Injective Compositition


by dijkarte
Tags: compositition, injective
dijkarte
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#1
Apr10-12, 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.
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dijkarte
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#2
Apr10-12, 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.
micromass
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#3
Apr10-12, 11:06 AM
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Quote Quote by dijkarte View Post
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
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#4
Apr10-12, 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!
micromass
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#5
Apr10-12, 11:51 AM
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Quote Quote by dijkarte View Post
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
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#6
Apr10-12, 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.
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#7
Apr10-12, 12:24 PM
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Quote Quote by dijkarte View Post
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
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#8
Apr10-12, 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.
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#9
Apr10-12, 01:19 PM
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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
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#10
Apr10-12, 02:19 PM
P: 200
Got it! Any good reference that helps with doing proper proofs?

Thanks.
micromass
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#11
Apr10-12, 02:25 PM
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The book "How to think like a mathematician" by Kevin Houston is a good book.


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