# Thread: g(f(x) is injective, is f injective?

1. ## g(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?

2. I know g doesn't have to be injective.

3. I'm think that it does but I am unsure of how to prove this.

4. 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?

5. Originally Posted by kathrynmath
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?
hint: $\displaystyle g(f(x))$ is another way of writing $\displaystyle (g \circ f)(x)$

(the statement is true) now you need to show that if $\displaystyle f(x_1) = f(x_2)$ then $\displaystyle x_1 = x_2$

6. Originally Posted by Jhevon
hint: $\displaystyle g(f(x))$ is another way of writing $\displaystyle (g \circ f)(x)$

(the statement is true) now you need to show that if $\displaystyle f(x_1) = f(x_2)$ then $\displaystyle x_1 = x_2$
Ok, so, I'm unsure on how to prove that $\displaystyle f(x_1)=f(x_2)$ that $\displaystyle x_1=x_2$. This is where I get stuck.

7. Originally Posted by kathrynmath
Ok, so, I'm unsure on how to prove that $\displaystyle f(x_1)=f(x_2)$ that $\displaystyle x_1=x_2$. This is where I get stuck.
yes, and i told you how to get around it. follow my hint

to start you off: assume $\displaystyle f(x_1) = f(x_2)$. then $\displaystyle g(f(x_1)) = g(f(x_2))$. but that means ...

8. Originally Posted by Jhevon
yes, and i told you how to get around it. follow my hint

to start you off: assume $\displaystyle f(x_1) = f(x_2)$. then $\displaystyle g(f(x_1)) = g(f(x_2))$. but that means ...
Ok, so I start off by assumming f($\displaystyle x_1$)=f($\displaystyle x_2$)? Then $\displaystyle g(f(x_1)=g(f(x_2)$. g(f) is injective, so f($\displaystyle x_1)=f(x_2)$.

9. Originally Posted by kathrynmath
Ok, so I start off by assumming f($\displaystyle x_1$)=f($\displaystyle x_2$)? Then $\displaystyle g(f(x_1)=g(f(x_2)$. g(f) is injective, so f($\displaystyle x_1)=f(x_2)$.
...that brings us back where we started. we know $\displaystyle f(x_1) = f(x_2)$, that's what we assumed! you are supposed to apply my hint where i left off

10. Originally Posted by Jhevon
...that brings us back where we started. we know $\displaystyle f(x_1) = f(x_2)$, that's what we assumed! you are to apply my hint there...
Ok, can I just say $\displaystyle x_1=x_2$ because $\displaystyle g(f(x_1))=g(f(x_2))$ and $\displaystyle g(f(x))$ is injective?

11. Originally Posted by kathrynmath
Ok, can I just say $\displaystyle x_1=x_2$ because $\displaystyle g(f(x_1))=g(f(x_2))$ and $\displaystyle g(f(x))$ is injective?
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

12. Originally Posted by Jhevon
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
Ok, that makes sense.