Iterates under Order Semiconjugacy¶
transform-stachurski #no-port-cons¶
Setup¶
Definition (Order Semiconjugacy). Let \((V, S)\) and \((\hat V, \hat S)\) be dynamical systems, where \(V\) and \(\hat V\) are posets. We call \((V, S)\) and \((\hat V, \hat S)\) order semiconjugate under \(F, G\) when there exist order-preserving maps \(F \colon V \to \hat V\) and \(G \colon \hat V \to V\) such that
\[ S = G \circ F \text{ on } V \qquad \text{and} \qquad \hat S = F \circ G \text{ on } \hat V. \tag{1} \]
When (1) holds, the following commuting identities are immediate:
Iterates Lemma¶
The commuting identities (2) extend naturally to iterates.
Lemma (Iterates under Semiconjugacy). Let \((V, S)\) and \((\hat V, \hat S)\) be order semiconjugate under \(F, G\). Then, for all \(n \in \mathbb{N}\),
\[ S^n = G \circ \hat S^{\,n-1} \circ F \qquad \text{and} \qquad \hat S^n = F \circ S^{\,n-1} \circ G, \tag{3} \]where we adopt the convention \(S^0 = I_V\) and \(\hat S^0 = I_{\hat V}\). Moreover, for all \(n \in \mathbb{N}\),
\[ F \circ S^n = \hat S^n \circ F \qquad \text{and} \qquad G \circ \hat S^n = S^n \circ G. \tag{4} \]
Proof. We prove the first identity in (3) by induction; the second follows by symmetry.
For \(n = 1\), we have \(S^1 = S = G \circ F = G \circ I_{\hat V} \circ F = G \circ \hat S^0 \circ F\), as required.
Now suppose \(S^n = G \circ \hat S^{n-1} \circ F\) for some \(n \geq 1\). Then
where the key step uses \(F \circ G = \hat S\) from (1).
The identities (4) follow from (3): \(F \circ S^n = F \circ G \circ \hat S^{n-1} \circ F = \hat S \circ \hat S^{n-1} \circ F = \hat S^n \circ F\), and similarly for the second identity. \(\blacksquare\)