Composite and one to one function

[brad sue]

Hi,

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

My answer is NO g doesn't have to be 1-1 function.

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.

Is my reasonning OK??

B

 

[Data]

The best way to look at "true/false" questions like this is often to try to prove whatever it's asking, and if you can't, then see if you can build a counterexample by paying attention to what "breaks" your proof.

Your counterexample doesn't work, since then you'll have, as you wrote, $(f\circ g)(3) = (f\circ g)(7)$, which obviously means $f \circ g$ is not 1-1.

So try to see if you can prove it. I'd suggest doing it by contradiction.

(And just to remind you, a function $h$ is 1-1 iff h(x) = g(y) implies x=y).

 

[brad sue]

The best way to look at "true/false" questions like this is often to try to prove whatever it's asking, and if you can't, then see if you can build a counterexample by paying attention to what "breaks" your proof.

Your counterexample doesn't work, since then you'll have, as you wrote, $(f\circ g)(3) = (f\circ g)(7)$, which obviously means $f \circ g$ is not 1-1.

So try to see if you can prove it. I'd suggest doing it by contradiction.

(And just to remind you, a function $h$ is 1-1 iff h(x) = g(y) implies x=y).

Ok thank I found the problem