Prove or give a counterexample Prove or give a counterexampl

Prove or give a counterexample...

Prove or give a counterexample: if f: X rightarrow Y and g: Y rightarrow X are functions such that g f = I_X and f g = I_Y, then f and g are both one-to-one and onto and g = f^-1.

Solution

If f: X Y and g: Y X are functions such that g_of = I_X and f_og = I_Y, then f and g are both one-one and onto and g = f^-1.

Definition of Invertible Mapping:

A function f: X Y is said to be invertible if there exists a mapping g: Y X such that g_of = I_X and f_og = I_Y. In this case g is said to be an inverse of f.

Proof: For the mappings f: X Y and g: Y X both the composites g_of: X X and f_og: Y Y are defined.

If possible let f be not one-one. Then there exist two distinct elements a, b X such that f(a) = f(b)

Therefore, g(f(a)) = g(f(b))

Implies, g_of(a) = g_of(b)

Implies, a = b (because g_of = I_X)

This contradicts that a b

Therefore, f is one-one

In the similar way we can prove that g is also one-one using the fact that f_og = I_Y

Now we will prove that g is onto.

Let x be an element of X. Therefore, g_of(x) = x since g_of = I_X

g_of(x) = g(f(x))

This implies that g(f(x)) = x

This shows that x has a pre-image f(x) in Y under the mapping g. Since, x is arbitrary, g is onto.

In the similar way we can prove that f is also onto using the fact that f_og = I_Y

Hence, we have proved that f and g are both one-one and onto.

Now, we will prove that f is invertible and that g = f^-1.

Proof: f: X Y be a bijection (i.e. one-one and onto)

Let y Y. Since f is a bijection, y has one and only one pre-image x in X.

Define a map g: Y X by g(y) = x (the pre-image of y under f), y Y

Therefore, g_of(x) = g(f(x)) = g(y) = x, x X and f_og(y) = f(g(y) = f(x) = y, y Y

This implies, g_of = I_X and f_og = I_Y, and therefore, f is invertible and that g = f^-1 (Proved)