# Mapping Proof

• Jan 31st 2008, 10:41 PM
Jhevon
Mapping Proof
For some reason i'm having trouble with this problem...:o

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

we define something like $\alpha (A)$ by $\alpha (A) = \{ \alpha (x) : x \in A \}$

Thanks guys
• Feb 1st 2008, 12:21 AM
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.
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, $
So $
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}, $\alpha (A \cap B) = \alpha (\oslash) = \oslash \not = \alpha (A) \cap \alpha (B)$.