Set Theory – Does the Intersection of Images Imply an Injective Function?

Question

I've read that $f(A \cap B) = f(A) \cap f(B)$ is true iff $f$ is an injective function' (and thus the statement is biconditional), and I've been trying to understand why this is true, but struggling.

I understand from the proofs the implication that if $f$ is injective $\implies f(A \cap B) = f(A) \cap f(B)$ . But it's the reverse implication I'm struggling with, as I seem to be finding counter examples; where am I going wrong?

Let $f$ be a function s.t. $f(A \cap B) = f(A) \cap f(B)$. Clearly, if $x\in (A \cap B)\implies f(x) \in ( f(A) \cap f(B) )$.

But for some functions, couldn't there also be a set $C \neq (A \cup B)$ such that $f(C) = f(A) \cap f(B)$? And hence, $f$ would be a many-one function, despite $f(A \cap B) = f(A) \cap f(B)$?

Example:

Consider the domain $D = \{1, 2, 3, 4, 5, 6, 7\}$ and the codomain $E = \{a, b\}$ and a function $f$ that maps $D$ to $E$ s.t. odd numbers in $D$ map to a and even numbers map to b.

Now, let the sets $A$, $B$, $C$ $\subset$ $D$ equal $\{1, 2\}$ and $\{1, 2, 3\}$ and $\{5, 6\}$ respectively. Then, $(A \cap B) = \{1, 2\}$ and $f(A) \cap f(B) = \{a, b\} = f(A \cap B)$.

But clearly now, $f(C) = \{a, b\} = f(A) \cap f(B)$ – the equality holds, but $f$ is many-one, not injective.

I must be wrong in my logic somewhere/not understood some basic facts about sets and their equality, but can't see what? Any thoughts?

Thanks very much, indeed.

Answer 1

Suppose that for each subsets $A$ and $B$ of the domain of $f$, it is true that $f(A\cap B)=f(A)\cap f(B)$. You want to deduce from this that $f$ is injective. Let $x$ and $y$ be distinct elements of the domain of $f$; you want to prove that $f(x)\neq f(y)$. Since $x\neq y$, $\{x\}\cap\{y\}=\emptyset$, and therefore$$f(\{x\})\cap f(\{y\})=f(\{x\}\cap\{y\})=\emptyset.$$But$$f(\{x\})\cap f(\{y\})=\bigl\{f(x)\bigr\}\cap\bigl\{f(y)\bigr\}$$and saying that this set is empty means precisely that $f(x)\neq f(y)$.