The pullback of a transitive relation is transitive Suppose $f : A \rightarrow B$ and $S$ is a relation on $B$. Define a relation $R$ on $A$ by:
$R = \{(x, y) \in A \times  A : (f(x), f(y)) \in S\}$
Prove that if $S$ is transitive then so is $R$.
Solution, given by professor: 
Let $x, y, z \in A$. 
Assume $_xR_y$ and $_yR_z$. Then $_{f(x)}S_{f(y)}$ and $_{f(y)}S_{f(z)}$. 
Since $S$ is transitive, $_{f(x)}S_{f(z)}$. 
By denition of $R$, this means $_xR_z$.
What I don't understand 
$R = \{(x, y) \in A \times  A : (f(x), f(y)) \in S\}$
Does the $:$ in this relation mean $(x, y) \in A \times  A $ $iff$ $(f(x), f(y)) \in S$ ? 
I'm unfamiliar with this notation, and am not sure I fully understand the relationship between the two relations. Any clarification would be appreciated. 

The colon : inside set brackets means "such that". Here, $(x,y)\in$$R$ iff $(x,y)\in$$A$$\times$$A$ and $(f(x),f(y))\in$$S$. 
