Prove that a function $f$ is injective if and only if it respects intersection.

Question

Let $f:X \to Y$ be a function. Prove that $f$ is injective if and only if $f(A\cap B)=f(A)\cap f(B),$ for any two sets $A, B \subseteq X.$

Answer 1

We want to prove the following result.

Theorem. Let $f:X \to Y$ be a function. Hence, $f$ is injective if and only if $f(A\cap B)=f(A)\cap f(B),$ for any two sets $A, B \subseteq X.$

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Sketch of the proof. Note that we want to prove an equivalence, i.e., a double implication. So, we must prove that $$f \text{ is injective} \implies f(A \cap B) = f(A) \cap f(B)$$ and $$f(A \cap B) = f(A) \cap f(B) \implies f \text{ is injective}.$$

First, let’s recall a few definitions. One of the hypothesis is that $f$ is injective.

Definition. Let $A$ and $B$ be any sets and $f:A \to B$ a function. We say that $f$ is injective if $$\forall x,y \in A, f(x) = f(y) \implies x = y.$$

The other hypothesis, states that the sets $f(A \cap B)$ and $f(A) \cap f(B)$ are equal.

Definition. Let $A$ and $B$ be any sets. We say that $A$ and $B$ are equal, and write $A = B$ if $A \subseteq B$ and $B \subseteq A.$

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Proof. $\implies.$ Suppose that $f$ is injective.

Let $x \in f(A \cap B).$ Hence, it exists $y \in A \cap B$ such that $f(y) = x.$ Since, $y \in A \cap B$ we know that $y \in A$ and $y \in B.$ It comes that $f(y) \in f(A)$ and $f(y) \in f(B).$ Then, $x \in f(A)$ and $x \in f(B)$ which means that $x \in f(A) \cap f(B).$ Therefore, $f(A \cap B) \subseteq f(A) \cap f(B).$

Now, let $z \in f(A) \cap f(B).$ Then $z \in f(A)$ and $z \in f(B).$ From here, we get that there exists $a_1 \in A$ and $a_2 \in B$ such that $f(a_1) = z$ and $f(a_2) = z.$ Since $f$ is injective, then $a_1 = a_2$ (otherwise it would contradict our hypothesis). Then, we have that exists $a_1 \in A$ and $a_1 \in B$ such that $z = f(a_1).$ Hence, it exists $a_1 \in A \cap B$ such that $f(a_1) = z,$ and so $f(a_1) \in f(A \cap B).$ This means that $z \in f(A \cap B).$ Therefore, $f(A) \cap f(B) \subseteq f(A \cap B).$

Since, $f(A) \cap f(B) \subseteq f(A \cap B) \subseteq f(A) \cap f(B),$ we conclude that $f(A) \cap f(B) = f(A \cap B).$

$\Longleftarrow.$ Suppose that $f(A \cap B) = f(A) \cap f(B).$ Let $x, y \in A,$ such that $f(x) = f(y).$ Suppose that $x \neq y$ and let $X = \{x\}$ and $Y = \{y\}.$ Since $x \neq y,$ we know that $X \cap Y = \emptyset$ and therefore $f(X \cap Y) = \emptyset.$ Although, note that $f(X) = \{f(x)\} = f(Y)$ and therefore $f(X) \cap f(Y) = \{f(x)\} \neq \emptyset.$ But this is absurd, it contradicts our hypothesis. The absurd came from assuming that $x \neq y.$ Hence, $x = y.$ Therefore, by definition, $f$ is injective. $\square.$

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Remark. Observe that in the $\implies$ part of the proof, we prove that $f(A \cap B) \subseteq f(A) \cap f(B)$ without using the hypothesis that $f$ is injective. That is because, for any two sets $A$ and $B$, $f(A \cap B) \subseteq f(A) \cap f(B).$

Answer 2

Let $f(a)=x = f(b)$ and write $A=\{a\}$, $B=\{b\}$. Then

$\{x\} = f(A) \cap f(B) = f(A\cap B)$ by hypothesis.

This means that $A\cap B\ne \emptyset$ and $A\cap B = \{a\}$ with $a=b$ as required.