Composite of Order Isomorphisms is Order Isomorphism

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Theorem

Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be ordered sets.

Let:

$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$

and:

$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$

be order isomorphisms.


Then $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is also an order isomorphism.


Proof

From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection, as, by definition, an order isomorphism is also a bijection.

From Inverse of Composite Bijection, the inverse of $\psi \circ \phi$ is given by:

$\paren {\psi \circ \phi}^{-1} = \phi^{-1} \circ \psi^{-1}$


By definition of composition of mappings:

$\map {\psi \circ \phi} x = \map \psi {\map \phi x}$

By definition of order isomorphism, we have:

$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$ is an increasing mapping

and:

$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$ is an increasing mapping.

Hence from Composite of Increasing Mappings is Increasing:

$\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an increasing mapping.


Similarly by definition of order isomorphism:

$\phi^{-1}: \struct {S_2, \preceq_2} \to \struct {S_1, \preceq_1}$ is an increasing mapping

and:

$\psi^{-1}: \struct {S_3, \preceq_3} \to \struct {S_2, \preceq_2}$ is an increasing mapping.

Hence from Composite of Increasing Mappings is Increasing:

$\phi^{-1} \circ \psi^{-1}: \struct {S_3, \preceq_3} \to \struct {S_1, \preceq_1}$ is an increasing mapping.


Hence we have that:

$\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an increasing mapping

and:

$\paren {\psi \circ \phi}^{-1}: \struct {S_3, \preceq_3} \to \struct {S_1, \preceq_1}$ is an increasing mapping

and it follows by definition that $\psi \circ \phi$ is an order isomorphism.

$\blacksquare$


Sources

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