Let $x \to y$ be an invertible function. Show that it has unique inverse.

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Solution

Let $f : x \to y$ be an invertible function.
Also, suppose f has two inverses (say $g_{1}$ and $g_{2}$ )
Then, for all $y ϵ Y$ , we have:
$f o g_{1} (y) = I Y (y) = f o g_{2} (y)$
$\Rightarrow f (g_{1} (y)) = f (g_{2} (y))$
$\Rightarrow g_{1} (y) = g_{2} (y)$ [f is invertible $\Rightarrow f$ is one - one]
$\Rightarrow g_{1} = g_{2}$ [g is one - one]
Hence, f has a unique inverse.

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Similar questions

Let f: X → Y be an invertible function. Show that f has unique inverse.

(Hint: suppose g₁ and g₂ are two inverses of f. Then for all y ∈ Y,

fog₁(y) = I_Y(y) = fog₂(y). Use one-one ness of f).

f : X \to Y

f^{- 1}

f

(f^{- 1})^{- 1} = f

f : N \to Y

f (x) = 4 x + 3

Y = {y \in N : y = 4 x + 3, x \in N}

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