If $g\circ f$ is onto, and $g$ is into, is true that $f$ is onto?

Question

I was faced by this question some time ago, and it keeps me thinking:

If $g\circ f$ is onto, and $g$ is into, prove that $f$ is onto.

In this link we can find something very close, but with $g$ being $1-1$, which is not my case here (if it was, it would be easy). Then, I was thinking if this is necessarilly true, trying to find some counter example.

Any help would be very appreciated! Thanks in advance!

Hi, @absolute0 I mean that $g$ is injective. (This is translated, the original language of the question is portuguese). — big_GolfUniformIndia, Apr 17 '21 at 15:21
@AlexProvost I thought that $1-1$ means bijective. Thanks for it! — big_GolfUniformIndia, Apr 17 '21 at 15:23

score 1 · Accepted Answer · answered Apr 17 '21 at 15:33

Let $y$ be a point in the codomain of $f$. Then $g(y)$ is defined, in the codomain of $g \circ f$ and so there is some $x$ in the domain of $f$ such that $g(f(x)) = g(y)$. It follows that $y=f(x)$ as $g$ is injective (or "into"). So $y$ is a value of $f$ and we're done.

Note that 1-1 = into = injective, so the quoted proof actually goes through.

If $g\circ f$ is onto, and $g$ is into, is true that $f$ is onto?

1 Answers1