**Isomorphism**: Let (P, Q) and (Q, R′) be two posets. A one-one onto map

*f *: P→Q is called an isomorphism if x R y ⇔*f* (x) R′* f *(x), x, y ∊P.

**Isotone**: *A map f : P→Q is called an isotone or order preserving if x≤y ⟹ f (x) ≤ f (y)*

Theorem: A mapping *f *: P→Q is an isomorphism iff is isotone and has an inverse.

**Proof:** Let *f *: P→Q be an isomorphism.

Thus being 1-1, onto, f^{-1}exists and is 1-1, onto.

Again by definition of isomorphism,* f *will be isotone. We show *f *^{-1} : Q→P is also isotone.

Let y_{1}, y_{2} ∊ Q where y_{1} ≤ y_{2}. Since f is onto, there exist x_{1}, x_{2} ∊ P, such that *f *(x_{1}) = y_{1}, *f* (x_{2}) = y_{2} ⇔ x_{1} = *f *^{–}^{1}(y_{1}), *f** ^{– }*

^{1}(y

_{2})

Now y_{1} ≤ y_{2} ⟹ *f *(x_{1}) ≤* f *(x_{2})

⟹ x_{1} ≤ x_{2}

⟹ *f ^{-1}*(y

_{1}) ≤

*f*

^{-1}

⟹* f *^{-1 }is isotone.

Conversely, let *f* be isotone such that *f *^{-1} is also isotone. Since *f*^{ -1} exists, *f* is one-one, onto.

Again, as *f* is isotone x_{1} ≤ x_{2} ⟹ f(x_{1}) ≤ f(x_{2}), x_{1}, x_{2} ∊ P

Also* f* ^{-1} is isotone implies

*f *(x_{1}) ≤ *f *(x_{2}) ⟹ *f*^{ -1}(*f *(x_{1})) ≤ *f* ^{-1}( *f*(x_{1}))

⟹ x_{1} ≤ x_{2}

Thus x_{1} ≤ x_{2} ⇔ *f *(x_{1}) ≤ *f *(x_{2})

*Hence f is an isomorphism.* **(Proved)**