.

Sources: 7 Sketches, Exercise 7.34 and Solution A.7.

Solution 7.38

example — Exercise 7.38

1. $\{a_1,a_2\}$. 2. $\{c_1\}$. 3. $\varnothing$. 4. E.g. send $a_2,e_1\mapsto a$; $c_1,b_1\mapsto b$; $b_2\mapsto c$; $b_3\mapsto d$; $e_2,a_1\mapsto e$ (eight elements over five points, fibers of sizes $2,2,1,1,2$).

Sources: 7 Sketches, Exercise 7.38 and Solution A.7.

Solution 7.40

example program — Exercise 7.40

1. Six sections $(a_i,b_j,c_1)$ for $i\in \{1,2\}$, $j\in \{1,2,3\}$.
2. None: the fiber over $d$ is empty, so ${\mathrm{Sec}}_f(V_2)=\varnothing$.
3. Also none, for the same reason — $|{\mathrm{Sec}}_f(V_3)|=2\cdot 3\cdot 0\cdot 2=0$. (The printed solution says $12$, forgetting the empty fiber over $d$.)

Julia

fibers = Dict("a"=>2, "b"=>3, "c"=>1, "d"=>0, "e"=>2)
nsections(U) = prod(fibers[u] for u in U)
nsections(["a","b","c"]), nsections(["a","b","c","d"]), nsections(["a","b","d","e"])   # (6, 0, 0)

Sources: 7 Sketches, Exercise 7.40 and Solution A.7.

Solution 7.42

example — Exercise 7.42

1. ${\mathrm{Sec}}_f(\{a,b,c\})=\{(a_1,b_1,c_1),(a_1,b_2,c_1),(a_1,b_3,c_1),(a_2,b_1,c_1),(a_2,b_2,c_1),(a_2,b_3,c_1)\}$; ${\mathrm{Sec}}_f(\{a,c\})=\{(a_1,c_1),(a_2,c_1)\}$.
2. Restriction drops the $b$-component: the first three sections go to $(a_1,c_1)$, the last three to $(a_2,c_1)$.

Sources: 7 Sketches, Exercise 7.42 and Solution A.7.

Solution 7.44

example — Exercise 7.44

1. $s_1=(a_1,b_1)$, $s_2=(b_2,e_1)$.
2. No: a section over $\{a,b,e\}$ has a single $b$-value, which would have to be both $b_1$ and $b_2$.
3. $h_1=(a_2,b_3)$, $h_2=(b_3,e_2)$.
4. Yes, uniquely: $h=(a_2,b_3,e_2)$. This is the sheaf condition in action.

Sources: 7 Sketches, Exercise 7.44 and Solution A.7.

Solution 7.47

annotation — Exercise 7.47

No. The set of all vector fields (over all opens) forms one sheaf, ${\mathrm{Sec}}_{\pi }$ for the tangent bundle $\pi :TM\to M$ (Sheaf of Sections). The sheaves on $M$ do not even form a set — they form a Topos $\mathbf{Shv}(M)$ — and ${\mathrm{Sec}}_{\pi }$ is one object of it.

Sources: 7 Sketches, Exercise 7.47 and Solution A.7.

Solution 7.49

example — Exercise 7.49

1. The chain $\varnothing \to \{1\}\to \{1,2\}$.
2. Three sets and two functions $F(\{1,2\})\to F(\{1\})\to F(\varnothing )$.
3. The only non-trivial cover is the empty cover of $\varnothing$ (7S Exercise 7.31), whose sheaf condition is $F(\varnothing )=\{()\}$ (Example 7.36).
4. Hence a sheaf is a set $F(\{1,2\})$, a set $F(\{1\})$ and a function between them; $\mathbf{Shv}(\text{Sierpinˊski})$ is equivalent to the arrow category ${\mathbf{Set}}^{\to }$.

Sources: 7 Sketches, Exercise 7.49 and Solution A.7.

Solution 7.52

example — Exercise 7.52

$X=\{1\}$ has opens $\varnothing$ and $\{1\}$; a sheaf $S$ on it has $S(\varnothing )=\{()\}$ forced, so its only data is the set $S(\{1\})$ — this is the identification $\mathbf{Set}\simeq \mathbf{Shv}(\{1\})$. Now $\Omega (\{1\})=\{\varnothing ,\{1\}\}$, a two-element set, corresponding to $\mathsf{false}$ and $\mathsf{true}$.

Sources: 7 Sketches, Exercise 7.52 and Solution A.7.

Solution 7.53

proof — Exercise 7.53

1. For $W\subseteq V\subseteq U$: $(U^{\prime }\cap V)\cap W=U^{\prime }\cap W$ since $W\subseteq V$; identities: $U^{\prime }\cap U=U^{\prime }$ for $U^{\prime }\subseteq U$.
2. Yes: a presheaf is just a functor ${\mathrm{Op}}^{\mathrm{op}}\to \mathbf{Set}$, and functoriality is all there is to check. (That it is moreover a Sheaf — the Subobject Classifier of $\mathbf{Shv}(X)$ — is verified separately.)

Sources: 7 Sketches, Exercise 7.53 and Solution A.7.

Solution 7.55

example program — Exercise 7.55

Write $\gamma =┌H┐$. Vertices: $\gamma (A)=\gamma (B)=\gamma (C)=V$ (present), $\gamma (D)=0$ (missing). Arrows: $\gamma (f)=(V,V;A)$ (present); $\gamma (g)=\gamma (h)=(V,V;0)$ (endpoints present, arrow missing); $\gamma (i)=(V,0;0)$ (source present, target missing). This is the unique homomorphism whose pullback of $\mathsf{true}$ is $H$.

Julia

using Catlab
Ω, _ = subobject_classifier(Graph)     # vertex 1 = V, 2 = 0; edges 1=(V,V;A) 2=(V,V;0) 3=(V,0;0) 4=(0,V;0) 5=(0,0;0)
G = @acset Graph begin V = 4; E = 4; src = [1, 1, 2, 3]; tgt = [2, 2, 3, 4] end   # A,B,C,D; f,g,h,i
γ = ACSetTransformation(G, Ω; V=[1, 1, 1, 2], E=[1, 2, 2, 3])
is_natural(γ)                           # true

Sources: 7 Sketches, Exercise 7.55 and Solution A.7.

Solution 7.59

example — Exercise 7.59

1. $\neg U$ is the interior of the complement $\{0\}$, which is $\varnothing$.
2. $\neg \neg U$ is the interior of $\mathbb{R}\setminus \varnothing =\mathbb{R}$, i.e. $\mathbb{R}$.
3. Yes. 4. No: $0\in \neg \neg U$ but $0\notin U$. Double negation is not the identity — the logic of a sheaf topos is intuitionistic.

Sources: 7 Sketches, Exercise 7.59 and Solution A.7.

Solution 7.60

proof — Exercise 7.60

1. Taking $U=X$: $\top \cap X=X$, and $\top \cap X=\top$, so $\top =X$.
2. $X\cup U=X$; $U\Rightarrow X=\bigcup \{R\mid R\cap U\subseteq X\}=X$; $X\Rightarrow U=\bigcup \{R\mid R\subseteq U\}=U$.
3. Taking $U=\varnothing$: $\perp \cup \varnothing =\varnothing$ so $\perp =\varnothing$.
4. $\varnothing \cap U=\varnothing$; $\varnothing \Rightarrow U=\bigcup \{R\mid R\cap \varnothing \subseteq U\}=X$.

Sources: 7 Sketches, Exercise 7.60 and Solution A.7.

Solution 7.62

example — Exercise 7.62

If $S$ is the sheaf of people (a section over an interval $U$ is a person alive throughout $U$), then a section of $\{S\mid p\}$ over $U$ is a person alive throughout $U$ who likes the weather throughout $U$ — i.e. a section $s\in S(U)$ with $p(s)=U$.

Sources: 7 Sketches, Exercise 7.62 and Solution A.7.

Solution 7.64

example — Exercise 7.64

Formal: $X$ the one-point space, $S=\mathbb{N}$, $p(s)$ = "$24\le s\le 28$", $q(s)$ = ”$s$ is not prime”. Informal: $X$ the surface of the Earth, $S$ the sheaf of wind vector fields, $p$ = “wind blows due east at 2–5 km/h”, $q$ = “wind blows at 1–5 km/h”; wherever $p$ holds, $q$ holds.

Sources: 7 Sketches, Exercise 7.64 and Solution A.7.

Solution 7.66

example program — Exercise 7.66

1. $\{0\}$ (only $0\le |z|$ for all $z$, since $z=0$ must work). 2. All of $\mathbb{N}$ (take $z=n$). 3. $\varnothing$ (no $z$ exceeds every $n$). 4. All of $\mathbb{Z}$ (take $n=0$).

Haskell

p :: Integer -> Integer -> Bool
p n z = n <= abs z
-- on finite windows: [n | n <- [0..5], all (p n) [-9..9]] == [0]
--                    [z | z <- [-9..9], all (`p` z) [0..20]] == []

Sources: 7 Sketches, Exercise 7.66 and Solution A.7.

Solution 7.67

example — Exercise 7.67

1. The largest open $V\subseteq U$ such that $p(s{|}_V,t)=V$ for every $t\in T(V)$: the largest interval throughout which $s$ is worried about every item in the news throughout that interval. For most people this is empty (there is always a happy kitten somewhere).
2. Yes, exactly.

Sources: 7 Sketches, Exercise 7.67 and Solution A.7.

Solution 7.68

example — Exercise 7.68

1. The union of all intervals $V_i\subseteq U$ for which some news item $t_i\in T(V_i)$ worries $s$ throughout $V_i$: all the time during which $s$ is worried about at least one thing — the thing being allowed to change.
2. Reasonable: “such a string of bad news, it’s like I’m always worried about something”. Someone who wants a single item to worry $s$ throughout is working in a different topos, with fewer coverings — it is the notion of covering that makes $\exists$ behave this way.

Sources: 7 Sketches, Exercise 7.68 and Solution A.7.

Solution 7.70

proof — Exercise 7.70

($\Leftarrow$) by reflexivity. ($\Rightarrow$) With $p:=j(q)$ we have $j(p)\le p$ by hypothesis and $p\le j(p)$ by inflation; $\Omega (U)$ is a poset, so $p=j(p)$, i.e. $j(q)=j(j(q))$.

Sources: 7 Sketches, Exercise 7.70 and Solution A.7.

Solution 7.72

example — Exercise 7.72

1. $p(s)$ = ”$s$ likes the weather”.
2. $U$ = January 2019; $p(s)\subseteq U$ is the sub-interval throughout which $s$ likes the weather.
3. $j(p(s))=(\text{Bob in SD})\Rightarrow p(s)$: the times at which either Bob is not in San Diego or $s$ likes the weather.
4. Yes, by 3.
5. Yes: “if Bob is in SD then (if Bob is in SD then $p$)” is equivalent to “if Bob is in SD then $p$“.
6. Yes, with $q$ = ”$s$ is happy”: “if Bob is in SD then ($p$ and $q$)” iff (“if Bob is in SD then $p$” and “if Bob is in SD then $q$”).

Sources: 7 Sketches, Exercise 7.72 and Solution A.7.

Solution 7.76

example — Exercise 7.76

1. $0<2\le 6<8$.
2. $[2,6]\in o_{[0,5]}$ would need $6<5$; $[2,6]\in o_{[4,8]}$ would need $4<2$. Neither holds.

Sources: 7 Sketches, Exercise 7.76 and Solution A.7.

Solution 7.77

proof — Exercise 7.77

Since $\mathbb{R}=\{[x,x]\}$, $o_{[a,b]}\cap \mathbb{R}=\{x\mid a<x<b\}=B(\frac{a+b}{2},\frac{b-a}{2})$. If $U=U^{\prime }\cap \mathbb{R}$ with $U^{\prime }={\bigcup }_i{o}_{[a_i,b_i]}$, then $U={\bigcup }_i(a_i,b_i)$ is a union of open balls, hence open. Conversely if $U={\bigcup }_{j}B(m_j,{ϵ}_j)$, put $a_j=m_j-{ϵ}_j$, $b_j=m_j+{ϵ}_j$; then $U=\left({\bigcup }_j{o}_{[a_j,b_j]}\right)\cap \mathbb{R}$ is open in the subspace topology.

Sources: 7 Sketches, Exercise 7.77 and Solution A.7.

Solution 7.80

proof — Exercise 7.80

1. Yes: for $V\subseteq U$ restrict $f$ along $V\cap R\subseteq U\cap R$; this is functorial.
2. Yes: given a cover $U={\bigcup }_i{U}_i$ and continuous $f_i:U_i\cap R\to X$ agreeing on overlaps, they glue to a unique continuous function on $U\cap R={\bigcup }_i(U_i\cap R)$ (continuity is local). So $H_X\in \mathbf{Shv}(\mathbb{IR})$ — a behavior type; with $R=\mathbb{R}$ this is $G_X$ of Example 7.79.

Sources: 7 Sketches, Exercise 7.80 and Solution A.7.