$g\circ f$ injective and $f\circ g$ surjective

Question

Let $f,g:\mathbb{R}\to\mathbb{R}$ such that $g\circ f$ is injective and $f\circ g$ is surjective. Prove that $f$ and $g$ are bijective.

What I have done:

$g\circ f=\text{injective}\Rightarrow f=\text{injective}$

$f\circ g=\text{surjective}\Rightarrow f=\text{surjective}$

$\Rightarrow f=\text{bijective}$

Now, how do I prove that g is bijective?

Answer 1

If $f$ is bijective, then in particular $f^{-1}$ is bijective, so then it follows that if $g\circ f$ is injective, so is $g \circ f \circ f^{-1} = g$. In the same manner, if $f\circ g$ is surjective then so is $f^{-1}\circ f \circ g = g$.

Answer 2

Suppose $g(x)=g(y)$. Since $f$ is surjective, $x=f(a)$ and $y=f(b)$ for some $a$ and $b$. Then $g(f(a))=g(f(b))$. But $g\circ f$ is injective, so $a=b$ and $x=f(a)=f(b)=y$. Then $g$ is injective.

Take any $x$ and consider $y=f(x)$. Since $f\circ g$ is surjective there exists $a$ such that $f(g(a))=y=f(x)$. Since $f$ is injective, $g(a)=x$. Then $g$ is surjective.