Q.Let A and B be sets. Show that $f: A \times B \rightarrow B \times A$ such that $f(a, b) = (b, a)$ is a bijective function.

Concept: Bijection Proof — show the function is both injective (one-to-one) and surjective (onto).

1. Injectivity: Suppose $f(a_1, b_1) = f(a_2, b_2)$. Then $(b_1, a_1) = (b_2, a_2)$, so $b_1 = b_2$ and $a_1 = a_2$. Hence $(a_1, b_1) = (a_2, b_2)$, so $f$ is injective. …

The idea is beautifully simple: swapping coordinates is a reversible operation. If you swap twice, you get back the original pair. That immediately tells us $f$ is its own inverse, which is the cleanest proof of bijectivity. But let's also walk through the standard approach to see why each property holds.

1. Injectivity (one-to-one)

   Suppose $f(a_1, b_1) = f(a_2, b_2)$. That means $(b_1, a_1) = (b_2, a_2)$. Two ordered pairs are equal exactly when their first coordinates match and their second coordinates match. So $b_1 = b_2$ and $a_1 = a_2$. Hence $(a_1, b_1) = (a_2, b_2)$. No two distinct inputs can give the same output.

2. Surjectivity (onto)

   Take any element of $B \times A$ — call it $(b, a)$. We need an input in $A \times B$ that maps to it. The natural candidate is $(a, b)$, because $f(a, b) = (b, a)$. Since $a \in A$ and $b \in B$, the pair $(a, b)$ is indeed in $A \times B$. So every element of the codomain is hit.

3. Inverse function

   Notice that applying $f$ twice gives:

So $f \circ f$ is the identity map on $A \times B$. That means $f$ is invertible with $f^{-1} = f$. A function with an inverse is automatically bijective. …

Method: Proving a structural map is a bijection (and spotting a self-inverse)

For maps defined by rearranging components — like $f(a,b)=(b,a)$ — prove bijectivity directly, and watch for a self-inverse shortcut.

Steps

Step 1: One-one — assume $f(u_1)=f(u_2)$, then equate the resulting tuples coordinate-by-coordinate to recover $u_1=u_2$. For pairs, $(b_1,a_1)=(b_2,a_2)$ gives $a_1=a_2$ and $b_1=b_2$. …

Common Mistakes

Mistake 1: Believing the swap map is a bijection only when $A$ and $B$ have equal size.

Why it's wrong: coordinate-swapping is a structural property of ordered pairs and works for any sets, finite or infinite. Correct approach: prove one-one and onto directly, which never uses cardinalities. …