$g\circ f=id_X$ implies $f$ is one-to-one and $g$ is onto

Question

I have $2$ functions, $f:X\to Y$ and $g:Y\to X$. If $g\circ f=id_X$, where $id_X$ is the identity function on $X$, that is $id_X(x)=x$ for any $x\in X$, then I have to prove that $f$ is one-to-one and $g$ is onto.

The first part is simple. Let $x_1, x_2\in X$ such that $f(x_1)=f(x_2)$. Then $g(f(x_1))=g(f(x_2))$, so $x_1=x_2$.

I couldn't figure out how the prove the second fact. I tried to assume it's not onto, which means there exists an $x_1\in X$ such that for all $y\in Y$, $g(y)\neq x_1$. Since $f$ is one-to-one, this leads to $f(g(y))\neq f(x_1)$. I have no idea how to continue.

Take an $x \in X$. You have $g \circ f(x)=x$. Can you find some $z$ such that $g(z)=x$ ? — TheSilverDoe, Jul 29 '21 at 11:21
Well, yes, that would be $f(x)$. Somehow I didn't think of that :) Thank you! — Wolfuryo, Jul 29 '21 at 11:24
I already got an answer from @TheSilverDoe, but that is helpful too! — Wolfuryo, Jul 29 '21 at 14:29

score -2 · Answer 1 · answered Jul 29 '21 at 14:29

$1)$
we have $id_x=(g\circ f)$ is bijective (one-to-one and onto )
let $(x_1.x_2)\in X^2$ where $f(x_1)=f(x_2)$\ $id_x(x_1)=id_x(x_2)\implies (g\circ f)(x_1)=(g\circ f)(x_2)\implies x_1=x_2$
this means $f$ is one-to-one
$2) $
let $z$ be a element on $X$ and $id_x=g\circ f $ is onto .so there is $x\in X$ where $z=g(f(x))$ but $f(x) \in Y$ so whataver $z\in X$ there is $y=f(x) \in X $ where $z=g(y)$ . this means $g$ is onto

$g\circ f=id_X$ implies $f$ is one-to-one and $g$ is onto

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