Injective Compositition

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

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

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

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.