The semigroup of cones

Proof. Suppose that $η = γ * f$ where $f$ is surjective. Since $γ$ is proper, Con-3 or equivalently, Equation (16) hold. Thus

U (c_{γ}) = ⋃_{c \in 𝖛 𝒞} im (U (γ (c)))

Since by Equation (18), $γ * f$ is the map sending $c \in 𝖛 𝒞$ to $γ (c) f$ , we have

U (γ * f) (c) = U (γ (c)) U (f) = U (γ (c)) \circ U (f) .

Let $y \in U (d)$ . Since the map $U (f) : U (c_{γ}) \to U (d)$ is surjective there is $x \in U (c_{γ})$ such that $x U (f) = y$ . Since the cone $γ$ is proper, by Cone-3, there is $c \in 𝖛 𝒞$ such that $x \in im U (γ (c))$ . This implies that $x = u U (γ (c))$ for some $u \in U (c)$ . Then $y = (u U (γ (c))) U (f)$ . Therefore $y \in U im (γ (c) f)$ .

\begin{array}{l} It follows that & U (d) \subseteq ⋃_{c \in 𝖛 𝒞} im U (γ (c) f) . \\ The inclusion & ⋃_{c \in 𝖛 𝒞} im U (γ (c) f) \subseteq U (d) is obvious. \\ Hence & U (d) = ⋃_{c \in 𝖛 𝒞} im U (γ (c) f) . \end{array}

Therefore by Cone-3, $γ * f$ is proper.

Suppose that the cone $γ * f$ is proper. By Cone-3 we have

U (d) = ⋃_{c \in 𝖛 𝒞} U (im (γ (c) f)) .

Hence, for $y \in U (d)$ there is $c \in 𝖛 𝒞$ such that $y \in U (im (γ (c) f))$ . Then there exist $x \in U (c)$ such that $y = x U (γ (c) f) = u U (f)$ where $u = x U (γ (c)$ ). Therefore $U (f) : U (c_{γ}) \to U (d)$ is surjective. The fact that $𝒞$ is an $𝒮$ -category now shows that $f : c_{γ} \to d$ is surjective. $□$

Proof. It is suﬃcient to show that the binary composition deﬁned by Equation (20) is associative. Let $α, β, γ \in p c n 𝒞$ and $c \in 𝖛 𝒞$ . Then

\begin{array}{l} (α (β γ)) (c) & = α (c) {((β γ) (c_{α}))}^{\circ} \\ = α (c) {((β (c_{α})) {((γ (c_{β})))}^{\circ})}^{\circ} \\ = α (c) (β (c_{α}) (γ (c_{β}) \circ \\ = α (c) (β (c_{α})) \circ (γ (c_{α β})) \circ \\ by (6), since & ȷ c_{β} im β (c_{α}) γ (c_{β}) & = γ (im β (c_{α})) = γ (c_{α β}) . \\ Similarly, & ((α β) γ) (c) & = (α β) (c) (γ (c_{α β})) \circ \\ = α (c) (β (c_{α})) \circ (γ (c_{α β})) \circ \\ Therefore & (α (β γ)) (c) & = ((α β) γ) (c) & for all c \in 𝖛 𝒞 . \end{array}

Thus $α (β γ) = (α β) γ$ . If $γ, η \in p c n 𝒞$ then by Proposition 5.1 $γ η \in p c n 𝒞$ . Thus $p c n 𝒞$ is a semigroup. $□$

Proof. Since $𝖛 𝒞 \subseteq S e t$ is an order embedding we see that the map $ȷ : c \mapsto ȷ Δ c$ is a cone from the base $𝖛 𝒞$ to the vertex $Δ$ . We prove that $ȷ$ is universal. Suppose that $α : c \mapsto α (c)$ is a cone from the base $𝖛 𝒞$ to the vertex $X$ . For each $x \in Δ$ deﬁne

x f = x α (c) if x \in c .

Since, by Equation (21a), $Δ$ is the union of sets $c \in 𝖛 𝒞$ , $x f$ is deﬁned for all $x \in Δ$ . If $x \in U (c_{i})$ , $i = 1, 2$ , then by the axiom (S:c) of Deﬁnition 3.1 there is $d \in 𝖛 𝒞$ such that $d \subseteq c_{i}$ , $i = 1, 2$ and $x \in d$ . Since, as noted above, $𝖛 𝒞 \subseteq S e t$ is inclusion preserving and $α$ is a cone on $𝖛 𝒞$ we have

x α (c_{1}) = x α (d) = x α (c_{2}) .

This shows that $f : Δ \to X$ is a map. The deﬁnition of $f$ shows that, for all $c \in 𝖛 𝒞$ ,

f ∣ c = ȷ Δ c \circ f = ȷ (c) \circ f = α (c) .

Thus the following diagram commutes

a = b

(23)

for all $c \in 𝖛 𝒞$ . To prove the uniqueness of $f$ , suppose that $g : Δ \to X$ is another map such that

\begin{matrix} ȷ (c) \circ g = α (c) for all c \in 𝖛 𝒞 . & Then we have \\ g ∣ c = α (c) = f ∣ c \end{matrix}

for all $c \in 𝖛 𝒞$ which implies that $f = g$ . Therefore the cone $ȷ$ is universal from the base $𝖛 𝒞$ to the vertex $Δ$ .

Let $η \in p c n 𝒞$ . Since $ȷ$ is universal from $𝖛 U$ to $Δ$ there is a unique map $\tilde{η} : Δ \to U (c_{η})$ making the digram Deﬁnition 24 commute or equivalently, Equation (22) holds.

a = b

(24)

□

Proof. We shall show that $ϕ : p c n 𝒞 \to 𝒯 Δ$ is a homomorphism. Let $α, β \in p c n 𝒞$ . By Equation (20),

(α β) (c) = α (c) β (c_{α}) for c \in 𝖛 𝒞 .

Hence, for all $x \in Δ$ , we have

\begin{array}{l} x ϕ (α β) & = x \tilde{α β} ȷ (c_{α β}) \\ = x (α β) (c) ȷ (c_{α β}) \\ = x α (c) {(β (c_{α}))}^{\circ} ȷ (c_{α β}) \\ by (20) . Since 𝖛 𝒞 \subseteq S e t is an order embedding this gives \\ x ϕ (α β) & = x α (c) β (c_{α})) ȷ (c_{β}) \\ = x \tilde{α} \tilde{β} ȷ (c_{β}) = x ϕ (α) ϕ (β) . \end{array}

by Equation (25). Therefore $ϕ (α β) = ϕ (α) ϕ (β)$ .

Assume that $α, β \in p c n 𝒞$ and $ϕ (α) = ϕ (β)$ . $□$

Proof. Let $g \in 𝒞 (c, d)$ and $h \in 𝒞 (d, u)$ . Then by Equations (27a) and (27b),

\begin{array}{l} (γ * f \circ) H (γ; g h) & = γ * (f g h) \circ \\ = ((γ * f \circ) H (γ; g)) H (γ; h) & for all γ * f \circ \in H ([; c)] γ . \\ Therefore H (γ; g h) & = (H (γ; g)) (H (γ; h)) . \end{array}

If $c \subseteq d$ then it follows readily from Equation (27b) that

H (γ; ȷ d c) = ȷ H (γ; d) H (γ; c) .

In particular, $H (γ; 1_{c}) = 1_{H (γ; c)}$ . In view of the remarks following Equation (27b), this shows that $H (γ; -) : 𝒞 \to S e t$ is an inclusion preserving covariant functor.‘

To prove the remaining statements, by Yoneda lemma see [MacLane, 1971, Section III.2] and Nambooripad [1994, section I.2], it is suﬃcient to show that $γ$ is universal element for $H (γ; -)$ in $H (γ; c_{γ})$ . Thus we must show that given any $γ * f \circ \in H (γ; c)$ , there is a unique $h : c_{γ} \to c$ such that

(γ) H (γ; h) = γ * f \circ .

By Equation (27b), the above equation holds for the choice $h = f$ . To see that this choice is unique, assum that $k : c_{γ} \to c$ also satisﬁes this equation. Then, again by Equation (27b), we have $γ * f \circ = γ * k \circ$ . Since $γ$ is proper, $D \neq \emptyset$ . Choose $c \in D$ . By Equation (18), we have

γ (c) f \circ = γ (c) k \circ which implies f \circ = k \circ

since $γ (c)$ is surjective (see Equation (17)). Hence $im f = im g$ and so, $f = f \circ ȷ c im f = k \circ ȷ c im k = k$ . This completes the proof. $□$

5. The semigroup of cones