Proof of Roger Heath-Brown: Interchanging Solutions of Involution On page 20 of "Proofs from THE BOOK" (click here) a proof of representing number as sum of two square is given. The proof is due to  Roger Heath-Brown (1971,appeared in 1984). The first part of the proof is -
I didn't get the below paragraph at all-

What do we get from the study of $f$ ? The main observation is that
  since $f$ maps the sets $T$ and $U$ to their complements, it also
  interchanges the elements in $T\setminus U$ with these in $U\setminus T$.
   That is, there is the same number of solutions in $U$ that are not
  in T as there are solutions in T that are not in U
  -so $T$ and $U$ have the same cardinality.

So my questions is- 
How does  $f$ interchanges the elements in $T\setminus U$ with these in $U\setminus T$?
 A: Notice that $\,S\,$ is partitioned into four pairwise disjoint sets:
 $$S = (T\cap U) \cup (T\setminus U) \cup (U\setminus T) \cup(S\setminus(T\cup U)). \tag{1}$$
We are given that the involution $\,f\,$ has no fixed points, and maps
$\,T\,$ to $\,S\setminus T\,$ and $\,U\,$ to $\,S\setminus U.\,$
Then it must map their intersection $\,T\cap U\,$ to
$$ (S\setminus T)\cap(S\setminus U) = S\setminus(T\cup U). \tag{2}$$
The exact same reasoning applies to their complements. This
means that $\,f\,$ maps $\,S\setminus T\,$ to $\,T\,$ and
$\,S\setminus U\,$ to $\,U.\,$ Thus their intersection as in
equation $(2)$ gets mapped to $\,T\cap U.\,$
The two remaining parts of $\,S\,$ are
 $\,T\setminus U\,$ and $\,U\setminus T.\,$
Each gets mapped to the other part since the first and
fourth parts are mapped by $\,f\,$ to each other and
since both $\,T\,$ and $\,U\,$ get mapped to their complements.
Equation $(1)$ is true for any set $\,S\,$
and with any given
$\,T\subseteq S\,$ and $\,U\subseteq S.\,$
The subsets $\,T\,$ and $\,U\,$ may or may not
be disjoint. Their union $\,T\cup U\,$ may or
may not be equal to $\,S.\,$ One of the subsets
may or may not be a subset of the other. In other
words, any of the four parts of the
partition in equation $(1)$ may or may not be
the empty set depending of the specific subsets
$\,T\,$ and $\,U.\,$
An example is for $\,p=13\,$ where $\,(1,1,-3)\in S\,$ 
and $\,(1,3,-1)\in S\,$ which are not in $\,T\cup U. $
