Prove $f(x) = x^3+a $ is bijective for any fixed $a\in \mathbb{R}$ by proving the existence of a surjective inverse function $f^{-1}$ I know the property:   If $f: \mathbb R \rightarrow \mathbb R$ is a bijection, $f(f^{-1}(y)) = y$ for any $y\in \mathbb R$ (i).
I think that is equivalent to:
If there exist an surjective inverse function $f^{-1}: \mathbb R \rightarrow \mathbb R$ of $f$ with
$f(f^{-1}(y)) = y$ for any $y\in \mathbb R$, then $f$ is a bijection (ii).
Statement:
$f: \mathbb R \rightarrow \mathbb R, f(x) = x^3+a $ is bijective for any fixed $a\in \mathbb{R}$
Proof:
Let $a\in \mathbb{R}$ be fixed but arbitrary.
Define $f^{-1}: \mathbb R \rightarrow \mathbb R$ as  $ f^{-1}(y)=(y-a)^{\frac13}$.
The function $f^{-1}$ is surjective because $ (y-a)^{\frac13}$ exists for every $y\in\mathbb R$ and $f(f^{-1}(y)) = y$.
$\overset{(ii)}{\Rightarrow}$ $f$ is a bijection
Edit
I want to prove it explicitly in the way I did (if my method is correct).
 A: Statement:
Let $A,B$ two nonempty sets and let $\,f:A\to B\,$ be a function.
If there exists a surjective function $\;g:B\to A\;$ such that $\;f[g(y)]=y\;$ for any $\;y\in B\;,\;$ then $\,f:A\to B\,$ is a bijection.
Proof:
The function $\;f:A\to B\;$ is surjective, indeed for any $\;y\in B\;$ there exists $\;x=g(y)\in A\;$ such that $\;f(x)=f[g(y)]=y\;.$
Now, we will prove that $\;f:A\to B\;$ is injective too.
Let $\;x_1,\,x_2\in A\;$ such that $\;f(x_1)=f(x_2)\;.$
Since $\;g:B\to A\;$ is a surjective function, then there exist $\,y_1,\,y_2\in B\;$ such that $\;g(y_1)=x_1\;$ and $\;g(y_2)=x_2\;,\;$ hence ,
$y_1=f[g(y_1)]=f(x_1)=f(x_2)=f[g(y_2)]=y_2\;$ and
$x_1=g(y_1)=g(y_2)=x_2\,.$
We have just proved that $\;x_1,x_2\!\in\!A,\,f(x_1)\!=\!f(x_2)\Rightarrow x_1\!=\!x_2$
i.e. $\;f:A\to B\;$ is injective.
Consequently, $\;f:A\to B\;$ is a bijection because it is injective and surjective.
Addendum:
Under the same hypotheses of the Statement, we can also prove that the function $\;g:B\to A\;$ is a bijection too and $\;g=f^{-1}.$
Proof:
Let $\;y_1,\,y_2\in B\;$ such that $\;g(y_1)=g(y_2)\;.$
It results that $\;y_1=f[g(y_1)]=f[g(y_2)]=y_2\;.$
We have just proved that $\;y_1,y_2\!\in\!B,\,g(y_1)\!=\!g(y_2)\Rightarrow y_1\!=\!y_2$
i.e. $\;g:B\to A\;$ is injective.
Consequently, $\;g:B\to A\;$ is a bijection because it is injective and surjective.
Let $\;f^{-1}:B\to A\;$ be the inverse function of $\;f:A\to B\;.$
Since $\;f[g(y)]=y=f[f^{-1}(y)]\;$ for any $\;y\in B\;,\;$ then
$g(y)=f^{-1}(y)\;$ for any $\;y\in B\;,\;$ i.e. $\;g=f^{-1}.$
