# Is this proof ok?

1. Mar 18, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

If f and g are both onto show g°f is onto.

f:X→Y and g:Y→Z

2. Relevant equations

3. The attempt at a solution

Since f is onto then there exists an x in X such that f(x) = y, for all y in Y.

Since g is onto then there exists a y in Y such that g(y) =z for all z in Z.

Hence, if g°f is not onto then there exists a z in Z such that there is no corresponding x in X. But since g and f are onto this is not possible. Therefore, g°f is onto.

2. Mar 18, 2012

### zooxanthellae

Your basic idea is correct. However, I would be careful with how exactly you write this out. Your first statement, "Since f is onto then there exists an x in X such that f(x) = y, for all y in Y" is not quite correct. The way you have it arranged, it reads as if there is one specific x in X that maps to all values of y in Y.

Of course this is not what you mean, but if you're starting out with proofs (which I'm assuming you are) then it is good to be precise at first. "For a given y in Y, there exists some x in X such that f(x) = y."

Hope that helps!

3. Mar 18, 2012

### charmedbeauty

Ok thanks that makes alot more sense, noted!

4. Mar 18, 2012

### SammyS

Staff Emeritus
Just pick an arbitrary z ∊ Z . It's pretty straight forward to show that there is some x ∊ X , such that g(f(x)) = z .
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