Example 9: $f(x)=2x$ on $\R$⁠: bijective

Basics you need

Solution

1. One-one: equal outputs force equal inputs

   Take arbitrary $x_1, x_2 \in \R$ with $f(x_1) = f(x_2)$⁠. Then

   $$\begin{aligned} f(x_1) = f(x_2) &\Rightarrow 2x_1 = 2x_2 \\ &\Rightarrow x_1 = x_2. \end{aligned}$$

   (The second implication divides both sides by 2.)

   So no two different real numbers have the same double, and $f$ is one-one.

   We start from the hypothesis of Definition 5, $f(x_1) = f(x_2)$⁠, and derive its conclusion, $x_1 = x_2$⁠. Because $x_1, x_2$ were arbitrary, the argument covers every pair of real numbers at once. Dividing by 2 is legitimate because $2 \ne 0$⁠.
   Cancellation law: if $c \ne 0$ and $ca = cb$⁠, then $a = b$⁠. The step is word for word the same as in Example 8: one-one does not care whether the inputs are natural or real.

2. Onto: find the candidate preimage by solving $f(x) = y$

   Take an arbitrary $y \in \R$⁠. We want some $x \in \R$ with $f(x) = y$⁠, that is $2x = y$⁠. Solving for $x$⁠:

   $$2x = y \quad\Longrightarrow\quad x = \frac{y}{2}.$$

   This is only a candidate: we found what $x$ must be if it exists. Next we check it.

   Working backwards from the target is the natural way to discover a preimage. But the definition of onto asks us to exhibit an $x$ in the domain with $f(x) = y$⁠, so discovery must be followed by verification. The textbook writes only the verification: "there exists $\frac{y}{2}$ in $\R$ such that $f(\frac{y}{2}) = 2 \cdot (\frac{y}{2}) = y$". This step shows where $\frac{y}{2}$ comes from.
   To solve $2x = y$⁠, divide both sides by 2 (allowed since $2 \ne 0$⁠).

3. Onto: check the candidate is in the domain and works

   (i) In the domain? $y$ is real, so $\tfrac{y}{2}$ is real: $\tfrac{y}{2} \in \R$⁠. (ii) Does it work?

   $$f\!\left(\frac{y}{2}\right) = 2 \cdot \frac{y}{2} = y.$$

   So every $y$ in the codomain $\R$ is the image of the element $\tfrac{y}{2}$ of the domain. Hence $f$ is onto. For example, $y = 3$ comes from $x = 1.5$⁠, and $y = -4$ comes from $x = -2$⁠.

   Part (i) is exactly where Example 8 failed: there the domain was $\N$ and $\tfrac12 \notin \N$⁠. Over $\R$⁠, halving never leaves the set, so the candidate is always admissible. Part (ii) confirms that the formula really produces $y$⁠, so we are not relying on algebra we have not checked.
   On a graph, $y_0$ is in the range exactly when the horizontal line $y = y_0$ meets the graph. Onto means every horizontal line meets it; one-one means no horizontal line meets it twice. Basics: one-one and onto

4. Conclude

   $f$ is one-one (Step 1) and onto (Step 3), so $f : \R \to \R,\ f(x) = 2x$ is one-one and onto, that is, bijective.

   The preimage found in Step 3 is unique (by Step 1), so the rule $y \mapsto \tfrac{y}{2}$ undoes $f$⁠. This is the inverse function of $f$⁠, studied in Section 1.4: $f^{-1}(y) = \tfrac{y}{2}$⁠.

   Definition 7 calls a function bijective exactly when it is both one-one and onto, so the two proved properties finish the question. The remark about the inverse previews why bijections matter: they are exactly the functions that can be reversed.
   Bijective $\iff$ every $y$ in the codomain has exactly one preimage: onto supplies at least one, one-one allows at most one.

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