Confused about how to prove a function is surjective/injective?

Question

Prove $f(x) = x^3$ is surjective. I've read about how to do these proofs online but I don't really understand it.

Obviously, the formula $y^{1/3} = x$ always maps to some x for any y, but I don't understand how to officially prove it. Someone said to let $y^{1/3} = x$ and $f(x) = x^3$ so $f(x) = y$.

I don't understand how that does anything. You're just taking the function of the inverse of the function which of course cancels out and results in $y = f(x)$

Answer 1

I'm going to do the mathematician wiggle and say this depends an what you have been taught so far and can take as a given.

If you have been told by God on high that for every $x \in \mathbb R$ then there $\sqrt[3]{x}$ exists then your proof will go like this:

$f(x) =x^3$ is surjective if for any $a \in \mathbb R$ there is exists $b \in \mathbb R$ so that $f(b) = a$. Let $b = \sqrt[3]{a}$ then $f(b) = f(\sqrt[3]{a}) = (\sqrt[3]{a})^3 = a$. So for every $a$ we can find a $b$ (namely the cube root of $a$) so that $f(b) = a$. So $f$ is surjective.

However you might not be allowed to use that God told you every number has a cube root. And you might need to prove it.

Were you taught the intermediate value theorem? Can you take it for granted that $f(x) = x^3$ is continuous?

If $a > 1$ then $a^3 > a$. ANd $1^3= 1 < a$. So $f(1) < a < f(a)$. That means there must be some $b: 1 < b < a$ so that $f(b) = a$.

And if $a = 1$ then $f(1) = 1^3 = 0 =a$. And if $0< a < 1$ then $f(0) = 0 < a^3 < 1=f(1)$ so there is a $b: 0< b< 1$ so that $f(b) =a$ and ... so on.

But you would have to prove the Intermediate value theorem the first time you see it.

And we'd have to prove $f(x) = x^3$ is continuous.

Should I go on.

Answer 2

Given $y$, you are actually defining the $x$ that you are searching for as $y^{1/3}$ (note that $t^{1/3}$ is well-defined for any $t\in\mathbb{R}$). Then you plug it in to obtain $f(x)=x^3=\left(y^{1/3}\right)^3=y$.

Answer 3

If a function has an inverse that is defined on the whole real line then the function is a bijection, hence injective (one to one) and surjective (onto).

Here we can define $g(x) = (\operatorname{sgn} x) \sqrt[3]{|x|}$ and check that $f(g(x)) = x$. Hence for any $y$, we know that there is some $x$ ($=g(y)$) such that $f(x) = y$.

Answer 4

A function $f:U\longrightarrow V$ always has $3$ ingredients. $U$ is the function's domain, $V$ its codomain, and $f$ is the relation that associates to each element $u \in U$ some element $v=f(u)\in V$.

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Now, a function need not attain all the values in $V$. For instance, take $f:\mathbb R \longrightarrow \mathbb R$ given by $f(x)=x^2$. If $v<0$, then there is no $u\in\mathbb R$ with $f(u)=v$.
The set of values of $V$ that are attained by $f$ is called the image $\text{Im}(f)$ of $f$. In our previous example, we had $\text{Im}(f)=[0,+\infty)$. Indeed, every nonnegative real number $v$ has a real square root $u=\sqrt a$, which satisfies $f(u)=v$.

When $V=\text{Im}(f)$, we say $f$ is surjective. Hence, while $f:\mathbb R \longrightarrow \mathbb R$ is not surjective, $f: \mathbb R\longrightarrow [0,+\infty)$ is.

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The association $v=f(u)$ in general need not be unique. This means that, given $v\in \text{Im}(f)$, there might be more than one $u\in U$ with $f(u)=v$.
For instance, take our previous example. Given any positive real number $v$, we have that both $u_1=\sqrt v$ and $u_2=-\sqrt v$ are solutions to $f(u)=v$.

When this does not occur -- that is, when the association is unique --, we say that $f$ is injective. Hence, while $f:\mathbb R \longrightarrow [0,+\infty)$ is not injective, $f: [0,+\infty) \longrightarrow [0,+\infty)$ is.

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When a function is both surjective and injective, we say it is bijective. In this case, we may define an inverse function $g=f^{-1}:V\longrightarrow U$ which associates to each $v\in V=\text{Im}(f)$ the unique $u\in U$ that solves $f(u)=v$.

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To prove that some map is bijective, you hence need to know not only the map's domain (where it's defined), but also its codomain. We've seen that even when the association is the same -- $f(x)=x^2$ in our examples --, changing the domain and codomain impact whether or not the function is surjective/injective.