# Injective Compositition

by dijkarte
Tags: compositition, injective
 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.
 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.
Mentor
P: 18,346
 Quote by dijkarte We need to show that g(f(a)) = g(f(b)) ==> a = b holds true for all a, b in A.
No, you don't need to show that, that's given.

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

 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!
Mentor
P: 18,346
 Quote by dijkarte But it's not a homework either
Doesn't really matter. It's the style of homework, so it belongs here. It's irrelevant whether it is actually homework.

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.
 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.
Mentor
P: 18,346
 Quote by dijkarte 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.
No, you don't need that g is an injection.
And if gf is an injection, then it does NOT imply that g is an injection.
 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.
 Mentor P: 18,346 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.
 P: 200 Got it! Any good reference that helps with doing proper proofs? Thanks.
 Mentor P: 18,346 The book "How to think like a mathematician" by Kevin Houston is a good book.

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