Is this correct?Is f onto if g∘f is onto and g is one-to-one?

[IProto]

Homework Statement

Let f: X$$\rightarrow$$Y and g: Y$$\rightarrow$$Z be functions. Prove or disprove the following: if g$$\circ$$f is onto and g is one-to-one then f is onto.

Homework Equations

N/A

The Attempt at a Solution

I'm honestly not sure what to do with this. I believe that the statement is true as I cannot think of an instance where it would be false, however actually proving it is another story. I know that:

since g$$\circ$$f is onto that $$\forall$$z$$\in$$Z, $$\exists$$x$$\in$$X so that g$$\circ$$f(x) = z

and I believe g must be onto so $$\forall$$z$$\in$$Z, $$\exists$$y$$\in$$Y so that g(y) = z.

and since g is one-to-one $$\forall$$y,z$$\in$$Y, if g(y) = g(z) then y=z.

I just don't know what to do with all of that. I've started by assuming y$$\in$$Y and x$$\in$$X but again I don't know what to do with those assumptions =\.

Since I believe the statement true I want to show $$\forall$$y$$\in$$Y, $$\exists$$x$$\in$$X si tgat f(x)=y.

Anyway I've been mashing my head against a wall over this to no avail so far. If anyone could help me I'd greatly appreciate it. This is for an assignment that's due tomorrow and it's the last one I'm unable to get. FYI the reason I said g is onto is from a previous question I had to show if g$$\circ$$f is onto then g must be onto as well.

Oh, and sorry for the horrible formatting, I'm not to good with the formula creator.

 

[Dick]

Prove it by contradiction. Pick a y in Y such that there is no x in X with f(x)=y. What next? What can you say about g(y)?

 

[IProto]

Before I do that I'm going to post what I was just working on, I may be onto something with a direct proof (no pun intended). Any comments appreciated.

Let f: X -> Y and g: Y-> Z be functions so that g is ont-to-one and gof is onto. Let z$$\in$$Z. Since g is onto $$\exists$$y$$\in$$Y such that g(y)=z. Also, since gof is onto $$\exists$$x$$\in$$X such that gof(x)=z. Since gof is one-to-one and gof(x) = g(f(x)) = z = g(y) it follows that f(x) = y for all y,z$$\in$$Y. This gives us $$\forall$$y$$\in$$Y,$$\exists$$x$$\in$$X such that f(x)=y, thus f is onto.

 

[Dick]

Sort of. But it's not too clear. You start out saying that there exists a y, and at the end it changes to for all y. You should start out by saying for ALL y (something, something) and end up saying, there EXISTS an x.

 

[HallsofIvy]

Before I do that I'm going to post what I was just working on, I may be onto something with a direct proof (no pun intended). Any comments appreciated.

Let f: X -> Y and g: Y-> Z be functions so that g is ont-to-one and gof is onto. Let z$$\in$$Z. Since g is onto $$\exists$$y$$\in$$Y such that g(y)=z.

You were NOT given that g is onto. The only condition on g is that it is one-to-one.

Also, since gof is onto $$\exists$$x$$\in$$X such that gof(x)=z. Since gof is one-to-one and gof(x) = g(f(x)) = z = g(y) it follows that f(x) = y for all y,z$$\in$$Y. This gives us $$\forall$$y$$\in$$Y,$$\exists$$x$$\in$$X such that f(x)=y, thus f is onto.