$ S = \{(x,y) \in B \times B | (\exists a,b \in A)[f(a)=x, f(b)=y, (a,b) \in R ] \} $ with R transitive, is S transitive?

Question

Be the funtion $ f: A \to B $ and $ R \subseteq A \times A $ a transitive relation. Be the relation $ S \subseteq B \times B $ defined as:

$ S = \{ (x,y) \in B \times B | (\exists a,b \in A)[f(a)=x, f(b)=y, (a,b) \in R ] \} $

Is S transitive? Which changes will be enough to S be transitive?

Answers:

I know S is not transitive, a counter-example is: $ A = \{ 1,2,3,4 \} $, $ B = \{ u,v,w \}$, $ f = \{ (1,u), (2,v), (3,v), (4,w) \} $ and $ R = \{ (1,2), (3,4) \}$

R is transitive. $ (u,v) \in S $, $ (v,w) \in S $, but $ (u,w) \notin S $

But don't have idea how make S transitive. If I change $ (x,y) \in B \times B $ to some relation transitive in B I tought that solved, but i'm not certain.

Answer 1

There are different ways to make S transitive:- 1. S will be always transitive if the function f is one-one and R is transitive. 2. If f is not injective, define R in such a way that functional image of first element of the ordered pair in R is not equal to the functional image of any second element of the other ordered pair. e.g. (from the example stated above) if R = {(1,2),(1,3)} => S={(u,v)} is transitive Or if R = {(2,3), (4,1)} then S = {{(v,v), (w,u)} is transitive.

Answer 2

Assume f in an injection.
If aSb, bSc, then exists x,y,z,t in A with
f(x) = a, f(y) = b, f(z) = b, f(t) = c.
Since y = z and xRy, zRt comes xRt and aSc.