Is $\Gamma = Inv(Pol(\Gamma))$? I'm reading A Rendezvous of Logic, Complexity and Algebra, my first introduction to the world of CSP. Let $\Gamma$ be a finite constraint language. It says in Pg. 9 that


  $Pol(\Gamma)$ is used to denote the set of all polymorphisms of $\Gamma$, that is,
  $$Pol(Γ) = \{f : \forall R \in \Gamma, \textrm{ f is a polymorphism of }R\}.$$
  Also, for a set of operations $O$, we use $Inv(O)$ to denote the set of relations having all operations in $O$ as a polymorphism, that is,
  $$Inv(O) = \{R : \forall f \in O, \textrm{ f is a polymorphism of } R\}.$$


The answer to my question is: No (in fact it is a subset of $\Gamma$). But I don't understand why: 


Isn't every $R\in \Gamma$ in $Inv(Pol(\Gamma))$?
Are the $R$'s from $Inv(O)$ allowed to come only from $\Gamma$?
The operation $\operatorname{Inv}(\operatorname{Pol}(\cdot))$ is a closure operation. This means that


$\Gamma \subseteq \operatorname{Inv}(\operatorname{Pol}(\Gamma))$.
$\operatorname{Inv}(\operatorname{Pol}(\Gamma)) = \operatorname{Inv}(\operatorname{Pol}(\operatorname{Inv}(\operatorname{Pol}(\Gamma))))$.


See for example lecture notes of Jin-Yi Cai.