Working in Martin-Löf intensional type theory, $\mathsf{ITT}$, with a universe $U$ within the theory closed under all the usual constructors, the univalence of the canonical family $\mathsf{EI}$ over $U$ can (via the Curry-Howard correspondence) be regarded as a type $\mathsf{isUnivalent}(U)$.



> 
> Is $\mathsf{isUnivalent}(U)$ a term of type $U$?
> 
> 
> 



Is the univalence of the canonical family over a universe small?
Let me first try to make the question precise. Suppose $U$ is a universe. The statement "$U$ is univalent" which you are asking about is
$$\prod\_{A,B : U} \mathsf{isEquiv}(\mathsf{idtoeq}\_{A,B}) \tag{1}$$
where
$$\mathsf{idtoeq}\_{A,B} : \mathsf{Id}\_U(A,B) \to \mathsf{Equiv}(\mathsf{El}(A), \mathsf{El}(B))$$
is the canonical map that turns paths in the universe $U$ to equivalences between the corresponding types, and $\mathsf{isEquiv}(f)$ is the statement that $f$ is an equivalence. Now, $\mathsf{isEquiv}(f)$ lives in the same universe as $f$, and so $\mathsf{isEquiv}(\mathsf{idtoeq}\_{A,B})$ lives in $U$. Thus the product (1) is a product over $U$ of types in $U$. This means that it is *not* an element of $U$, unless $U$ is *impredicative*, i.e., closed under prdoucts over $U$.


The question the is whether we can have an impredicative univalent universe. I do not know the answer.


