Problem involving Indentity, Coposition functoin

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April 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
April 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 $a, b \in X$ such that a is not equal to b and $f(a) = f(b) = c \in Y$ . By definition, $g(c) = a$ and $g(c) = b$ . But in order for g to be a function, $a = b$ , which is a contradiction.
April 29th 2008, 06:50 PM
892king

Quote:

Originally Posted by icemanfan

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

Thank you for your help.
Can I prvoe g's being surjective by same way?
Like,

Suppose g is not surjective. Then there exists $a \in Y$ 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).

892King