Prove that function f: X->Y is INJECTIVE if...

student127

New member
Joined
Nov 18, 2017
Messages
2
Hello,

I have a discreet math problem that I cannot figure out. Not even where to start from.

The problem: Let X,Y,Z be non empty sets. Prove that function f:X->Y is injective if whatever functions g,h:Z->X we get that (f o g) = (f o h) from that follows that g = h.

I know that (f o g), a composition of functions means that it's basicly f(g(x)).
I know that a function is injective if f(x1) == f(x2) -> x1 == x2.


How would I go about proving this?

Best regards
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,858
Hello,

I have a discreet math problem that I cannot figure out. Not even where to start from.

The problem: Let X,Y,Z be non empty sets. Prove that function f:X->Y is injective if whatever functions g,h:Z->X we get that (f o g) = (f o h) from that follows that g = h.

I know that (f o g), a composition of functions means that it's basicly f(g(x)).
I know that a function is injective if f(x1) == f(x2) -> x1 == x2.


How would I go about proving this?

Best regards
Did you copy the problem word for word? The wording is a little awkward, so I'd want to clarify it first. I think you are to prove this theorem:

Let X,Y,Z be non-empty sets, and function f:X->Y. If, for any pair of functions g,h:Z->X, (f o g) = (f o h) implies that g = h, then f is injective.

You might start by supposing that f(x1) = f(x2), and trying to use the given fact to conclude that x1 = x2. Perhaps you can define some specific functions g and h such that (f o g) = (f o h), that is, f(g(x)) = f(h(x)) for any x, but also such that g = h would imply that x1 = x2.

What I've said here is just a natural starting point for any proof like this: express the goal clearly, and then try to fit the givens into that context. Having said that, I do see that this beginning can lead to a good proof. (It isn't very hard; in fact, it may seem so simple to you that you will initially miss it.)
 
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