Domain and range can't be negative

1. Jul 31, 2010

EV33

1. The problem statement, all variables and given/known data
Given f:A→B and g:B→C, let h=g(f(a))

If h is injective, then g is injective.

Give a counter example.

2. Relevant equations
Injection: Let f:A→B

For all f(x1)=f(x2) implies x1=x2

3. The attempt at a solution

f(a)=$$\sqrt{a}$$ from [0, infinity]→[0,infinity]

g(b)=b2 from [all real numbers]→[0,infinity]

h(a)=a from [0, infinity]→[0,infinity]

Assuming h is injective and g is not, then this is a counter example. My problem is I am not sure if the square root messes things up here. I know that the square root of a squared is plus or minus a, but because the domain and range can't be negative ( I think), then this works.

So is this correct?

2. Jul 31, 2010

Coto

Re: Injection

I believe you are correct.