TITLE: $f(x)=x^{3}+1$ - Injective and Surjective?
QUESTION [12 upvotes]: Dr. Pinter's "A Book of Abstract Algebra" presents this exercise:

Prove/disprove whether the following function $f$ (with inputs/outputs of real numbers) is injective and/or surjective:
  $$f(x)=x^{3}+1.$$

Using this helpful answer, I solved for $$f(a)=f(b)\colon$$
$$f(a)=a^{3}+1$$ 
$$f(b)=b^{3}+1$$
$$a^{3}+1=b^{3}+1$$
$$a^{3}=b^{3} \text{ subtract 1 from each side}$$
$$(a^{3})^{1/3}=(b^{3})^{1/3} \text{ take the cubic root of each side}$$
$$a=b$$
Therefore, $f$ is injective.
Again, following that helpful answer, solve for $x$, and then plug into $f(x)$:
$$y=x^{3}+1$$
$$y-1=x^{3}$$
$$x=(y-1)^{\frac{1}{3}}$$
Now, plug $x$ into $f$, and check if result equals $y$, which would prove the surjective property.
$$f(x)=x^{3}+1$$
$$y=((y-1)^{1/3})^{3}+1 \text{ use $y$ rather than $f(x)$}$$
$$y=y-1+1$$
$$y=y$$
Therefore, it's also surjective.
Is this right?

REPLY [8 votes]: As in the answer you linked to (my answer), I would try to structure your proofs more clearly:
Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+1$ is injective.
Proof. Let $x_1,x_2\in\mathbb{R}$ and suppose $f(x_1)=f(x_2)$. Then
$$
f(x_1)=f(x_2)\\[0.5em]
x_1^3+1=x_2^3+1\\[0.5em]
x_1^3=x_2^3\\[0.5em]
x_1=x_2.
$$
Thus, the mapping is injective. $\blacksquare$
Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+1$ is surjective.
Proof. Suppose $y\in\mathbb{R}$. Then let $x=\sqrt[3]{y-1}$. We have the following:
\begin{align}
f(x) &= x^3+1\\[0.5em]
&= (\sqrt[3]{y-1})^3+1\\[0.5em]
&= (y-1)+1\\[0.5em]
&= y.
\end{align}
Thus, the mapping is surjective. $\blacksquare$

Note: As you have seen, writing up proofs in math is not only about squeezing out an accurate proof. You want to write it clearly, concisely, etc. The goal is to be concise but not at the cost of clarity. Oftentimes this is a fine line to walk.