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

• Nov 9th 2008, 06:02 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, 06:17 PM
kathrynmath
I know g doesn't have to be injective.
• Nov 9th 2008, 06:20 PM
kathrynmath
I'm think that it does but I am unsure of how to prove this.
• Nov 9th 2008, 06:36 PM
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?
• Nov 9th 2008, 06:56 PM
Jhevon
Quote:

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\$
• Nov 9th 2008, 06:59 PM
kathrynmath
Quote:

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.
• Nov 9th 2008, 07:01 PM
Jhevon
Quote:

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 ...
• Nov 9th 2008, 07:05 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 \$\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)\$.
• Nov 9th 2008, 07:08 PM
Jhevon
Quote:

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
• Nov 9th 2008, 07:10 PM
kathrynmath
Quote:

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?
• Nov 9th 2008, 07:17 PM
Jhevon
Quote:

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
• Nov 9th 2008, 07:23 PM
kathrynmath
Quote:

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.