MHB How do you prove that a function is bijective?

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To prove that the function f: (0, ∞) → (0, ∞) defined by f(x) = x^2 is bijective, it is first established as injective by showing that if f(a) = f(b), then a must equal b. For surjectivity, it is necessary to demonstrate that for every y in (0, ∞), there exists an x in (0, ∞) such that y = x^2, which can be achieved by defining a right-inverse function g(x) = √x. This approach emphasizes the importance of the function's domain and co-domain in determining its properties. The discussion also highlights how changing these parameters can affect the function's injectivity and surjectivity. Understanding these concepts is essential for proving a function's bijectiveness.
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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 $y=x^2$?
 
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I believe that follows from the definition of the function.
 
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".
 
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