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Prove if (\alpha) is a one-to-one mapping of a set (s) onto a set (t) and (\beta) is a one-to-one mapping of (t) onto a set (u), then ((\alpha\beta)^{-1} = \beta^{-1} \cdot \alpha^{-1}).

User Aduchate
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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.

User Gaetanm
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