# 4.3. Categories of profunctors#

## 4.3.1. Composing profunctors#

Definition 4.21.

This definition uses inconsistent variable names; see a comment in the errata.

Exercise 4.22.

import numpy as np

return np.min(A[:,:,None] + B[None,:,:], axis=1)


The function add_min_dot is from custom matrix multiplication with numpy - Stack Overflow. See also Matrix multiplication. For multiplying an M×N and N×P matrix, the broadcasting version constructs an M×N×P matrix. The M×N matrix is duplicated P times so you have P copies of all M rows (M·P rows). The N×P matrix is duplicated so you have M copies of all P columns (M·P columns). Then we take the equivalent of the dot product across all those columns and rows: Interpreted in terms of the comments above Eq. (4.20), one can see the N dimension in the intermediate A[:,:,None] + B[None,:,:] as documenting the costs of all possible paths between some fixed m and p, which we then compress into a single answer. Said another way, the equivalent of the dot product is finding the minimum cost between some fixed m and p.

import pandas as pd

inf = float("inf")

Φ = pd.DataFrame(
index=["A","B","C","D"],
columns=["x","y","z"],
data=np.array(
[[17, 20, 20],
[11, 14, 14],
[14, 17, 17],
[12,  9, 15]])
)
display(Φ)

MY2 = pd.DataFrame(
index=["x","y","z"],
columns=["x","y","z"],
data=np.array(
[[0, 4, 3],
[3, 0, 6],
[7, 4, 0]])
)
display(MY2)

MZ1 = pd.DataFrame(
index=["p","q","r","s"],
columns=["p","q","r","s"],
data=np.array(
[[0, 2, inf, inf],
[inf, 0, 2, inf],
[inf, inf, 0, 1],
[1, inf, inf, 0]])
)
display(MZ1)
MZ3 = pd.DataFrame(
index=["p","q","r","s"],
columns=["p","q","r","s"],
)
display(MZ3)

MΨ = np.array([[inf, inf, inf, inf], [inf, inf, 0, inf], [4, inf, inf, 4]])

Ψ = pd.DataFrame(
index=["x","y","z"],
columns=["p","q","r","s"],
)
display(Ψ)

ΦΨ = pd.DataFrame(
index=["A","B","C","D"],
columns=["p","q","r","s"],
)
display(ΦΨ)

x y z
A 17 20 20
B 11 14 14
C 14 17 17
D 12 9 15
x y z
x 0 4 3
y 3 0 6
z 7 4 0
p q r s
p 0.0 2.0 inf inf
q inf 0.0 2.0 inf
r inf inf 0.0 1.0
s 1.0 inf inf 0.0
p q r s
p 0.0 2.0 4.0 5.0
q 4.0 0.0 2.0 3.0
r 2.0 4.0 0.0 1.0
s 1.0 3.0 5.0 0.0
p q r s
x 6.0 8.0 4.0 5.0
y 2.0 4.0 0.0 1.0
z 4.0 6.0 4.0 4.0
p q r s
A 22.0 24.0 20.0 21.0
B 16.0 18.0 14.0 15.0
C 19.0 21.0 17.0 18.0
D 11.0 13.0 9.0 10.0

## 4.3.2. The categories 𝓥-Prof and Feas#

Exercise 4.26. Exercise 4.30.

For part 1. of this question, the first step (an =) is justified by the unitality of the underlying symmetric monoidal preorder structure (see (b) of Definition 2.2).

The second step (a ≤) is justified by the fact that this is a 𝓥-category so that $$I$$ < 𝓟$$(p,p)$$ as well as the monotonicity property of a symmetric monoidal preorder (see (a) of Definition 2.2.).

The third step (a ≤) is justified based on the definition of a join (Definition 1.81). Clearly 𝓟$$(p,p)$$ ⊗ Φ$$(p,q)$$ is a member of the set defined by 𝓟$$(p,p₁)$$ ⊗ Φ$$(p₁,q)$$ for all p₁ ∈ 𝓟, so it should be ≤ the join of that set.

The last step (an =) is justified by both Eq. 4.25 (meaning $$U_\mathcal{P}(x,y) := \mathcal{P}(x,y)$$ in this context) and the definition of the composite of profunctors in Definition 4.21.

For part 2. we only need to address the second and third steps, since the first and fourth are already equalities. The author uses the term “directly” here because for Bool we can make this more direct proof without needing part 3.. We only generalize the result to non-Bool cases by completing part 3..

For the second step, because $$I = true$$ and the monoidal product ⊗ is ∧ (logical and), as well that $$\mathcal{P}(p,p)$$ is true by definition, we have an equality. We could replace both these terms with Φ$$(p,q)$$.

For the third step we only need to consider two situations: that Φ$$(p,q)$$ is true and that it is false. If it’s true then the join must also be true, since Φ$$(p,q)$$ is a member of the set and it’s true. If Φ$$(p,q)$$ is false, then we know that any Φ$$(p₁,q)$$ where p₁ ≥ p is also false (and any x ⊗ false is false, so all those terms are false in the join). And if p₁ ≤ p, then we know that $$\mathcal{P}(p,p₁)$$ is also false (and any x ⊗ false is false, so all those terms are false in the join). This is exhaustive, so the join is also false.

For part 3., the first step (an =) is justified by the unitality of the underlying symmetric monoidal preorder structure (see (b) of Definition 2.2).

The second step (a ≤) is justified by the fact that this is a 𝓥-category so that $$I$$ < 𝓟$$(p,p)$$ as well as the monotonicity property of a symmetric monoidal preorder (see (a) of Definition 2.2.).

The last step (a ≤) is justified by the relationship shown in Exercise 4.9.

Exercise 4.31.

Starting from Definition 4.21:

$(\Phi⨟\Psi)(p,r)=\bigvee_{q\in Q}\big(\Phi(p,q)\otimes\Psi(q,r)\big)$

We can extend this to:

$((\Phi⨟\Psi)⨟\Upsilon)(p,s) = \ \bigvee_{r\in R}\Big(\bigvee_{q\in Q}\Phi(p,q)\otimes\Psi(q,r)\Big)\otimes\Upsilon(r,s)$

Take Eq. (2.88) and replace the isomorphism with an equality because 𝓥 is skeletal, and apply:

$((\Phi⨟\Psi)⨟\Upsilon)(p,s) = (\Phi⨟\Psi⨟\Upsilon)(p,s) = \ \bigvee_{r\in R}\bigvee_{q\in Q}\Phi(p,q)\otimes\Psi(q,r)\otimes\Upsilon(r,s)$

We can similarly produce the same for the right-hand side of the equation in Lemma 3.41.

## 4.3.3. Fun profunctor facts: companions, conjoints, collages#

It may be helpful to think of a profunctor as “proto functor” (generalization of a functor).

$\newcommand{\ovs}{\overset{\smile}{#1}} \newcommand{\ovf}{\overset{\frown}{#1}}$

Exercise 4.36.

The identity function id is defined as $$F(p) = p$$. Since $$\ovf{F} = \mathcal{Q}(F(p), q) = \mathcal{Q}(p, q)$$, and $$\mathcal{Q} = \mathcal{P}$$ for the identity functor, we have that $$\mathcal{Q}(p, q) = \mathcal{P}(p, q) = U_{\mathcal{P}}(p, q)$$.

Exercise 4.38.

The conjoint $$\ovs{+}: ℝ ⇸ (ℝ × ℝ × ℝ)$$ would be the function that sends $$(a, b, c, d)$$ to true if $$a ≤ b + c + d$$ and false otherwise.

Exercise 4.41.

For part 1. of this question, let’s start with the definition of: $$\ovf{F}(p,q) = \mathcal{Q}(F(p),q)$$ from Definition 4.34. Notice this already matches the right-hand side of Eq. (4.40).

For the definition of $$\ovs{G}(p,q)$$, see Definition 4.34 again. Applying the definition is somewhat confusing because the order of the arguments $$p$$ and $$q$$ are reversed, as well as the order of the letters in the signature $$G: \mathcal{Q} → \mathcal{P}$$, but you should end up with $$\ovs{G}(p,q) = \mathcal{P}(p, G(q))$$. This matches the left-hand side of Eq. (4.40).

Eq. (4.40) expresses an isomorphism but given we are working with a skeletal quantale we can infer an equality.

For part 2. let’s first show that $$\ovf{id}$$ and $$\ovs{id}$$ are adjoints:

$\mathcal{P}(p, G(q)) = \mathcal{Q}(F(p),q)$

We have that $$\mathcal{P} = \mathcal{Q}$$:

$\mathcal{P}(p, G(q)) = \mathcal{P}(F(p),q)$

Then if $$F = G = id$$ we have:

\begin{split} \begin{align} \mathcal{P}(p, F(q)) & = \mathcal{P}(F(p),q) \\ \mathcal{P}(p, q) & = \mathcal{P}(p,q) \end{align} \end{split}

Since they are adjoints we have that $$\ovf{F} = \ovs{G}$$ and therefore that $$\ovf{id} = \ovs{id}$$.

Exercise 4.44. 