If g(f(x) is injective I need to determine if f is injective. If it is false, I need to give a counterexample.

I was told that using arrow diagrams helps.

Can some one help me get started?

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- Nov 9th 2008, 06:02 PMkathrynmathg(f(x) is injective, is f injective?
If g(f(x) is injective I need to determine if f is injective. If it is false, I need to give a counterexample.

I was told that using arrow diagrams helps.

Can some one help me get started? - Nov 9th 2008, 06:17 PMkathrynmath
I know g doesn't have to be injective.

- Nov 9th 2008, 06:20 PMkathrynmath
I'm think that it does but I am unsure of how to prove this.

- Nov 9th 2008, 06:36 PMkathrynmath
So g(f(x)) is 1-1, so g(f($\displaystyle a_1)$=g(f($\displaystyle a_2)$. Then this implies f($\displaystyle a_1)$=f($\displaystyle a_2)$.

Where do I go from here? Would that last statement prove that $\displaystyle a_1=a_2$? therefore, f is one to one? - Nov 9th 2008, 06:56 PMJhevon
- Nov 9th 2008, 06:59 PMkathrynmath
- Nov 9th 2008, 07:01 PMJhevon
- Nov 9th 2008, 07:05 PMkathrynmath
- Nov 9th 2008, 07:08 PMJhevon
- Nov 9th 2008, 07:10 PMkathrynmath
- Nov 9th 2008, 07:17 PMJhevon
assume $\displaystyle f(x_1) = f(x_2)$. then $\displaystyle g(f(x_1)) = g(f(x_2))$. but that means $\displaystyle (g \circ f)(x_1) = (g \circ f)(x_2)$. since $\displaystyle g \circ f$ is injective, this means $\displaystyle x_1 = x_2$. thus, $\displaystyle f$ is injective

- Nov 9th 2008, 07:23 PMkathrynmath