Why Is Pudlak's relation on the family of one- or two-element subsets of a set transitive?

Question

The following comes from Definition 2 in Pavel Pudlak, "A new proof of the congruence lattice representation theorem," Algebra Universalis 6 (1976), 269-275.

Let $X$ be a set. Let $F$ be a family of functions from $X$ to itself containing the identity map and closed under composition.

Define a binary relation on the family of one- or two-element subsets of $X$ as follows. Let $a,b,c,d\in X$. We will say that $\{a,b\}$ dominates $\{c,d\}$ if there are $n\in\mathbb N_0$, $u_0,\dots,u_n\in X$, and $f_1,\dots,f_n\in F$ such that $u_0=c$, $u_n=d$, and $\{f_i(a),f_i(b)\}=\{u_{i-1},u_i\}$ for $i=1,\dots,n$.

Why is domination transitive?

Answer 1

I don’t see what the problem is supposed to be; you just compose the functions witnessing the two domination relations in the obvious way:

Fix $c=u_0,\dots,u_n=d$ and $f_1,\dots,f_n\in F$ such that $\{f_i(a),f_i(b)\}=\{u_{i-1},u_i\}$ for each $i=1,\dots,n$.

Fix $e=v_0,\dots,v_m=f$ and $g_1,\dots,g_m\in F$ such that $\{g_j(c),g_j(d)\}=\{v_{j-1},v_j\}$ for each $j=1,\dots,m$.

Then the fact that $\{a,b\}$ dominates $\{e,f\}$ is witnessed by the sequence $$e=v_0=w_{1,0},w_{1,1},\dots,w_{1,n}=v_2=w_{2,0},w_{2,1},\dots,w_{m,n}=v_m=f,$$ where $$w_{j,i}=\begin{cases}g_j(u_i)&\text{if }g_j(c)=v_{j-1}\text{ and }g_j(d)=v_j,\\ g_j(u_{n-i})&\text{otherwise.}\end{cases}$$ We have $\{w_{j,i-1},w_{j,i}\}=\{h_{j,i}(a),h_{j,i}(b)\}$, where $$h_{j,i}=\begin{cases}g_j\circ f_i&\text{if }g_j(c)=v_{j-1}\text{ and }g_j(d)=v_j,\\ g_j\circ f_{n+1-i}&\text{otherwise}\end{cases}$$ is in $F$ as it is closed under composition.