
Mapping Proof
For some reason i'm having trouble with this problem...:o
Prove that a mapping $\displaystyle \alpha : S \to T$ is onetoone iff $\displaystyle \alpha (A \cap B) = \alpha (A) \cap \alpha (B)$ for every pair of subsets $\displaystyle A$ and $\displaystyle B$ of $\displaystyle S$.
we define something like $\displaystyle \alpha (A)$ by $\displaystyle \alpha (A) = \{ \alpha (x) : x \in A \}$
Thanks guys

a mapping $\displaystyle \alpha : S \to T$ is onetoone if $\displaystyle \alpha (A \cap B) = \alpha (A) \cap \alpha (B)$ for every pair of subsets A and B of S.
The contrapositive of this is:
if $\displaystyle \alpha (A \cap B) \not = \alpha (A) \cap \alpha (B)$ for every pair of subsets A and B of S then $\displaystyle \alpha$ is not one to one.
intuitively, $\displaystyle
\alpha (A \cap B) \subseteq \alpha (A) \cap \alpha (B) $
So $\displaystyle
\alpha (A \cap B) \not = \alpha (A) \cap \alpha (B) $ implies that there is some element in $\displaystyle \alpha (A) \cap \alpha (B)$ that is not in $\displaystyle \alpha (A \cap B)$. So there is an element a in A and an element b in B such that $\displaystyle \alpha (a) = \alpha (b)$ but $\displaystyle a,b \not \in A \cap B$ so $\displaystyle a \not = b$ so $\displaystyle \alpha$ is not onetoone.
To go the other way, we know that the function is not onetoone so for some $\displaystyle a \not = b, $$\displaystyle \alpha (a) = \alpha (b)$. using A={a} and B = {b}, $\displaystyle \alpha (A \cap B) = \alpha (\oslash) = \oslash \not = \alpha (A) \cap \alpha (B)$.