Let \(X\) and \(Y\) be finite sets and \(f: X \to Y\) be a function. Which one…

Concept: The bars \(|S|\) denote the cardinality (number of elements) of a set \(S\), so a statement written with \(|\cdot|\) compares set SIZES, while a statement written without bars — such as \(f(A \cap B) = f(A) \cap f(B)\) or \(f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)\) — compares the sets THEMSELVES. For any function \(f: X \to Y\) and any subsets \(S\) and \(T\) of \(Y\), the preimage operation always distributes over intersection: \(f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)\), regardless of whether \(f\) is injective or surjective. The forward image operation behaves differently — for subsets \(A\) and \(B\) of \(X\), \(f(A \cap B) \subseteq f(A) \cap f(B)\), but equality can fail whenever \(f\) is not injective, because two different elements of \(X\) may map to the same element of \(Y\).

Proof that f⁻¹(S ∩ T) = f⁻¹(S) ∩ f⁻¹(T):

1. If \(x \in f^{-1}(S \cap T)\), then \(f(x) \in S \cap T\), so \(f(x) \in S\) and \(f(x) \in T\). Hence \(x \in f^{-1}(S)\) and \(x \in f^{-1}(T)\), so \(x \in f^{-1}(S) \cap f^{-1}(T)\).

2. Conversely, if \(x \in f^{-1}(S) \cap f^{-1}(T)\), then \(f(x) \in S\) and \(f(x) \in T\), so \(f(x) \in S \cap T\), which means \(x \in f^{-1}(S \cap T)\).

Both inclusions hold for every \(S, T \subseteq Y\), so \(f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)\) — exactly the statement that is always true.

Why the other statements fail:

• \(|f(A \cup B)| = |f(A)| + |f(B)|\) fails when the images overlap: in general \(|f(A \cup B)| = |f(A)| + |f(B)| - |f(A) \cap f(B)|\). Example: \(X = \{1, 2\}, Y = \{a\}, f(1) = f(2) = a, A = \{1\}, B = \{2\}\) gives \(|f(A \cup B)| = 1\) but \(|f(A)| + |f(B)| = 2\).

• \(f(A \cap B) = f(A) \cap f(B)\) need not hold — only the inclusion \(f(A \cap B) \subseteq f(A) \cap f(B)\) is guaranteed, and it can be strict once \(f\) sends distinct elements to the same value. Same example as above: \(f(A \cap B) = \emptyset\) while \(f(A) \cap f(B) = \{a\}\).

• \(|f(A \cap B)| = \min\{|f(A)|, |f(B)|\}\) need not hold — only \(|f(A \cap B)| \le \min\{|f(A)|, |f(B)|\}\) is guaranteed. Example: \(X = \{1, 2\}, Y = \{a, b\}, f(1) = a, f(2) = b, A = \{1\}, B = \{2\}\) gives \(|f(A \cap B)| = 0\) but \(\min\{|f(A)|, |f(B)|\} = 1\).