7S Chapter 6 Solutions

solution

Solutions to the exercises of 7 Sketches, Chapter 6: 7S Chapter 6 Exercises. Index: Map of Content.

Solution 6.3

example — Exercise 6.3

1. The Discrete Preorder on $A$ (only $a\le a$ and $b\le b$): neither $a\le b$ nor $b\le a$, so no element has a morphism to every other.
2. The Walking Arrow $a\le b$: $a$ is the unique initial object.
3. The Codiscrete Preorder ($a\le b$ and $b\le a$): both $a$ and $b$ are initial. Note that they are isomorphic, as 7S Exercise 6.10 predicts.

Sources: 7 Sketches, Exercise 6.3 and Solution A.6.

Solution 6.6

example — Exercise 6.6

The objects of a free category are the vertices and the morphisms are paths, so an initial object is a vertex with exactly one path to every vertex (including itself).

1. Yes: $a$, whose only path to itself is the empty path.
2. Yes: $a$ has a unique path to $a$, $b$, $c$.
3. No: there is no path from $a$ to $b$ nor from $b$ to $a$.
4. No: $a$ has infinitely many paths to itself (the loop iterated $n$ times), so the hom-set $\mathcal{C}(a,a)$ is not a singleton.

Sources: 7 Sketches, Exercise 6.6 and Solution A.6.

Solution 6.7

proof — Exercise 6.7

1. $f(1_R)=1_S$ and $f(r_1{\ast }_R{r}_2)=f(r_1){\ast }_{S}f(r_2)$.
2. The Natural Numbers rig $(\mathbb{N},0,+,1,\ast )$. Given any rig $R$, a homomorphism $f:\mathbb{N}\to R$ must send $0\mapsto 0_R$, $1\mapsto 1_R$, and by additivity

$$f(m)=f(1+\cdots +1)=1_R{+}_R\cdots {+}_R{1}_R(m\text{ summands}),$$

so $f$ is determined. This formula also preserves multiplication: by distributivity, $(1_R+\cdots +1_R)$ ($m$ times) ${\ast }_R$ $(1_R+\cdots +1_R)$ ($n$ times) expands to the sum of $mn$ copies of $1_R$, which is $f(m\ast n)$. Hence there is exactly one rig homomorphism $\mathbb{N}\to R$, i.e. $\mathbb{N}$ is initial.

LeanHaskell

import Mathlib
-- ℕ is the initial semiring: the unique ring hom is the canonical cast.
#check @Nat.castRingHom            -- (R : Type) [NonAssocSemiring R] : ℕ →+* R
#check @RingHom.eq_natCast         -- every f : ℕ →+* R equals Nat.cast

-- the unique rig homomorphism from Nat into any semiring
fromNat :: Num r => Integer -> r
fromNat 0 = 0
fromNat n = 1 + fromNat (n - 1)

Sources: 7 Sketches, Exercise 6.7 and Solution A.6.

Solution 6.8

annotation — Exercise 6.8

The initial object $\varnothing \in \mathcal{C}$ is the universal thing. Since the property quantifies over all objects of $\mathcal{C}$, every object $c$ counts as a “comparable object”. The Universal Property then reads: for every $c\in \mathcal{C}$ there is a unique morphism $\varnothing \to c$.

Sources: 7 Sketches, Exercise 6.8 and Solution A.6.

Solution 6.10

proof — Exercise 6.10

Since $c_1$ is initial there is a unique $f:c_1\to c_2$; since $c_2$ is initial there is a unique $g:c_2\to c_1$. Now $c_1\to c_1$ has a unique morphism because $c_1$ is initial, and both ${\mathrm{id}}_{c_1}$ and $f;g$ are such morphisms, so $f;g={\mathrm{id}}_{c_1}$. Symmetrically $g;f={\mathrm{id}}_{c_2}$. Hence $f$ is an isomorphism, and it is the unique isomorphism between them: initial objects are unique up to unique isomorphism.

Lean

import Mathlib
open CategoryTheory Limits
#check @Limits.initialIsoIsInitial   -- IsInitial X → (⊥_ C ≅ X)
#check @Limits.IsInitial.uniqueUpToIso

Sources: 7 Sketches, Exercise 6.10 and Solution A.6.

Solution 6.13

proof — Exercise 6.13

A preorder is a category with at most one morphism between any two objects, so every diagram commutes. Unfolding the definition of coproduct: $p+q$ is an element with $p\le p+q$ and $q\le p+q$ (the inclusions), such that whenever $p\le x$ and $q\le x$ we have $p+q\le x$ (the unique copairing). That is exactly the least upper bound $p\vee q$.

Dually Products are Meets, and the Initial Object is the bottom element.

Sources: 7 Sketches, Exercise 6.13 and Solution A.6.

Solution 6.16

example — Exercise 6.16

$A\bigsqcup B$	apple${}_1$	banana${}_1$	pear${}_1$	cherry${}_1$	orange${}_1$	apple${}_2$	tomato${}_2$	mango${}_2$
$[f,g]$	a	b	p	c	o	e	o	o

The Coproduct in $\mathbf{Set}$ is the disjoint union, so the two apples are distinct elements with different images.

JuliaHaskell

using Catlab
A = FinSet(5); B = FinSet(3); T = FinSet(26)            # letters as 1..26
letter(c) = Int(c) - Int('a') + 1
f = FinFunction(letter.(['a','b','p','c','o']), 26)     # first letters
g = FinFunction(letter.(['e','o','o']), 26)             # last letters
cp = coproduct(A, B)
h = copair(cp, f, g)                                    # [f, g] : 8 → 26
collect(h)                                              # [1, 2, 16, 3, 15, 5, 15, 15]

copair :: (a -> t) -> (b -> t) -> Either a b -> t
copair = either
-- either (head) (last) :: Either String String -> Char

Sources: 7 Sketches, Exercise 6.16 and Solution A.6.

Solution 6.17

proof — Exercise 6.17

1–2. These are exactly the two commuting triangles in the diagram defining the copairing $[f,g]$. 3. Both $[f,g];h$ and $[f;h,g;h]$ are morphisms $A+B\to D$ whose precompositions with ${\iota }_A,{\iota }_B$ are $f;h$ and $g;h$ (by 1–2). The Universal Property says such a morphism is unique, so they are equal. 4. ${\mathrm{id}}_{A+B}$ satisfies ${\iota }_A;\mathrm{id}={\iota }_A$ and ${\iota }_B;\mathrm{id}={\iota }_B$; by uniqueness of the copairing, $[{\iota }_A,{\iota }_B]={\mathrm{id}}_{A+B}$.

LeanHaskell

import Mathlib
open CategoryTheory Limits
#check @Limits.coprod.inl_desc   -- coprod.inl ≫ coprod.desc f g = f
#check @Limits.coprod.inr_desc
#check @Limits.coprod.desc_comp  -- coprod.desc f g ≫ h = coprod.desc (f ≫ h) (g ≫ h)
#check @Limits.coprod.desc_inl_inr

-- either f g . Left  == f
-- either f g . Right == g
-- h . either f g == either (h . f) (h . g)
-- either Left Right == id

Sources: 7 Sketches, Exercise 6.17 and Solution A.6.

Solution 6.18

proof — Exercise 6.18

1. On objects take the coproduct; on a morphism $(f,g):(A,B)\to (C,D)$ set $f+g:=[f;{\iota }_C,g;{\iota }_D]:A+B\to C+D$. Identities are preserved: ${\mathrm{id}}_A+{\mathrm{id}}_B=[{\iota }_A,{\iota }_B]={\mathrm{id}}_{A+B}$ by 7S Exercise 6.17 (4). Composition is preserved because both $(f+g);(h+k)$ and $(f;h)+(g;k)$ equal $[f;h;{\iota }_E,g;k;{\iota }_F]$ by uniqueness of copairing.
2. Let ${!}_A:\varnothing \to A$ be the unique map. The copairing $[{\mathrm{id}}_A,{!}_A]:A+\varnothing \to A$ is inverse to ${\iota }_A$: ${\iota }_A;[{\mathrm{id}}_A,{!}_A]={\mathrm{id}}_A$, and $[{\mathrm{id}}_A,{!}_A];{\iota }_A=[{\iota }_A,{!}_A;{\iota }_A]=[{\iota }_A,{\iota }_{\varnothing }]={\mathrm{id}}_{A+\varnothing }$ (using that ${!}_A;{\iota }_A$ and ${\iota }_{\varnothing }$ are both maps out of the initial object). Symmetrically for $[{!}_A,{\mathrm{id}}_A]:\varnothing +A\to A$.
3. (a) $\alpha =[[{\iota }_A,{\iota }_B;{\iota }_{B+C}],{\iota }_C;{\iota }_{B+C}]$ with inverse $[{\iota }_A;{\iota }_{A+B},[{\iota }_B;{\iota }_{A+B},{\iota }_C]]$. (b) $\sigma =[{\iota }_A^{\prime },{\iota }_B^{\prime }]$ where ${\iota }_A^{\prime }:A\to B+A$, ${\iota }_B^{\prime }:B\to B+A$ are the inclusions of the other coproduct; its inverse is the analogous map $B+A\to A+B$, and $\sigma ;{\sigma }^{-1}=\mathrm{id}$ by 7S Exercise 6.17 (3–4).

The same argument dualised shows that finite Products give a symmetric monoidal structure (Cartesian Category).

JuliaLeanHaskell

using Catlab
f = FinFunction([1, 2], 3); g = FinFunction([1], 2)
fg = oplus(f, g)                       # f + g : 3 → 5 in the prop FinSet
collect(fg)                            # [1, 2, 4]

import Mathlib
open CategoryTheory
-- Mathlib packages exactly this: coproducts give a monoidal structure.
#check @CategoryTheory.monoidalOfHasFiniteCoproducts

import Data.Bifunctor (bimap)          -- bimap f g :: Either a b -> Either c d
assoc :: Either (Either a b) c -> Either a (Either b c)
assoc = either (either Left (Right . Left)) (Right . Right)
swap :: Either a b -> Either b a
swap = either Right Left

Sources: 7 Sketches, Exercise 6.18 and Solution A.6.

Solution 6.24

proof — Exercise 6.24

1. A span $B\leftarrow A\to C$ in ${\mathbf{Disc}}_S$ consists of identities, so $A=B=C$, and the square of identities on $A$ is a pushout: any cocone consists of two equal maps $A\to T$ (both identities, so $T=A$), and the identity is the unique mediating map.
2. Exactly when $|S|=1$. If $S=\varnothing$ there is no object at all; if $s\ne s^{\prime }$ are two objects there is no morphism $s\to s^{\prime }$.

Sources: 7 Sketches, Exercise 6.24 and Solution A.6.

Solution 6.26

example program — Exercise 6.26

The pushout is the set of connected components of $\underline{5}\bigsqcup \underline{3}$ under the relation generated by $f(a)\sim g(a)$: $1\sim 1^{\prime }$, $3\sim 1^{\prime }$, $5\sim 2^{\prime }$, $5\sim 3^{\prime }$. The classes are $\{1,1^{\prime },3\}$, $\{2\}$, $\{4\}$, $\{5,2^{\prime },3^{\prime }\}$: the pushout is $\underline{4}$, with $\underline{5}\to \underline{4}$ given by $(1,2,1,3,4)$ and $\underline{3}\to \underline{4}$ given by $(1,4,4)$.

JuliaHaskell

using Catlab
f = FinFunction([1, 3, 5, 5], 5)
g = FinFunction([1, 1, 2, 3], 3)
P = pushout(f, g)
ob(P)                     # FinSet(4)
collect.(legs(P))         # ([1, 2, 1, 3, 4], [1, 4, 4])

-- pushout of finite functions via union-find on the disjoint union
import Data.List (nub)
pushout :: Int -> Int -> [Int] -> [Int] -> [[Int]]   -- classes of 5 ⊔ 3, elements of Y offset
pushout nx ny f g = classes (zip f (map (+ nx) g)) [1 .. nx + ny]
  where classes rel xs = nub [ closure [x] | x <- xs ]
          where step cs = nub (cs ++ [ b | (a,b) <- rel, a `elem` cs ] ++ [ a | (a,b) <- rel, b `elem` cs ])
                closure cs = let cs' = step cs in if length cs' == length cs then cs else closure cs'
-- pushout 5 3 [1,3,5,5] [1,1,2,3] gives 4 classes

Sources: 7 Sketches, Exercise 6.26 and Solution A.6; Kittenlab lecture on colimits (union-find).

Solution 6.28

proof — Exercise 6.28

1. The square $X\leftarrow \varnothing \to Y$, $X\to X+Y\leftarrow Y$ commutes because there is only one map $\varnothing \to X+Y$; so $f;{\iota }_X=g;{\iota }_Y$.
2. Given $x:X\to T$, $y:Y\to T$ (the square with $\varnothing$ commutes automatically), the universal property of the coproduct gives a unique $[x,y]:X+Y\to T$ with ${\iota }_X;[x,y]=x$ and ${\iota }_Y;[x,y]=y$.
3. Conversely, if the pushout $X{+}_{\varnothing }Y$ exists then for any $x,y$ as above, the outer square commutes (again because $\varnothing$ is initial), so the pushout property gives a unique $t:X{+}_{\varnothing }Y\to T$ with ${\iota }_X;t=x$, ${\iota }_Y;t=y$. This is exactly the universal property of the coproduct.

Sources: 7 Sketches, Exercise 6.28 and Solution A.6.

Solution 6.35

proof — Exercise 6.35

Suppose $T$ is a Cocone on the original diagram: maps from $X,Y,Z$ to $T$ making the two squares with $A$ and $B$ commute. Since $Q$ is the pushout of $X\leftarrow A\to Y$ there is a unique $Q\to T$ compatible with $X,Y$; since $R$ is the pushout of $Y\leftarrow B\to Z$ there is a unique $R\to T$ compatible with $Y,Z$. Both agree with the given map on $Y$, so $(Q,R,T)$ forms a cocone on $Q\leftarrow Y\to R$, and the pushout $S$ gives a unique $S\to T$ making everything commute. Hence $S$ is the colimit. This is the mechanism behind Finite Colimits in Set: an initial object and pushouts give all finite colimits.

Lean

import Mathlib
open CategoryTheory Limits
-- finite colimits from an initial object and pushouts
#check @CategoryTheory.Limits.hasFiniteColimits_of_hasInitial_and_pushouts

Sources: 7 Sketches, Exercise 6.35 and Solution A.6.

Solution 6.41

proof — Exercise 6.41

Theorem 6.37 says the colimit is the quotient of $X\bigsqcup N\bigsqcup Y$ by the equivalence relation generated by $x\sim n$ if $x=f(n)$ and $y\sim n$ if $y=g(n)$. Every $n\in N$ is related to $f(n)\in X$, so each class contains an element of $X\bigsqcup Y$; thus the quotient equals the quotient of $X\bigsqcup Y$ by the relation generated by $f(n)\sim g(n)$, which is exactly Example 6.25.

Sources: 7 Sketches, Exercise 6.41 and Solution A.6.

Solution 6.48

example — Exercise 6.48

The monoidal product of cospans $A\to N\leftarrow B$ and $B\to P\leftarrow C$ is the cospan $A+B\to N+P\leftarrow B+C$: one simply stacks the two wiring pictures vertically (disjoint union of apices and of feet). See Hypergraph Category for the general structure.

Julia

using Catlab
c1 = Cospan(FinFunction([1, 1], 2), FinFunction([2], 2))       # A=2 → N=2 ← B=1
c2 = Cospan(FinFunction([1], 3), FinFunction([1, 2, 3], 3))   # B=1 → P=3 ← C=3
c12 = Cospan(oplus(left(c1), left(c2)), oplus(right(c1), right(c2)))
apex(c12)                                                     # FinSet(5) = N + P

Sources: 7 Sketches, Exercise 6.48 and Solution A.6.

Solution 6.49

annotation — Exercise 6.49

Concatenate the two wire diagrams along the shared foot $B$. (i) The apex of the composite has one element for each connected component of the concatenated picture (this is the Pushout as a set of connected components, cf. Colimits and Connection). (ii) Each element of the outer feet $A$, $C$ is wired to the element representing the component it belongs to.

Sources: 7 Sketches, Exercise 6.49 and Solution A.6.

Solution 6.57

example — Exercise 6.57

By Theorem 6.55 (the spider theorem: a connected Frobenius diagram is determined by its number of inputs and outputs), two connected diagrams with the same inputs and outputs are equal, and a disconnected diagram is determined by its partition of the ports. Morphisms 1, 4 and 6 are equal (a single connected spider $2\to 3$); morphisms 3 and 5 are equal (the same disconnected pattern); morphism 2 is not equal to any other.

Sources: 7 Sketches, Exercise 6.57 and Solution A.6.

Solution 6.59

example — Exercise 6.59

1. $B$: the wire into $h$ comes from a spider whose other legs are labelled $B$, and spiders connect only wires of the same type.
2. $D$, since $h:B\to D\otimes D$ and $g$‘s output shares a spider with $h$‘s outputs.
3. $D$ as well, for the same reason.

Sources: 7 Sketches, Exercise 6.59 and Solution A.6.

Solution 6.62

example — Exercise 6.62

	cospan	wiring
multiplication $\mu$	$\underline{2}\stackrel{[\mathrm{id},\mathrm{id}]}{\to }\underline{1}\stackrel{\mathrm{id}}{\leftarrow }\underline{1}$	two wires merging into one
unit $\eta$	$\underline{0}\stackrel{!}{\to }\underline{1}\stackrel{\mathrm{id}}{\leftarrow }\underline{1}$	a wire starting from nothing
comultiplication $\delta$	$\underline{1}\stackrel{\mathrm{id}}{\to }\underline{1}\stackrel{[\mathrm{id},\mathrm{id}]}{\leftarrow }\underline{2}$	one wire splitting into two
counit $ϵ$	$\underline{1}\stackrel{\mathrm{id}}{\to }\underline{1}\stackrel{!}{\leftarrow }\underline{0}$	a wire ending in nothing

The empty set is depicted as blank space.

Julia

using Catlab
μ = Cospan(FinFunction([1, 1], 1), FinFunction([1], 1))
η = Cospan(FinFunction(Int[], 1), FinFunction([1], 1))
δ = Cospan(FinFunction([1], 1), FinFunction([1, 1], 1))
ε = Cospan(FinFunction([1], 1), FinFunction(Int[], 1))

Sources: 7 Sketches, Exercise 6.62 and Solution A.6.

Solution 6.63

proof — Exercise 6.63

The composite $\delta ;\mu$ is the cospan $X\stackrel{\mathrm{id}}{\to }X\stackrel{[\mathrm{id},\mathrm{id}]}{\leftarrow }X+X\stackrel{[\mathrm{id},\mathrm{id}]}{\to }X\stackrel{\mathrm{id}}{\leftarrow }X$, so it suffices to show that the square with $X+X$ at the top left, two copies of $[\mathrm{id},\mathrm{id}]:X+X\to X$, and two identities $X\to X$ is a Pushout. It commutes trivially. Given $f,g:X\to T$ with $[\mathrm{id},\mathrm{id}];f=[\mathrm{id},\mathrm{id}];g$, precomposing with ${\iota }_1:X\to X+X$ and using ${\iota }_1;[\mathrm{id},\mathrm{id}]=\mathrm{id}$ (7S Exercise 6.17) gives $f=g$. Then $f$ is the unique map from the apex $X$ making the cocone commute. Hence the pushout apex is $X$ with identity legs, i.e. $\delta ;\mu ={\mathrm{id}}_X$.

Julia

using Catlab
X = FinSet(2)
δ = Cospan(id(X), FinFunction([1, 2, 1, 2], 2))
μ = Cospan(FinFunction([1, 2, 1, 2], 2), id(X))
P = pushout(right(δ), left(μ))
ob(P)                                # FinSet(2): the special law holds

Sources: 7 Sketches, Exercise 6.63 and Solution A.6.

Solution 6.67

proof — Exercise 6.67

The cup is $\eta ;\delta :I\to X\otimes X$ and the cap is $\mu ;ϵ:X\otimes X\to I$. The snake equation $(\eta ;\delta )\otimes \mathrm{id};\mathrm{id}\otimes (\mu ;ϵ)={\mathrm{id}}_X$ is proved by the chain: apply the Frobenius law (6.53) to move the $\delta$ past the $\mu$, obtaining $\mathrm{id}\otimes \eta ;\mu ;\delta ;ϵ\otimes \mathrm{id}$; the missing diagram is this middle step, in which the unit and counit are attached to the multiplication and comultiplication respectively. Then the unit law $(\mathrm{id}\otimes \eta );\mu =\mathrm{id}$ and its opposite $\delta ;(ϵ\otimes \mathrm{id})=\mathrm{id}$ reduce it to ${\mathrm{id}}_X$.

Sources: 7 Sketches, Exercise 6.67 and Solution A.6.

Solution 6.70

proof — Exercise 6.70

Let $A\subseteq S$, $B\subseteq T$. Then

$${\phi }_{S^{\prime },T^{\prime }}({\mathrm{im}}_f(A),{\mathrm{im}}_g(B))=\{f(a)\mid a\in A\}\times \{g(b)\mid b\in B\}=\{(f(a),g(b))\mid a\in A,b\in B\}={\mathrm{im}}_{f\times g}(A\times B)={\mathrm{im}}_{f\times g}({\phi }_{S,T}(A,B)),$$

so the square commutes.

Sources: 7 Sketches, Exercise 6.70 and Solution A.6.

Solution 6.78

annotation — Exercise 6.78

Take $F:\mathcal{C}\to \mathbf{Set}$ constant at the singleton $\{\ast \}$; it is lax symmetric monoidal (all structure maps are the unique map into $\{\ast \}$). A morphism in ${\mathbf{Cospan}}_F$ is a cospan $X\to N\leftarrow Y$ together with an element of $F(N)=\{\ast \}$, i.e. no extra choice. Composition is the usual pushout composition since the decoration is forced. Hence ${\mathbf{Cospan}}_F\cong {\mathbf{Cospan}}_{\mathcal{C}}$ via the identity-on-objects functor that decorates each cospan with $\ast$; category theorists happily call these “equal”.

Sources: 7 Sketches, Exercise 6.78 and Solution A.6.

Solution 6.79

example program — Exercise 6.79

$V=\{ul,ur,dl,dr\}$, $A=\{r_1,r_2,r_3,c_1,i_1\}$, with

	$r_1$	$r_2$	$r_3$	$c_1$	$i_1$
$s$	dl	ul	ur	ul	dl
$t$	ul	ur	dr	ur	dr
$\ell$	$1\Omega$	$2\Omega$	$1\Omega$	$3F$	$1H$

A $\mathcal{C}$-circuit is precisely an edge-labelled Graph (C-Set on the schema of graphs with a label attribute).

JuliaHaskell

using Catlab
@present SchCircuit <: SchGraph begin
  Label::AttrType
  label::Attr(E, Label)
end
@acset_type Circuit(SchCircuit, index=[:src, :tgt])
# vertices 1=ul 2=ur 3=dl 4=dr
c = @acset Circuit{String} begin
  V = 4; E = 5
  src = [3, 1, 2, 1, 3]; tgt = [1, 2, 4, 2, 4]
  label = ["1Ω", "2Ω", "1Ω", "3F", "1H"]
end

data Circuit v l = Circuit { arrows :: [(v, v, l)] }   -- (s a, t a, ℓ a)
c :: Circuit String String
c = Circuit [("dl","ul","1Ω"),("ul","ur","2Ω"),("ur","dr","1Ω"),("ul","ur","3F"),("dl","dr","1H")]

Sources: 7 Sketches, Exercise 6.79 and Solution A.6.

Solution 6.80

example — Exercise 6.80

The decoration functor $\mathrm{Circ}:\mathbf{FinSet}\to \mathbf{Set}$ acts on a function $f$ by relabelling vertices: $\mathrm{Circ}(f)(V,A,s,t,\ell )=(V^{\prime },A,f;s,f;t,\ell )$. So $\mathrm{Circ}(f)(c)$ has vertices $\{1,2\sim 3,4\}$ and the $3\Omega$ resistor now runs from the merged vertex $2\sim 3$ to $4$; the wire from $1$ ends at $2\sim 3$.

Julia

using Catlab
# Circ(f) is the pushforward of the vertex set: Σ-migration along f on V
@present SchCircuit <: SchGraph begin Label::AttrType; label::Attr(E, Label) end
@acset_type Circuit(SchCircuit, index=[:src, :tgt])
c = @acset Circuit{String} begin V = 4; E = 1; src = [3]; tgt = [4]; label = ["3Ω"] end
f = [1, 2, 2, 3]                                    # 4 → 3, identifies 2 and 3
c′ = @acset Circuit{String} begin V = 3; E = 1; src = f[c[:src]]; tgt = f[c[:tgt]]; label = c[:label] end

Sources: 7 Sketches, Exercise 6.80 and Solution A.6.

Solution 6.82

example — Exercise 6.82

${\psi }_{V,V^{\prime }}$ takes the disjoint union of labelled graphs: ${\psi }_{2,2}(b,s)$ is the 4-vertex circuit consisting of the battery between vertices $1,2$ and the switch between vertices $3,4$, with no connection between them. This laxator is what makes $\mathrm{Circ}$ a lax Monoidal Functor, and thus ${\mathbf{Cospan}}_{\mathrm{Circ}}$ a Hypergraph Category (Decorated Cospan).

Julia

using Catlab
@present SchCircuit <: SchGraph begin Label::AttrType; label::Attr(E, Label) end
@acset_type Circuit(SchCircuit, index=[:src, :tgt])
b = @acset Circuit{Symbol} begin V = 2; E = 1; src = [1]; tgt = [2]; label = [:battery] end
s = @acset Circuit{Symbol} begin V = 2; E = 1; src = [1]; tgt = [2]; label = [:switch] end
ψ = ob(coproduct(b, s))          # ψ₂,₂(b, s): 4 vertices, 2 edges

Sources: 7 Sketches, Exercise 6.82 and Solution A.6.

Solution 6.84

example — Exercise 6.84

The cospan is $\underline{1}\stackrel{f}{\to }\underline{2}\stackrel{g}{\leftarrow }\underline{1}$ with $f(1)=1$, $g(1)=2$. The decoration is the $\mathcal{C}$-circuit $(\underline{2},\{a\},s,t,\ell )$ with $s(a)=1$, $t(a)=2$, $\ell (a)=\mathrm{battery}$: an open battery whose two terminals are exposed on the left and right. See the Julia snippet in Decorated Cospan.

Sources: 7 Sketches, Exercise 6.84 and Solution A.6.

Solution 6.86

example program — Exercise 6.86

The first is the cospan $\underline{1}\stackrel{f}{\to }V\stackrel{g}{\leftarrow }\underline{2}$, $f(1)=ul$, $g(1)=g(2)=ur$, decorated by the circuit $C$ of 7S Exercise 6.79. The second is $\underline{2}\stackrel{f^{\prime }}{\to }{V}^{\prime }\stackrel{g^{\prime }}{\leftarrow }\underline{2}$ with $V^{\prime }=\{l,r,d\}$, $f^{\prime }(1)=l$, $f^{\prime }(2)=d$, $g^{\prime }(1)=g^{\prime }(2)=r$, decorated by $C^{\prime }=(V^{\prime },\{r_1^{\prime },r_2^{\prime }\},s^{\prime },t^{\prime },{\ell }^{\prime })$ with $r_1^{\prime }:l\to r$ ($5\Omega$), $r_2^{\prime }:r\to d$ ($8\Omega$).

Composing: the Pushout of $V\stackrel{g}{\leftarrow }\underline{2}\stackrel{f^{\prime }}{\to }{V}^{\prime }$ identifies $ur\sim l\sim d$ into one vertex $m$, giving $V^{\prime \prime }=\{ul,dl,dr,m,r\}$ (five vertices), and the composite cospan $\underline{1}\to V^{\prime \prime }\leftarrow \underline{2}$ has $1\mapsto ul$ and $(1,2)\mapsto (r,m)$. The decoration is $\mathrm{Circ}$ of the pushout applied to $\psi (C,C^{\prime })$: arrows $r_1:dl\to ul$, $r_2:ul\to m$, $r_3:m\to dr$, $c_1:ul\to m$, $i_1:dl\to dr$, $r_1^{\prime }:m\to r$, $r_2^{\prime }:r\to m$. This matches Eq. (6.74).

Julia

using Catlab
@present SchCircuit <: SchGraph begin Label::AttrType; label::Attr(E, Label) end
@acset_type Circuit(SchCircuit, index=[:src, :tgt])
const OpenCircuitOb, OpenCircuit = OpenACSetTypes(Circuit, :V)
# vertices 1=ul 2=ur 3=dl 4=dr
C = @acset Circuit{String} begin V = 4; E = 5
  src = [3, 1, 2, 1, 3]; tgt = [1, 2, 4, 2, 4]; label = ["1Ω", "2Ω", "1Ω", "3F", "1H"] end
C′ = @acset Circuit{String} begin V = 3; E = 2       # 1=l 2=r 3=d
  src = [1, 2]; tgt = [2, 3]; label = ["5Ω", "8Ω"] end
x = OpenCircuit{String}(C, FinFunction([1], 4), FinFunction([2, 2], 4))
y = OpenCircuit{String}(C′, FinFunction([1, 3], 3), FinFunction([2, 2], 3))
xy = compose(x, y)
nparts(apex(xy), :V), nparts(apex(xy), :E)          # (5, 7)

Sources: 7 Sketches, Exercise 6.86 and Solution A.6.

Solution 6.88

example — Exercise 6.88

$\eta ;x$ identifies the two left terminals of $x$ into one vertex (a wire closing the left side); $\eta ;x;ϵ$ then also identifies the two right terminals. The result is a cospan $\underline{0}\to N\leftarrow \underline{0}$ decorated with the circuit of $x$ in which the left pair and the right pair of terminals are each merged: a closed loop with no exposed terminals. Such closed circuits are the scalars $I\to I$ of the Hypergraph Category.

Julia

using Catlab
@present SchCircuit <: SchGraph begin Label::AttrType; label::Attr(E, Label) end
@acset_type Circuit(SchCircuit, index=[:src, :tgt])
const OpenCircuitOb, OpenCircuit = OpenACSetTypes(Circuit, :V)
x = @acset Circuit{Symbol} begin V = 4; E = 2; src = [1, 3]; tgt = [2, 4]; label = [:battery, :resistor] end
ox = OpenCircuit{Symbol}(x, FinFunction([1, 3], 4), FinFunction([2, 4], 4))
empty1 = @acset Circuit{Symbol} begin V = 1 end
η = OpenCircuit{Symbol}(empty1, FinFunction(Int[], 1), FinFunction([1, 1], 1))
ε = OpenCircuit{Symbol}(empty1, FinFunction([1, 1], 1), FinFunction(Int[], 1))
closed = compose(η, ox, ε)
nparts(apex(closed), :V), nparts(apex(closed), :E)   # (2, 2): a closed loop

Sources: 7 Sketches, Exercise 6.88 and Solution A.6.

Solution 6.96

example program — Exercise 6.96

1. Two inner circles with two ports each, an outer circle with two ports, and three links: one port of the first circle is wired to a port of the second; the remaining port of each inner circle is wired to an outer port.
2. Three inner circles with two ports each, no outer ports; the links connect the circles in a chain.
3. $g{\circ }_{1}f$ substitutes $f$ into the first circle of $g$; the operad composition is a pushout of the apices along the shared foot $\underline{2}$. Its arity is $(2,2,2,2;0)$: four inner circles and no outer ports.
4. The drawing shows $f$‘s two circles sitting where the first circle of $g$ used to be — literally substitution of one wiring diagram into a circle of another.

Julia

using Catlab
f = UndirectedWiringDiagram(2)        # outer circle with 2 ports
add_box!(f, 2); add_box!(f, 2)        # two inner circles, 2 ports each
add_junctions!(f, 3)                  # apex 3
set_junction!(f, [1, 2, 2, 3])        # box ports → junctions
set_junction!(f, [1, 3], outer=true)  # outer ports → junctions
g = UndirectedWiringDiagram(0)
add_box!(g, 2); add_box!(g, 2); add_box!(g, 2)
add_junctions!(g, 3)
set_junction!(g, [1, 2, 2, 3, 3, 1])
h = ocompose(g, 1, f)                 # substitute f into box 1 of g
nboxes(h), length(ports(h, outer=true)), njunctions(h)   # (4, 0, 4)

Sources: 7 Sketches, Exercise 6.96 and Solution A.6.