I'm trying to prove that for sets $A,B \subset S$ a function $f: S \to Y$ is injective if and only if $f(A \cap B) = f(A) \cap f(B)$.

The first inclusion is clear and always true, but I'm trying to understand a proof of the latter direction: assuming $f(A \cap B) = f(A) \cap f(B)$ and proving that $f$ is injective. I can understand most of the steps of the proof, reproduced below.

Suppose $f(a) = f(b)$ for $a,b \in S$, and let $A = \{a\}$ and $B = \{b\}$. Then $f(A) = f(B) = \{f(a)\} = \{f(b)\}$. In particular, $f(A) \cap f(B) = \{f(a)\}$. By assumption, $f(A \cap B) = f(A) \cap f(B)$, so $f(a) \in f(A \cap B)$. So $a \in A \cap B$, so $a \in B$, so $a = b$.

The step I don't understand is where we argue that $f(a) \in f(A \cap B)$ implies $a \in A \cap B$. $f(a)$ is just some element $y$ of $Y$, so we could have $y = f(x)$ for some $x \in A \cap B$ where $x \neq a$. The sets are defined so narrowly that I can see why this can't be the case, by why does this imply, directly, that $a \in A \cap B$?

## Best Answer

As you mention this is not true in general. I think the better argument is as follows.

Here we have that $$f(A \cap B) = f(A) \cap f(B) = \{ f(a) \} \neq \emptyset.$$

On the other hand, if $a \neq b$, then $A \cap B = \emptyset$ and then $f(A \cap B) = f(\emptyset) = \emptyset$. Thus $A \cap B \neq \emptyset$. Hence since $A$ and $B$ are both sets with a single elements, these elements must be equal, as they must be the single element in $A \cap B$.