# 1-1 function

• Sep 17th 2006, 12:20 PM
1-1 function
Hi,

I have this question:
if f and f o g are 1 to 1 functions, does it follow that g is 1-1 function?

I tried for example ,

if g(3)=4 and g(7)=4,
then f( g(3) )= f( G(7) )=f(4) will produce the 1-1 function property.( it does not work)

I have tried other things but still , it doesn't work can someone help me?

B
• Sep 17th 2006, 12:28 PM
ThePerfectHacker
Let,
f:Y--> Z be one-to-one
g:X--> Y be a function
f o g: X-->Z be one-to-one
Let,
h = f o g
Then,
we need to show that if g(x1)=g(x2) then x1=x2
We know that if,
h(x1)=h(x2) then x1=x2 thus,
f(g(x1))=f(g(x2))
then,
g(x1)=g(x2)
I cannot see how you can conclude that g is one-to-one from these statements.
I did not think of a counterexample but I would guess no.
• Sep 17th 2006, 12:44 PM
Plato
Actually g has to be one-to-one.
Suppose that g(a)=g(b).
The by function definition f(g(a))=f(g(b)) or fog(a)=fog(b).
But we are given that fog is one-to-one, therefore a=b.
If g(a)=g(b) then a=b; g is one-to-one.
• Sep 17th 2006, 02:12 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
Actually g has to be one-to-one.
Suppose that g(a)=g(b).
The by function definition f(g(a))=f(g(b)) or fog(a)=fog(b).
But we are given that fog is one-to-one, therefore a=b.
If g(a)=g(b) then a=b; g is one-to-one.

Good explanation.