Prove the following result by contradiction: Let f : X TO Y be a mapping. Suppose f (A INTERSECTION B) = f (A) INTERSECTION f (B) for all subsets A, B PROPER SET X , f (PHI) = PHI. Then f is a 1-1 mapping.

 Suppose it weren't necessary that $f$ is injective. Then there would exist a non-injective function $f$ such that for any two subsets $A,B⊆X$ we get $f(A∩B)=f(A)∩f(B)$.

Let $a,b\in X$

$f(\{a\})\cap f(\{b\})=\{f(a)=f(b)\}\ne f(\{a\}\cap\{b\})\\ =f(\emptyset)=\emptyset$

Contradiction.

Suppose it weren't necessary that $f$ is surjective.

Now assume $a \not\in f^{−1}(\{a\})$. Then by definition of inverse image of set, $f(a) \not\in \{a\}$ meaning $a\ne f(a)$. Contradiction.

''injective + surjective = 1-1''