7S Chapter 1 Solutions

solution

Solutions to the exercises of 7 Sketches, Chapter 1: 7S Chapter 1 Exercises. Index: Map of Content.

Solution 1.1

Exercise 1.1

• Order: $f(x)=x+5$ preserves order; $g(x)=-x$ does not ($1\le 2$ but $-1\nleqq -2$).
• Metric: $f(x)=x+5$ preserves the metric; $g(x)=2x$ does not ($|1-2|=1$ but $|2-4|=2$).
• Addition: $f(x)=3x$ preserves addition; $g(x)=x+1$ does not ($g(0+0)=1\ne 2=g(0)+g(0)$).

The moral: “asking which aspects of $X$ one wants to preserve under the observation $f$ becomes the question what category are you working in?”

Sources: 7 Sketches, Exercise 1.1 and Solution A.1.

Solution 1.4

Exercise 1.4

Take the transitive closure of the union of connections: $11\sim 12$ and $12\sim 13$ give $\{11,12,13\}$, and the second system adds nothing to the bottom row, so the join is $\{11,12,13\},\{21,22,23\}$.

Sources: 7 Sketches, Exercise 1.4 and Solution A.1.

Solution 1.6

Exercise 1.6

1. $(1)(2)\le (12)$: two elements, one arrow.
2. The 15 partitions of $\{1,2,3,4\}$ in four rows: bottom $(1)(2)(3)(4)$; then the six with one pair $(12)(3)(4),(13)(2)(4),(14)(2)(3),(1)(23)(4),(1)(24)(3),(1)(2)(34)$; then the seven with a triple or two pairs $(123)(4),(124)(3),(134)(2),(1)(234),(12)(34),(13)(24),(14)(23)$; top $(1234)$. Choose $A=(12)(3)(4)$, $B=(13)(2)(4)$.
3. $A\vee B=(123)(4)$.
4. Yes.
5. $C\in \{(123)(4),(1234)\}$.
6. Yes: $(123)(4)\le (123)(4)$ and $(123)(4)\le (1234)$.

Sources: 7 Sketches, Exercise 1.6 and Solution A.1.

Solution 1.7

Exercise 1.7

$\mathsf{true}$, $\mathsf{true}$, $\mathsf{true}$, $\mathsf{false}$ — the join in $\mathbb{B}$ is OR.

Sources: 7 Sketches, Exercise 1.7 and Solution A.1.

Solution 1.10

Exercise 1.10

1. True. 2. False: $0\in \mathbb{N}$ but $0\notin \{n\ge 1\}$. 3. True: no integer lies strictly between $1$ and $2$.

Sources: 7 Sketches, Exercise 1.10 and Solution A.1.

Solution 1.11

Exercise 1.11

1. $\varnothing ,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}$.
2. E.g. $\{1,2,3\}\cup \{1\}=\{1,2,3\}$.
3. $(h,1),(h,2),(h,3),(1,1),(1,2),(1,3)$.
4. $(h,1),(1,1),(1,2),(2,2),(3,2)$ (tagged by which set they come from).
5. $h,1,2,3$.

Sources: 7 Sketches, Exercise 1.11 and Solution A.1.

Solution 1.16

proof — Exercise 1.16

1. If $A_p=A_{p_1^{\prime }}^{\prime }=A_{p_2^{\prime }}^{\prime }$ then $A_{p_1^{\prime }}^{\prime }\cap A_{p_2^{\prime }}^{\prime }=A_{p_1^{\prime }}^{\prime }\ne \varnothing$; by the partition axiom distinct labels have disjoint parts, so $p_1^{\prime }=p_2^{\prime }$.
2. Pick $a\in A_{p^{\prime }}^{\prime }$ (parts are nonempty). Since $A={\bigcup }_p{A}_p$ there is $p$ with $a\in A_p$, and by assumption $A_p=A_{p^{\prime \prime }}^{\prime }$ for some $p^{\prime \prime }$. Then $a\in A_{p^{\prime }}^{\prime }\cap A_{p^{\prime \prime }}^{\prime }$, so $p^{\prime }=p^{\prime \prime }$ and $A_p=A_{p^{\prime }}^{\prime }$.

Hence “same partition up to relabeling” is a well-defined notion.

Sources: 7 Sketches, Exercise 1.16 and Solution A.1.

Solution 1.17

Exercise 1.17

$(11,11),(11,12),(12,11),(12,12),(13,13),(21,21),(22,22),(22,23),(23,22),(23,23)$.

Sources: 7 Sketches, Exercise 1.17 and Solution A.1.

Solution 1.20

proof — Exercise 1.20

1. Connected subsets are nonempty by definition.
2. Suppose $a\in A_p\cap A_q$. For $a^{\prime }\in A_p$, connectedness gives $a\sim a^{\prime }$, and closedness of $A_q$ gives $a^{\prime }\in A_q$; symmetrically $A_q\subseteq A_p$. So $A_p=A_q$, contradicting $p\ne q$.
3. For $a\in A$ let $X:=\{a^{\prime }\mid a^{\prime }\sim a\}$. $X$ is closed (if $a^{\prime }\in X$ and $b\sim a^{\prime }$ then $b\sim a$ by transitivity/symmetry), connected (if $b,c\in X$ then $b\sim c$), and contains $a$ (reflexivity). So $a$ lies in some part.

Sources: 7 Sketches, Exercise 1.20 and Solution A.1.

Solution 1.24

Exercise 1.24

1. The unique function $\varnothing \to \{1\}$. 2. The unique function $\{a,b\}\to \{1\}$. 3–4. The second and third relations are not functions (the second is not deterministic — one element related to two — and neither is total). The first is a function that is neither injective nor surjective; the fourth is a Bijection.

Sources: 7 Sketches, Exercise 1.24 and Solution A.1.

Solution 1.25

proof — Exercise 1.25

By Definition 1.22, $f$ is a subset $F\subseteq A\times \varnothing$ such that for every $a\in A$ there is a unique $b\in \varnothing$ with $(a,b)\in F$. There are no $b\in \varnothing$, so there can be no $a\in A$: $A=\varnothing$. (Categorically: $\varnothing$ is initial and strict in $\mathbf{Set}$ — any map into it is an isomorphism.)

Sources: 7 Sketches, Exercise 1.25 and Solution A.1.

Solution 1.27

Exercise 1.27

partition	surjection onto
$(\bullet )(\ast )(\circ )$	$\{p_1,p_2,p_3\}$, $\bullet \mapsto p_1,\ast \mapsto p_2,\circ \mapsto p_3$
$(\bullet \ast )(\circ )$	$\{p_1,p_2\}$, $\bullet ,\ast \mapsto p_1$, $\circ \mapsto p_2$
$(\bullet \circ )(\ast )$	$\{p_1,p_2\}$, $\bullet ,\circ \mapsto p_1$, $\ast \mapsto p_2$
$(\bullet )(\ast \circ )$	$\{p_1,p_2\}$, $\bullet \mapsto p_1$, $\ast ,\circ \mapsto p_2$
$(\bullet \ast \circ )$	$\{p_1\}$, everything $\mapsto p_1$

Sources: 7 Sketches, Exercise 1.27 and Solution A.1.

Solution 1.38

Exercise 1.38

$a:1\to 2$, $b:1\to 3$, $c:1\to 3$, $d:2\to 2$, $e:2\to 3$.

Sources: 7 Sketches, Exercise 1.38 and Solution A.1.

Solution 1.40

Exercise 1.40

$P=\{1,2,3,4\}$ with $p\le q$ iff there is a path $p\to q$: $1\le 1,1\le 2,1\le 3,2\le 2,2\le 3,3\le 3,4\le 4$. The parallel arrows $b,c$ and the loop $d$ are “useless” from a preorder point of view but do no harm.

Sources: 7 Sketches, Exercise 1.40 and Solution A.1.

Solution 1.41

Exercise 1.41

Yes: it is the Hasse diagram of the discrete order $x\le y$ iff $x=y$ (a graph with no arrows).

Sources: 7 Sketches, Exercise 1.41 and Solution A.1.

Solution 1.42

Exercise 1.42

Writing $\alpha =(\bullet )(\circ )(\ast )$, $\beta =(\bullet \circ )(\ast )$, $\gamma =(\bullet \ast )(\circ )$, $\delta =(\bullet )(\circ \ast )$, $\omega =(\bullet \circ \ast )$: the five reflexive pairs $\alpha \le \alpha ,\dots ,\omega \le \omega$; $\alpha \le \beta ,\alpha \le \gamma ,\alpha \le \delta ,\alpha \le \omega$; $\beta \le \omega ,\gamma \le \omega ,\delta \le \omega$.

Sources: 7 Sketches, Exercise 1.42 and Solution A.1.

Solution 1.44

Exercise 1.44

Almost: every element is comparable with itself. A discrete preorder is one where $x$ and $y$ are comparable iff $x=y$.

Sources: 7 Sketches, Exercise 1.44 and Solution A.1.

Solution 1.46

Exercise 1.46

No: e.g. $4\nmid 6$ and $6\nmid 4$, so $4$ and $6$ are incomparable.

Sources: 7 Sketches, Exercise 1.46 and Solution A.1.

Solution 1.48

Exercise 1.48

Yes: for all $a,b\in \mathbb{R}$, either $a\le b$ or $b\le a$. See Real Numbers.

Sources: 7 Sketches, Exercise 1.48 and Solution A.1.

Solution 1.51

Exercise 1.51

$\mathcal{P}(\varnothing )=\{\varnothing \}$: a single point. $\mathcal{P}\{1\}$: $\varnothing \to \{1\}$. $\mathcal{P}\{1,2\}$: a square $\varnothing \to \{1\},\{2\}\to \{1,2\}$.

Sources: 7 Sketches, Exercise 1.51 and Solution A.1.

Solution 1.53

Exercise 1.53

Coarsest: the unique map $!:S\to \{1\}$. Finest: the identity ${\mathrm{id}}_S:S\to S$. See Preorder of Partitions.

Sources: 7 Sketches, Exercise 1.53 and Solution A.1.

Solution 1.55

proof — Exercise 1.55

Every subset $U\subseteq X$ is an upper set: if $p\in U$, the only $q$ with $p\le q$ is $p$ itself, which is in $U$. So $\mathcal{U}(X)$ contains all subsets and is ordered by inclusion, i.e. $\mathcal{U}(X)=\mathcal{P}(X)$.

Sources: 7 Sketches, Exercise 1.55 and Solution A.1.

Solution 1.57

Exercise 1.57

The product has six elements: $(a,1)$ at the bottom; $(c,1),(a,2),(b,1)$ above it; $(c,2)$ above $(c,1),(a,2)$; $(b,2)$ above $(a,2),(b,1)$ (diagram in Product Preorder).

Upper sets (14 of them, ordered by inclusion): $\varnothing$; $\{(c,2)\}$, $\{(b,2)\}$; $\{(c,1),(c,2)\}$, $\{(b,2),(c,2)\}$, $\{(b,1),(b,2)\}$; $\{(b,2),(c,1),(c,2)\}$, $\{(a,2),(b,2),(c,2)\}$, $\{(b,1),(b,2),(c,2)\}$; $\{(a,2),(b,2),(c,1),(c,2)\}$, $\{(a,2),(b,1),(b,2),(c,2)\}$; $\{(a,2),(b,1),(b,2),(c,1),(c,2)\}$; and all six elements.

Sources: 7 Sketches, Exercise 1.57 and Solution A.1.

Solution 1.63

Exercise 1.63

$\mathcal{P}(X)$ is the cube in Power Set; the chain is $0\to 1\to 2\to 3$; $|\cdot |$ sends $\varnothing \mapsto 0$, singletons $\mapsto 1$, pairs $\mapsto 2$, $X\mapsto 3$ — each level of the cube to the corresponding element of the chain. It is a Monotone Map.

Sources: 7 Sketches, Exercise 1.63 and Solution A.1.

Solution 1.65

Exercise 1.65

$\mathcal{U}(\mathbb{B})=\varnothing \to \{\mathsf{true}\}\to \{\mathsf{true},\mathsf{false}\}$ maps into the square $\mathcal{P}(\mathbb{B})$ by inclusion, missing only $\{\mathsf{false}\}$. See Upper Set, Booleans.

Sources: 7 Sketches, Exercise 1.65 and Solution A.1.

Solution 1.66

proof — Exercise 1.66

See Yoneda Lemma for Preorders for the proofs. For 4: $\uparrow a=\{a,b,c\}$, $\uparrow b=\{b\}$, $\uparrow c=\{c\}$; in $\mathcal{U}(P)$, $\{b\},\{c\}\subseteq \{b,c\}\subseteq \{a,b,c\}$, and $\uparrow$ sends the bottom element $a$ of $P$ to the top element of $\mathcal{U}(P)$.

Sources: 7 Sketches, Exercise 1.66 and Solution A.1.

Solution 1.67

proof — Exercise 1.67

If $p_1{\le }_P{p}_2$ then $p_1=p_2$ (discreteness), so $f(p_1)=f(p_2)$, and hence $f(p_1){\le }_{Q}f(p_2)$ by reflexivity.

Sources: 7 Sketches, Exercise 1.67 and Solution A.1.

Solution 1.69

Exercise 1.69

Let $X=\mathbb{Z}$, $Y=\{n,z,p\}$, $f$ sending negatives to $n$, zero to $z$, positives to $p$. With $P=(nz)(p)$ and $Q=(np)(z)$: $f^{\ast }(P)=\{\{x\le 0\},\{x\ge 1\}\}$ and $f^{\ast }(Q)=\{\{0\},\{x\ne 0\}\}$.

Sources: 7 Sketches, Exercise 1.69 and Solution A.1.

Solution 1.71

proof — Exercise 1.71

1. ${\mathrm{id}}_P(p)=p$, so $p_1\le p_2$ implies $\mathrm{id}(p_1)\le \mathrm{id}(p_2)$.
2. $p_1\le p_2\Rightarrow f(p_1)\le f(p_2)\Rightarrow g(f(p_1))\le g(f(p_2))$, which is monotonicity of $f;g$. See Category of Preorders.

Sources: 7 Sketches, Exercise 1.71 and Solution A.1.

Solution 1.73

proof — Exercise 1.73

Skeletal: $p_1\le p_2$ and $p_2\le p_1$ imply $p_1=p_2$. Dagger: $p_1\le p_2$ implies $p_2\le p_1$. Hence $p_1\le p_2$ implies $p_1=p_2$, which is the definition of discrete. So such a preorder “can be identified with” its underlying set.

Sources: 7 Sketches, Exercise 1.73 and Solution A.1.

Solution 1.77

proof — Exercise 1.77

Let $P\le Q$ be partitions, i.e. $P$ is finer: $x{\sim }_{P}y$ implies $x{\sim }_{Q}y$. If $\Phi (P)=\mathsf{true}$ then $\bullet {\sim }_P\ast$, hence $\bullet {\sim }_Q\ast$, so $\Phi (Q)=\mathsf{true}$. Thus $\Phi (P)\le \Phi (Q)$. It nonetheless has a Generative Effect.

Sources: 7 Sketches, Exercise 1.77 and Solution A.1.

Solution 1.79

proof — Exercise 1.79

Let $u:Q\to \mathbb{B}$ classify $U$, i.e. $u(q)=\mathsf{true}$ iff $q\in U$. Then $(f;u)(p)=\mathsf{true}$ iff $f(p)\in U$ iff $p\in f^{-1}(U)$, so $f;u$ classifies $f^{-1}(U)=f^{\ast }(U)$. This is the preorder version of preimage as precomposition (Kittenlab Lecture 14).

Sources: 7 Sketches, Exercise 1.79 and Solution A.1.

Solution 1.80

proof — Exercise 1.80

1. $0\le \frac{1}{n+1}$ for all $n$.
2. Suppose $b$ is a lower bound with $0<b$. Pick $n$ with $1/b<n+1$; then $\frac{1}{n+1}<b$, contradicting that $b$ is a lower bound. So every lower bound is $\le 0$.

Sources: 7 Sketches, Exercise 1.80 and Solution A.1.

Solution 1.85

proof — Exercise 1.85

1. $p\le a$ for the only $a=p$; and if $q\le p$ then $q\le p$: so $p$ is a Meet. Any other meet $q$ satisfies $q\le p$ and $p\le q$, so $q\cong p$.
2. In a partial order $p\cong q$ implies $p=q$.
3. Yes; replace $\le$ by $\ge$ and “meet” by “join” throughout.

Sources: 7 Sketches, Exercise 1.85 and Solution A.1.

Solution 1.90

Exercise 1.90

$4\wedge 6=2$ and $4\vee 6=12$: the meet is the greatest common divisor, the join the least common multiple.

Sources: 7 Sketches, Exercise 1.90 and Solution A.1.

Solution 1.94

proof — Exercise 1.94

Since $a\le a\vee b$ and $b\le a\vee b$, monotonicity gives $f(a)\le f(a\vee b)$ and $f(b)\le f(a\vee b)$. So $f(a\vee b)$ is an upper bound of $\{f(a),f(b)\}$, and the join $f(a)\vee f(b)$ is the least one: $f(a)\vee f(b)\le f(a\vee b)$.

Sources: 7 Sketches, Exercise 1.94 and Solution A.1.

Solution 1.98

proof — Exercise 1.98

The right adjoint is $\lfloor -/3\rfloor :\mathbb{R}\to \mathbb{Z}$; we must show $3z\le r$ iff $z\le \lfloor r/3\rfloor$. If $z\le \lfloor r/3\rfloor$ then $3z\le 3\lfloor r/3\rfloor \le r$. If $3z\le r$ then $z\le r/3$, and since $z$ is an integer below $r/3$ it is below the greatest such, $\lfloor r/3\rfloor$.

Sources: 7 Sketches, Exercise 1.98 and Solution A.1.

Solution 1.99

Exercise 1.99

1. $f=(1\mapsto 1,2\mapsto 1,3\mapsto 3)$, $g=(1\mapsto 2,2\mapsto 2,3\mapsto 3)$. Checking all nine pairs, $f(p)\le q$ iff $p\le g(q)$ holds (for $(p,q)=(3,1),(3,2)$ both sides fail; otherwise both hold), so $f⊣g$.
2. Here $f(2)=1$ but $2\nleqq g(1)$, so $f$ is not left adjoint to $g$. In pictures of total orders, adjoint pairs are exactly those whose bent arrows do not cross (Remark 1.100).

Sources: 7 Sketches, Exercise 1.99 and Solution A.1.

Solution 1.101

proof — Exercise 1.101

Suppose $L⊣\lceil -/3\rceil$. Then $L(z)\le r$ iff $z\le \lceil r/3\rceil$. Take $z=1$, $r=0.01$: $\lceil 0.01/3\rceil =1\ge 1$, so $L(1)\le 0.01$; similarly $L(1)\le r$ for every $r>0$, so $L(1)\le 0$. But then $1\le \lceil 0/3\rceil =0$, a contradiction. So there is no left adjoint. (Equivalently: $\lceil -/3\rceil$ does not preserve meets — ${\bigwedge }_{r>0}\lceil r/3\rceil =1\ne 0=\lceil {\bigwedge }_{r>0}r/3\rceil$; see Adjoint Functor Theorem for Preorders.)

Sources: 7 Sketches, Exercise 1.101 and Solution A.1.

Solution 1.103

Exercise 1.103

$g_{!}((1)(2)(3)(4))=(12)(3)(4)$; $g_{!}((12)(3)(4))=(12)(3)(4)$; $g_{!}((13)(2)(4))=(123)(4)$; $g_{!}((1)(2)(34))=(12)(34)$; $g_{!}((14)(23))=(1234)$; $g_{!}((1)(234))=(1234)$. In general merge $1,2$ into $12$ and take the transitive closure.

Sources: 7 Sketches, Exercise 1.103 and Solution A.1.

Solution 1.105

Exercise 1.105

$s_1\sim s_2$ iff $g(s_1)\sim g(s_2)$; since $g(1)=g(2)$, elements $1,2$ are always identified: $(12)(3)(4)$, $(123)(4)$, $(124)(3)$, $(12)(34)$, $(1234)$.

Sources: 7 Sketches, Exercise 1.105 and Solution A.1.

Solution 1.106

Exercise 1.106

Take $c=(13)(2)(4)$; then $g_{!}(c)=(123)(4)$. Let $d=(1234)$ (coarser) and $e=(12)(34)$ (not coarser). Then $g^{\ast }(d)=(1234)$ and $g^{\ast }(e)=(12)(34)$. Indeed $c\le g^{\ast }(d)$, but $c\nleqq g^{\ast }(e)$ since $1{\sim }_{c}3$ while $1\nsim 3$ in $(12)(34)$ — consistent with the Galois Connection formula $g_{!}(c)\le d⟺c\le g^{\ast }(d)$.

Sources: 7 Sketches, Exercise 1.106 and Solution A.1.

Solution 1.109

proof — Exercise 1.109

1. Apply the definition with $p:=g(q)$ to the reflexivity fact $g(q)\le g(q)$: $f(g(q))\le q$.
2. If $p\le g(q)$, apply $f$: $f(p)\le f(g(q))\le q$. If $f(p)\le q$, apply $g$: $p\le g(f(p))\le g(q)$.

Sources: 7 Sketches, Exercise 1.109 and Solution A.1.

Solution 1.110

proof — Exercise 1.110

1. Using $p\le g^{\prime }(f(p))$ with $p=g(q)$ and monotonicity of $g^{\prime }$ applied to $f(g(q))\le q$: $g(q)\le g^{\prime }(f(g(q)))\le g^{\prime }(q)$. Symmetrically $g^{\prime }(q)\le g(q)$.
2. Yes, by the dual argument. (Categorically: adjoints are unique up to unique Natural Isomorphism.)

Sources: 7 Sketches, Exercise 1.110 and Solution A.1.

Solution 1.112

proof — Exercise 1.112

Let $f⊣g$, $A\subseteq P$ with join $j$. Monotonicity gives $f(a)\le f(j)$ for all $a\in A$, so $f(j)$ is an upper bound of $f(A)$. If $b$ is another upper bound, $f(a)\le b$ for all $a$, so by adjunction $a\le g(b)$ for all $a$, hence $j\le g(b)$, hence $f(j)\le b$. So $f(j)=\bigvee f(A)$.

Sources: 7 Sketches, Exercise 1.112 and Solution A.1.

Solution 1.114

Exercise 1.114

$p$	$q$	$f(p)\le q$	$p\le g(q)$
1	1	yes	yes
1	2	no	no
1	4	yes	yes
2	1	no	no
2	2	yes	yes
2	4	yes	yes
3.9	1	no	no
3.9	2	no	no
3.9	4	yes	yes
4	1	no	no
4	2	no	no
4	4	yes	yes

All agree, so $f⊣g$; yet $g$ does not preserve joins (Right Adjoints Preserve Meets but not joins).

Sources: 7 Sketches, Exercise 1.114 and Solution A.1.

Solution 1.118

Exercise 1.118

$X=\{a_1,c_1,c_2\}$, $Y=\{a,b,c\}$, $f$ “projects down” ($a_1\mapsto a$, $c_i\mapsto c$). 1. $f^{\ast }\{a,b\}=\{a_1\}$, $f^{\ast }\{c\}=\{c_1,c_2\}$. 2. $f_{!}\varnothing =\varnothing$, $f_{!}\{a_1,c_1\}=\{a,c\}$. 3. $f_{\ast }\varnothing =\{b\}$, $f_{\ast }\{a_1,c_1\}=\{a,b\}$.

Sources: 7 Sketches, Exercise 1.118 and Solution A.1.

Solution 1.119

proof — Exercise 1.119

1. This is the unit inequality of Proposition 1.107.
2. $\ge$: apply (1) to $g(f(p))$. $\le$: the counit gives $f(g(f(p)))\le f(p)$; apply the monotone $g$ to get $g(f(g(f(p))))\le g(f(p))$.

Sources: 7 Sketches, Exercise 1.119 and Solution A.1.

Solution 1.124

Exercise 1.124

$\mathrm{Rel}(\{1,2\})=\mathcal{P}(\{1,2\}\times \{1,2\})$ is the Power Set of a four-element set: a 4-dimensional cube with 16 vertices, from $\varnothing$ at the bottom to the total relation $\{(1,1),(1,2),(2,1),(2,2)\}$ at the top.

Sources: 7 Sketches, Exercise 1.124 and Solution A.1.

Solution 1.125

Exercise 1.125

1. Take $1\le 2\le 3$: $U(\le )=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$.
2. $Q=\{(1,2)\}$, $Q^{\prime }=\{(2,1)\}$.
3. $\mathrm{Cl}(Q)=\{(1,1),(2,2),(3,3),(1,2)\}\subseteq U(\le )$.
4. $\mathrm{Cl}(Q^{\prime })\ni (2,1)\notin U(\le )$. This illustrates the adjunction $\mathrm{Cl}⊣U$ of Reflexive Transitive Closure.

Sources: 7 Sketches, Exercise 1.125 and Solution A.1.