Question

If f:S--> T is a function from S into T, show that the following are equivalent
a) f is one-to-one
b) For every y in T, the set f^{-1({y}) contains at most one point.
c) f(D1∩D2)=f(D1)∩f(D2) for all subsets D1, D2 of S
Please, help part c--> a

Answer 1

Assuming \(f\) was not one-to-one. Consider \(x_1\in D_1\) and \(x_2 \in D_2\). Then, you'd have the case where \(y=f(x_1)=f(x_2)\) but \(x_1\ne x_2\) since we assumed not one-to-one. Considering \(D_1=\{x_2\}\) and \(D_2=\{x_2\}\), then \(D_1 \cap D_2 = \emptyset \implies f(D_1 \cap D_2)=f(\emptyset)=\emptyset\). However, \(f(D_1)\cap f(D_2)=\{y\}\). And do we get a contradiction of our assumption, so \(f\) must be one-to-one.

Answer 2

Thank you so much.

Answer 3

=]