Problem involving Indentity, Coposition functoin

• Apr 29th 2008, 11:49 AM
892king
Problem involving Indentity, Coposition functoin
Hey, I don't know how to prove this.
So... help me (Bow)

Let f: X -> Y and g: Y -> X be functions so that g o f = 1x(which means Identity function). Prove that f is injective and g is surjective. Need either be bijective?

On hints sheet, This has 3 parts. Justify your answer to last part w/ a proof or counterexample.

Thank you
• Apr 29th 2008, 12:13 PM
icemanfan
Quote:

Originally Posted by 892king
Hey, I don't know how to prove this.
So... help me (Bow)

Let f: X -> Y and g: Y -> be functions so that g o f = 1x(which means Identity function). Prove that f is injective and g is surjective. Need either be bijective?

On hints sheet, This has 3 parts. Justify your answer to last part w/ a proof or counterexample.

Thank you

For Part 1: Suppose f is not injective. Then there exist $\displaystyle a, b \in X$ such that a is not equal to b and $\displaystyle f(a) = f(b) = c \in Y$. By definition, $\displaystyle g(c) = a$ and $\displaystyle g(c) = b$. But in order for g to be a function, $\displaystyle a = b$, which is a contradiction.
• Apr 29th 2008, 06:50 PM
892king
Quote:

Originally Posted by icemanfan
For Part 1: Suppose f is not injective. Then there exist $\displaystyle a, b \in X$ such that a is not equal to b and $\displaystyle f(a) = f(b) = c \in Y$. By definition, $\displaystyle g(c) = a$ and $\displaystyle g(c) = b$. But in order for g to be a function, $\displaystyle a = b$, which is a contradiction.

Suppose g is not surjective. Then there exists $\displaystyle a \in Y$ such that a is not contained in the image of set Y. By definition, g(a) doesn't exist. But in order for g to be a function, g(a) should be defined(not sure about this part).