For any $f:A\to A$ and any relation $R$ on $A$, define $S$ on $A$ by $aSb$ iff $(f(a),(f(b))\in R$. Does $S$ reflexive or symmetric imply it for $R$? Let $A = \{1,2,3,4\}$. For any function $f : A \rightarrow A$ and any relation $R$ on $A$, we define the relation $S$ on $A$ by: for any $a,b \in A,   aSb$ iff $(f(a),(f(b)) \in R$.
Prove / Disprove:
(a) $\forall$ functions $f : A \rightarrow A$ and all relations $R$ on $A$, if $S$ is reflexive then $R$ is reflexive.
(b) $\forall$ functions $f : A \rightarrow A$ and all relations $R$ on $A$, if $S$ is symmetric then $R$ is symmetric.

For both (a) and (b) I think they are false, but I am not sure on this as I do not know exactly how to start my proofs.
For (a) could I just give an example where the function $f(a)$ equals a constant for every element in $A$? i.e. $f(a) = 1,  \forall a \in A$. Then go from there to show $S$ is reflexive, but $R$ is not reflexive.
For (b) I was thinking of doing the same thing, and assigning $f(a) = 1, \forall a \in A$ and showing an example where $S$ is symmetric but $R$ isn't.

A counterexample is a perfectly valid means of approach, and both of those functions work. Take $R$ to be the relation $\{(1,1),(1,2)\}$ for both of them.
In the first case, $S$ is reflexive since $(f(a),f(a))=(1,1) \in R$ for every $a \in A$. However, $(2,2) \not \in R$, so $R$ isn't reflexive.
In the second case, $S$ is symmetric trivially since $S = A \times A$. However, $(1,2) \in R$ doesn't imply $(2,1) \in R$, so $R$ is not symmetric.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
