Problem 19: Solution
First we recall the definition of inverse function:
Definition: If F : X → Y is a bijective function, then the function F −1 : Y →
X is defined as follows: for every ȳ ∈ Y , F −1 (ȳ) = x̄ ∈ X such that F (x̄) = ȳ.
Then we recall a useful theorem about inverse functions:
Theorem: If F : X → Y and G : Y → X are functions such that F ◦ G = IdY
and G ◦ F = IdX , then G = F −1 .
Now we can state the problem we want to solve:
Problem 19: If f : X → Y and g : Y → Z are bijective functions, then prove
that:
(g ◦ f )−1 = f −1 ◦ g −1 .
We show two different solutions of this problem. Either one is good.
Solution 1: using the definitions. To prove (g ◦ f )−1 = f −1 ◦ g −1 , we must
show that for any element z̄ ∈ Z,
(g ◦ f )−1 (z̄) = f −1 ◦ g −1 (z̄).
These are two elements of X that we wish to be equal. We use the definition to
describe each of them.
(g ◦ f )−1 (z̄) = x̄ ∈ X such that (g ◦ f )(x̄) = z̄.
¯ ∈ X such that f (x̄
¯) = ȳ.
f −1 ◦ g −1 (z̄) = f −1 (ȳ ∈ Y such that g(ȳ) = z̄) =x̄
¯ = x̄ we must show
Since the first of these two lines defines x̄, to show that x̄
¯
that (g ◦ f )(x̄) = z̄.
But
¯) = g(f (x̄
¯)) = g(ȳ) = z̄.
(g ◦ f )(x̄
¯ = x̄, we showed the two functions (g ◦ f )−1 and f −1 ◦ g −1
Having proved that x̄
assign the same output to every input, and therefore they are the same function.
Solution 2: using the theorem. If we assume the theorem, in order to show
that (g ◦ f )−1 = f −1 ◦ g −1 it suffices to show that
(g ◦ f ) ◦ f −1 ◦ g −1 = IdZ
and
f −1 ◦ g −1 ◦ (g ◦ f ) = IdX .
For any z ∈ Z
(g ◦ f ) ◦ f −1 ◦ g −1 (z) = g(f (f −1 (g −1 (z)))) = g(g −1 (z)) = z,
therefore the first equality is verified.
For any x ∈ X,
f −1 ◦ g −1 ◦ (g ◦ f )(x) = f −1 (g −1 (g(f (x)))) = f −1 (f (x)) = x,
therefore the first equality is verified and the proof concluded.