In summary, if both α and β are one-to-one mappings, then the inverse of their composite function is the composition of their inverses, i.e., (αβ)⁻¹ = β⁻¹ ∘ α⁻¹.
Proof:
One-to-one mappings: Define α : S → T and β : T → U as one-to-one mappings. This means that for any distinct elements s1, s2 ∈ S, α(s1) ≠ α(s2), and for any distinct elements t1, t2 ∈ T, β(t1) ≠ β(t2).
Inverse mappings: Since α and β are one-to-one, they have unique inverse mappings α⁻¹ : T → S and β⁻¹ : U → T, respectively. These mappings satisfy the following conditions:
α⁻¹(α(s)) = s for all s ∈ S,
β⁻¹(β(t)) = t for all t ∈ T.
Composite function: Consider the composite function αβ : S → U, defined as:
(αβ)(s) = β(α(s)) for all s ∈ S.
Inverse of the composite function: We need to show that (αβ)⁻¹ = β⁻¹ ∘ α⁻¹. To achieve this, we will prove that for any element u ∈ U,
(β⁻¹ ∘ α⁻¹)(u) = (αβ)⁻¹(u).
Let u ∈ U: Since β is onto, there exists a unique element t ∈ T such that β(t) = u.
Substitute u with β(t):
(β⁻¹ ∘ α⁻¹)(u) = (β⁻¹ ∘ α⁻¹)(β(t)).
Apply the properties of inverse mappings:
β⁻¹(β(t)) = t,
α⁻¹(t) = s, where s is the unique element in S such that α(s) = t.
Chain the equations:
β⁻¹ ∘ α⁻¹(β(t)) = β⁻¹(t) = α⁻¹(t) = s.
Apply the definition of composite function:
α(α⁻¹(t)) = t = α(s).
Since α is one-to-one:
α⁻¹(α(s)) = s.
Combine the results:
β⁻¹(α(s)) = s.
This implies:
(αβ)⁻¹(β(t)) = (αβ)⁻¹(u) = s.
Therefore:
(αβ)⁻¹(u) = (β⁻¹ ∘ α⁻¹)(u) for all u ∈ U.
Conclusion:
Since the equality holds for all elements in U, we can conclude that (αβ)⁻¹ = β⁻¹ ∘ α⁻¹.
In summary, if both α and β are one-to-one mappings, then the inverse of their composite function is the composition of their inverses, i.e., (αβ)⁻¹ = β⁻¹ ∘ α⁻¹.