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

• Nov 9th 2008, 07:02 PM
kathrynmath
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?
• Nov 9th 2008, 07:17 PM
kathrynmath
I know g doesn't have to be injective.
• Nov 9th 2008, 07:20 PM
kathrynmath
I'm think that it does but I am unsure of how to prove this.
• Nov 9th 2008, 07:36 PM
kathrynmath
So g(f(x)) is 1-1, so g(f( $a_1)$=g(f( $a_2)$. Then this implies f( $a_1)$=f( $a_2)$.

Where do I go from here? Would that last statement prove that $a_1=a_2$? therefore, f is one to one?
• Nov 9th 2008, 07:56 PM
Jhevon
Quote:

Originally Posted by kathrynmath
So g(f(x)) is 1-1, so g(f( $a_1)$=g(f( $a_2)$. Then this implies f( $a_1)$=f( $a_2)$.

Where do I go from here? Would that last statement prove that $a_1=a_2$? therefore, f is one to one?

hint: $g(f(x))$ is another way of writing $(g \circ f)(x)$

(the statement is true) now you need to show that if $f(x_1) = f(x_2)$ then $x_1 = x_2$
• Nov 9th 2008, 07:59 PM
kathrynmath
Quote:

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

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

Ok, so, I'm unsure on how to prove that $f(x_1)=f(x_2)$ that $x_1=x_2$. This is where I get stuck.
• Nov 9th 2008, 08:01 PM
Jhevon
Quote:

Originally Posted by kathrynmath
Ok, so, I'm unsure on how to prove that $f(x_1)=f(x_2)$ that $x_1=x_2$. This is where I get stuck.

yes, and i told you how to get around it. follow my hint

to start you off: assume $f(x_1) = f(x_2)$. then $g(f(x_1)) = g(f(x_2))$. but that means ...
• Nov 9th 2008, 08:05 PM
kathrynmath
Quote:

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

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

Ok, so I start off by assumming f( $x_1$)=f( $x_2$)? Then $g(f(x_1)=g(f(x_2)$. g(f) is injective, so f( $x_1)=f(x_2)$.
• Nov 9th 2008, 08:08 PM
Jhevon
Quote:

Originally Posted by kathrynmath
Ok, so I start off by assumming f( $x_1$)=f( $x_2$)? Then $g(f(x_1)=g(f(x_2)$. g(f) is injective, so f( $x_1)=f(x_2)$.

...that brings us back where we started. we know $f(x_1) = f(x_2)$, that's what we assumed! you are supposed to apply my hint where i left off
• Nov 9th 2008, 08:10 PM
kathrynmath
Quote:

Originally Posted by Jhevon
...that brings us back where we started. we know $f(x_1) = f(x_2)$, that's what we assumed! you are to apply my hint there...

Ok, can I just say $x_1=x_2$ because $g(f(x_1))=g(f(x_2))$ and $g(f(x))$ is injective?
• Nov 9th 2008, 08:17 PM
Jhevon
Quote:

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

assume $f(x_1) = f(x_2)$. then $g(f(x_1)) = g(f(x_2))$. but that means $(g \circ f)(x_1) = (g \circ f)(x_2)$. since $g \circ f$ is injective, this means $x_1 = x_2$. thus, $f$ is injective
• Nov 9th 2008, 08:23 PM
kathrynmath
Quote:

Originally Posted by Jhevon
assume $f(x_1) = f(x_2)$. then $g(f(x_1)) = g(f(x_2))$. but that means $(g \circ f)(x_1) = (g \circ f)(x_2)$. since $g \circ f$ is injective, this means $x_1 = x_2$. thus, $f$ is injective

Ok, that makes sense.