Set theory proof question with functions Let $f:A\rightarrow B$ be a function and let $C_1,C_2\subset A$
Prove that
$f(C_1\cap C_2)=f(C_1)\cap f(C_2) \leftrightarrow$ $f$ is injective
Attempt:
$(\leftarrow)$ Let $f(x)\in f(C_1\cap C_2)$. Then there exists $x\in C_1\cap C_2$ because $f$ is injective. So $x\in C_1$ and $x\in C_2$. So
$f(x)\in f(C_1)$ and $f(x)\in f(C_2)$. So $f(x)\in f(C_1)\cap f(C_2)$. Thus $f(C_1\cap C_2)=f(C_1)\cap f(C_2)$ I got to this but don't have an idea on how to prove the other direction..
 A: $\Leftarrow$:
If $f$ is injective, then $f(C_1) \cap f(C_2) = \{y: \exists x_1 \in C_1, x_2 \in C_2: f(x_1) = f(x_2) = y\} = \{y: \exists! x \in C_1 \cap C_2: f(x) = y\} = f(C_1 \cap C_2)$ where the second (and third) equality follows from $f$ being injective and thus $x_1 = x_2$.
$\Rightarrow$: Consider sets $C_1$ and $C_2$ consisting of single elements.
A: This is one of those situations where we really, really shouldn't omit quantifiers. So lets state the claim as follows.
Proposition. Suppose $f:A \rightarrow B$ is a function. Then TFAE.


*

*For all subsets $X$ and $Y$ of $A,$ we have $f(X \cap Y) = f(X) \cap f(Y).$

*$f$ is injective.


Proof. The backward direction has been proved on this site about 100x already. So lets prove the forward direction. Let $f : A \rightarrow B$ denote a function satisfying Condition 1 (above), and consider fixed but arbitrary $a$ and $a'$ in $A$ with $f(a)=f(a')$. Now assume for a contradiction that $a \neq a'$. Then the first statement in the following list is true, and each statement in the list implies the next.


*

*$\{a\} \cap \{a'\} = \emptyset_A,$

*$f(\{a\} \cap \{a'\}) = f(\emptyset_A),$

*$f(\{a\}) \cap f(\{a'\}) = \emptyset_B,$

*$\{f(a)\} \cap \{f(a')\} = \emptyset_B$

*$f(a) \neq f(a').$


But this contradicts the earlier statement $f(a)=f(a')$.
