Expert: Paul Klarreich - 2/9/2007

Question
Hi,

i'm studying discrete math

can you please help me with the following questions:

1)If f : X -> Y has an inverse g, show that g is a bijection.

2)Show that if f : X ->Y is a bijection, then it has exactly one inverse.

Answer
Questioner:   Maya
Category:  Advanced Math

Subject:  Functions
Question:  Hi,

i'm studying discrete math

can you please help me with the following questions:

1)If f : X -> Y has an inverse g, show that g is a bijection.

2)Show that if f : X ->Y is a bijection, then it has exactly one inverse.
..............................................................
Hi, Maya,

1) Assuming that by writing

f : X -> Y has an inverse FUNCTION g,

you mean that:

a. the domain of f is X and
b. the domain of g is Y,
c. f is a function, i.e. the image of any x in X is unique.
d. g is the inverse function of f, i.e. the inverse image of any y in Y is unique and

f(x) = y implies g(y) = x.

To prove g is a bijection, (a 1-1 correspondence) you must show:

A. x in X implies there exists y in Y such that g(y) = x
B. y1 /= y2 implies  g(y1) /= g(y2)

For A, since the domain of f is X, for all x, there exists y such that f(x) = y, therefore, since  g is the inverse function,  g(y) = x.

For B, Let x1 = g(y1), x2 = g(y2).  Since f and g are inverses, f(x1) = y1  and f(x2)

= y2.  Suppose  g(y1) = g(y2).  Then  x1 = x2, and f(x1) = f(x2), since f is a function.  Therefore y1 = y2.

 
2) If f is a bijection, it is 1-1.  Then given y in Y, the inverse image of y is unique.  Since f is ONTO y, the inverse image of y is not empty.  So an inverse function exists.

Suppose g1 and g2 are both inverses of f.  Then, given y in Y, g1(y) = x1 and g2(y) = x2.  

But then  f(x1) = y and f(x2) = y, which implies (since f is 1-1) that x1 = x2.  

Therefore g1 and g2 are the same function.