
#1
Apr1012, 09:20 AM

P: 200

Given two functions:
f:A > B g:B > C How to show that if the (g ° f) is injection, then f is injection? I tried this: We need to show that g(f(a)) = g(f(b)) ==> a = b holds true for all a, b in A. But there's nothing said about function g. 



#2
Apr1012, 10:26 AM

P: 200

I've tried using function mapping diagrams and actually it showed this proposition is wrong.
(g ° f) injective ==> g and f are injective. 



#3
Apr1012, 11:06 AM

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P: 16,562

You need to show that f is an injection. That is: f(a)=f(b) ==> a=b. That is what you need to show. 



#4
Apr1012, 11:48 AM

P: 200

Injective Compositition
You are absolutely right, my bad expressing the problem...
And yeah my post should have been moved under elementary school math ;) But it's not a homework either, it's a question my professor did not have time to clarify well! 



#5
Apr1012, 11:51 AM

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P: 16,562

So, got any ideas?? You have f(a)=f(b) and you need to prove a=b. Convert it to g(f(a))=g(f(b)) in some way. 



#6
Apr1012, 12:22 PM

P: 200

But I think in order to show that f(a) = f(b) ==> a = b, g has to be given as injection as well, though I could prove that both functions g and f are injections using function mapping diagram.




#7
Apr1012, 12:24 PM

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P: 16,562

And if gf is an injection, then it does NOT imply that g is an injection. 



#8
Apr1012, 01:05 PM

P: 200

Ok I could prove it by contradiction. Assuming f(x) is not injection, then
Then there's the case where f(a) = f(b) and a != b for some a, b Then g(f(a)) = g(f(b)) where a != b, which contradicts the given argument. 



#9
Apr1012, 01:19 PM

Mentor
P: 16,562

That is ok. But there is no need for a contradiction argument.
If f(a)=f(b). Taking g of both sides, we get g(f(a))=g(f(b)). By hypothesis, this implies a=b. 



#10
Apr1012, 02:19 PM

P: 200

Got it! Any good reference that helps with doing proper proofs?
Thanks. 


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