MHB How do you prove that a function is bijective?

[Guest2]

Suppose I want to prove that the function $f: (0, \infty) \to (0, \infty)$ defined by $f(x) = x^2$ is bijective.

Let $a, b \in (0, \infty)$ and $f(a) = f(b)$. Then $a^2 = b^2 \implies a = b$ since everything is non-negative we can simply take square roots. Therefore $f$ is injective. To prove that $f$ is surjective, let $y \in (0, \infty)$. How do I prove that there's $x \in (0, \infty)$ such that $y=x^2$?

 

[Greg]

I believe that follows from the definition of the function.

 

[Deveno]

One way to prove a function $f:A \to B$ is surjective, is to define a function $g:B \to A$ such that $f\circ g = 1_B$, that is, show $f$ has a right-inverse.

Equivalently, we must show for all $b \in B$, that $f(g(b)) = b$. Beware! This does NOT mean that $g(f(a)) = a$, in fact this is usually untrue (unless $f$ is injective).

In this case, we may take $g(x) = \sqrt{x}$, which works BECAUSE OF THE DOMAINS of $f$ and $g$ (non-negative reals).

Investigate which (injectivity or surjectivity, or both) fails if we change the domain, or co-domain, like so:

$f: \Bbb R \to [0,\infty)$ given by $f(x) = x^2$

$f: \Bbb R \to \Bbb R$ given by $f(x) = x^2$.

Perhaps this will persuade you that the properties of a function do not depend just on its "rule of computation".