# Is this proof ok?

## Homework Statement

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

f:X→Y and g:Y→Z

## 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.

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!

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!

Ok thanks that makes alot more sense, noted!

SammyS
Staff Emeritus