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 December 6th, 2007, 12:15 PM #1 Newbie   Joined: Dec 2007 Posts: 4 Thanks: 0 compostion and injectivity I need help with the following problem... If the composition function g compose f : A-> C is an injection, then the function f : A->B is an injection.
December 6th, 2007, 01:15 PM   #2
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Re: compostion and injectivity

Quote:
 Originally Posted by jblaine271 I need help with the following problem... If the composition function g compose f : A-> C is an injection, then the function f : A->B is an injection.
Have a look here
or here.

 December 6th, 2007, 01:37 PM #3 Newbie   Joined: Dec 2007 Posts: 4 Thanks: 0 Great! Thanks a lot! Now... how about this one If the composition function g compose f: A-> C is an injection, then the function g : B->C is an injection.
December 6th, 2007, 01:42 PM   #4
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Quote:
 Originally Posted by jblaine271 Great! Thanks a lot! Now... how about this one If the composition function g compose f: A-> C is an injection, then the function g : B->C is an injection.
This is not true in general.
Example:
f: {0}->{0,1} f(0)=0
g: {0,1}->{0,1} g(0)=g(1)=0
Their composition is
h: {0} -> {0,1} h(0)=0.
Note that the composition is injective, but g is not injective.

In case you wanted to write surjective instead of injective, the proof can be found again at planetmath.

 December 6th, 2007, 01:48 PM #5 Newbie   Joined: Dec 2007 Posts: 4 Thanks: 0 Nah, that's what the problem stated. However the prompt was "prove or disprove", so I'll just use a counterexample. Could you maybe explain yours a little further?

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