Q.Let $A = \mathbb{R} - \{3\}$, $B = \mathbb{R} - \{1\}$. Let $f : A \to B$ be defined by $f(x) = \dfrac{x - 2}{x - 3}$, $\forall\, x \in A$. Then show that $f$ is bijective.

Concept: Bijection Proof — show injectivity (one-to-one) and surjectivity (onto).

Step 1 – Injectivity:

Assume $f(x_1) = f(x_2)$. Then

Cross-multiply: $(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$.

Expand and simplify: $x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6$

$\Rightarrow -3x_1 - 2x_2 = -3x_2 - 2x_1 \Rightarrow x_1 = x_2$. Hence $f$ is injective.

Step 2 – Surjectivity:

For any $y \in B$ ($y \neq 1$), solve $y = \frac{x - 2}{x - 3}$ for $x$.

$y(x - 3) = x - 2 \Rightarrow yx - 3y = x - 2 \Rightarrow yx - x = 3y - 2$ …

The core idea: a rational function of this form is a Möbius transformation — it’s essentially a linear function in disguise. For such functions, injectivity follows from the fact that if two inputs give the same output, cross-multiplying forces the inputs to be equal. Surjectivity follows because we can solve $y = f(x)$ for $x$ in terms of $y$, and the only value $y$ cannot take is the one that makes the denominator zero in the expression for $x$.

Let’s walk through it step by step.

1. Injectivity (one-to-one) Suppose $f(x_1) = f(x_2)$ for some $x_1, x_2 \in A$. That means

Cross-multiply (valid since denominators are never zero in $A$):

Expand both sides:

Cancel $x_1 x_2$ and $6$ from both sides:

Bring terms together:

Hence $x_1 = x_2$. So $f$ is injective.

2. Surjectivity (onto) We need to show: for every $y \in B$ (i.e., $y \in \mathbb{R}$, $y \neq 1$), there exists some $x \in A$ (i.e., $x \in \mathbb{R}$, $x \neq 3$) such that $f(x) = y$. Start with the equation:

Solve for $x$ in terms of $y$:

Bring $x$ terms together:

Since $y \neq 1$ (because $y \in B$), we can divide by $y - 1$:

This gives a candidate $x$ for each $y$. We must check two things: …

Method: The two-step bijection proof for a rational function

Use this for any 'show $f$ is bijective' question, especially $f(x)=\dfrac{ax+b}{cx+d}$ type maps between sets with points removed.

Steps

Step 1: Prove one-one algebraically.

Assume $f(x_1)=f(x_2)$, cross-multiply (safe because the denominators are non-zero on the domain), expand, and cancel. The mixed $x_1x_2$ terms cancel, leaving a linear equation that forces $x_1=x_2$.

Step 2: Prove onto by solving $y=f(x)$ for $x$.

Take an arbitrary $y$ in the codomain and rearrange $y=\dfrac{x-2}{x-3}$ to isolate $x$: collect the $x$-terms, factor, and divide. This yields a formula $x=g(y)$ — the candidate preimage.

Step 3: Check the candidate actually lies in the domain. …

Common Mistakes

Mistake 1: Proving only one of one-one and onto and calling it bijective.

Why it's wrong: a bijection needs BOTH properties; one alone (e.g. only injectivity) is insufficient. Correct approach: give a separate argument for one-one and for onto.

Mistake 2: Forgetting to check the preimage lies in the domain.

Why it's wrong: after solving $x=\dfrac{3y-2}{y-1}$ you must confirm $x\neq 3$, otherwise the 'preimage' is outside $A=\mathbb{R}\setminus\{3\}$. Correct approach: show $x=3$ leads to the contradiction $-2=-3$, so $x$ is always a valid element of $A$. …