Proving Injection in Composite Functions

[dijkarte]

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.

 

[dijkarte]

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

 

[micromass]

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.

 

[dijkarte]

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!

 

[micromass]

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.

 

[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.

 

[micromass]

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.

 

[dijkarte]

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.

 

[micromass]

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.

 

[dijkarte]

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

Thanks.

 

[micromass]

The book "How to think like a mathematician" by Kevin Houston is a good book.