Products and coproducts in the cyclic category

Question

Does the category $\Lambda$ of finite cyclically ordered sets and monotone functions between sets have binary products and coproducts?

Answer 1

Big edit : we use betweenness relation

The category $\Lambda$
An object of $\Lambda$ is a cyclically ordered set, that is to say a set $A$ together with a ternary relation $R$, the betweenness relation, that satisfies the following properties :

• Cyclicity :For any $a, b, c \in A$, $R(a,b,c) \Longrightarrow R(b, c, a)$
• Asymetry : For any $a, b, c \in A$, $R(a,b,c) \Longrightarrow \neg R(c, b, a)$
• Transitivity : For any $a, b, c, d \in A$, if $R(a, b, c)$ and $R(a, c, d)$ then $R(a, b, d)$
• Totality : For any distinct $a, b, c \in A$, either $R(a, b, c)$ or $R(c, b, a)$ holds.

An arrow between two objects $A, B$ of $\Lambda$ is a map $f$ that preserves the betweenness : for all $a, b, c \in A, \ R(a, b, c) \Longrightarrow R\big(f(a), f(b), f(c)\big)$.

Let $A=\{a_1, a_2, a_3\}$ and $B=\{b_1, b_2, b_3\}$ be cyclically ordered sets where $R(a_1, a_2, a_3)$ and $R(b_1, b_2, b_3)$. Then there is no product of $A$ and $B$ in $\Lambda$ :

Assume otherwise (for contradiction), and suppose we have a cyclically ordered set $X$ and monotone maps $\pi_A : X \mapsto A$ and $\pi_B : X \mapsto B$ satisfying the following universal property :

universal property for product
For any cyclically ordered set $Y$ and monotone functions $f : Y \mapsto A$ and $g : Y \mapsto B$ there is a unique function $h : Y \mapsto X$ such that $\pi_A \circ h = f$ and $\pi_B \circ h = g$.

Let $Y = \{1, 2, 3\} \in \Lambda$, where $R(1, 2, 3)$.

We consider two pairs of arrows, $(f_A, f_B)$ and $(g_A, g_B)$ :

• $f_A : i \mapsto a_{i+1\textrm{ mod } 3}$ and $f_B : i \mapsto b_i$. Call $F$ the function $Y \mapsto X$ given by the universal property. We have $\pi_A(F(1)) = a_2$ and $\pi_B(F(1)) = b_1$.
• $g_A : i \mapsto a_i$ and $g_B : i \mapsto b_{i+1\textrm{ mod } 3}$. Call $G$ the function $Y \mapsto X$ given by the universal property. We have $\pi_A(G(1)) = a_1$ and $\pi_B(G(1)) = b_2$.

The elements $G(1), F(1), F(3)$ are pairwise distinct. Hence, by totality we have either $R\big(G(1), F(1), F(3)\big)$ or $R\big(F(3), F(1), G(1)\big)$.
Suppose $R\big(G(1), F(1), F(3)\big)$, then we must have $R\big(\pi_B(G(1)), \pi_B(F(1)), \pi_B(F(3))\big)$ i.e. $R\big(b_2, b_1, b_3\big)$ which is absurd.
Suppose $R\big(F(3), F(1), G(1)\big)$, then we must have $R\big(\pi_A(F(3)), \pi_A(F(1)), \pi_A(G(1))\big)$ ie $R\big(a_1, a_2, a_1\big)$ which is absurd.

We have derived the desired contradiction, hence there is no product $A \times B$ in $\Lambda$.