Results 1 to 5 of 5

Thread: If gof is injective, then f is injective

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    5

    If gof is injective, then f is injective

    Let $\displaystyle f:A \rightarrow B$ and $\displaystyle g: B \rightarrow C$ be functions.

    Show that if $\displaystyle g \circ f$ is injective, then $\displaystyle f$ is injective.

    Here is what I did.

    $\displaystyle Proof$. Spse. $\displaystyle g \circ f$ is injective and $\displaystyle f$ is not injective. Then $\displaystyle \exists x_1,x_2 \in A \ni f(x_1)=f(x_2)$ but $\displaystyle x_1 \neq x_2$.
    But then $\displaystyle g \circ f(x_1)=g \circ f(x_2)$ would imply $\displaystyle x_1 \neq x_2$ thus $\displaystyle g \circ f$ is not injective. A contradiction.
    Therefore if $\displaystyle g \circ f$ is injective, then $\displaystyle f$ is injective.

    Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

    Thanks

    dubito
    Last edited by dubito; Oct 17th 2009 at 10:49 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by dubito View Post
    Let $\displaystyle f:A \rightarrow B$ and $\displaystyle g: B \rightarrow C$ be functions.

    Show that if $\displaystyle g \circ f$ is injective, then $\displaystyle f$ is injective.

    Here is what I did.

    $\displaystyle Proof$. Spse. $\displaystyle g \circ f$ and $\displaystyle f$ is not injective. Then $\displaystyle \exists x_1,x_2 \in f \ni f(x_1)=f(x_2)$ but $\displaystyle x_1 \neq x_2$.
    But then $\displaystyle x_1,x_2 \in g \circ f$ would imply $\displaystyle x_1 \neq x_2$ thus $\displaystyle g \circ f$ is not injective. A contradiction.
    Therefore if $\displaystyle g \circ f$ is injective, then $\displaystyle f$ is injective.

    Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

    Thanks

    dubito
    Seems you have confused yourself.
    $\displaystyle
    \exists x_1,x_2 \in f \ni f(x_1)=f(x_2)
    $



    etc make no sense to me.

    gof is not a 'set', so how can you say x1,x2 belong to gof? What does it mean??
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by dubito View Post
    Let $\displaystyle f:A \rightarrow B$ and $\displaystyle g: B \rightarrow C$ be functions.

    Show that if $\displaystyle g \circ f$ is injective, then $\displaystyle f$ is injective.

    Here is what I did.

    $\displaystyle Proof$. Spse. $\displaystyle g \circ f$ and $\displaystyle f$ is not injective. Then $\displaystyle \exists x_1,x_2 \in f \ni f(x_1)=f(x_2)$ but $\displaystyle x_1 \neq x_2$.
    But then $\displaystyle x_1,x_2 \in g \circ f$ would imply $\displaystyle x_1 \neq x_2$ thus $\displaystyle g \circ f$ is not injective. A contradiction.
    Therefore if $\displaystyle g \circ f$ is injective, then $\displaystyle f$ is injective.

    Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

    Thanks

    dubito
    Assume that $\displaystyle f$ is not injective; that is, $\displaystyle \exists~x\neq y$ such that $\displaystyle f(x)=f(y)$. Then $\displaystyle g(f(x))=g(f(y))$, but since $\displaystyle x\neq y$, this implies that $\displaystyle g\circ f$ is not injective either. $\displaystyle \Rightarrow\Leftarrow$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2009
    Posts
    5
    Thanks guys! I reread my proof and I am like . @ aman_cc, I meant $\displaystyle x_1,x_2 \in A$ sry about that (if I remember correctly though I think my teacher used to do the same thing). @ redsoxfan325 .

    dubito
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by dubito View Post
    Thanks guys! I reread my proof and I am like . @ aman_cc, I meant $\displaystyle x_1,x_2 \in A$ sry about that (if I remember correctly though I think my teacher used to do the same thing). @ redsoxfan325 .

    dubito
    Another quick way

    If f(x1)= f(x2) =>
    gof(x1) = g(f(x1))=g(f(x2)) = gof(x2) =>
    x1 = x2 (as gof is injective)

    Hence done
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. f: [0,1]-->]0,1[, f injective
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Oct 1st 2010, 09:08 AM
  2. Injective
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 1st 2010, 03:15 AM
  3. B & AB injective but A not injective
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 24th 2009, 10:33 PM
  4. g(f(x) is injective, is f injective?
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: Nov 9th 2008, 07:23 PM
  5. Injective
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 23rd 2006, 11:09 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum