Mapping Proof

a mapping $\alpha : S \to T$ is one-to-one if $\alpha (A \cap B) = \alpha (A) \cap \alpha (B)$ for every pair of subsets A and B of S.

The contrapositive of this is:
if $\alpha (A \cap B) \not = \alpha (A) \cap \alpha (B)$ for every pair of subsets A and B of S then $\alpha$ is not one to one.

intuitively, $<br /> \alpha (A \cap B) \subseteq \alpha (A) \cap \alpha (B)$
So $<br /> \alpha (A \cap B) \not = \alpha (A) \cap \alpha (B)$ implies that there is some element in $\alpha (A) \cap \alpha (B)$ that is not in $\alpha (A \cap B)$ . So there is an element a in A and an element b in B such that $\alpha (a) = \alpha (b)$ but $a,b \not \in A \cap B$ so $a \not = b$ so $\alpha$ is not one-to-one.

To go the other way, we know that the function is not one-to-one so for some $a \not = b,$ $\alpha (a) = \alpha (b)$ . using A={a} and B = {b}, $\alpha (A \cap B) = \alpha (\oslash) = \oslash \not = \alpha (A) \cap \alpha (B)$ .