Proof of Preimages of Unions, Intersections and Complements

By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii), the composite $g\circ V$ of Relations, Domain, Range, Inverse and Composition (v) is a map $I\to\mathcal{P}(S)$, and its value at $i\in I$ is $g(V_{i})=u^{-1}(V_{i})$, as $V_{i}\in\mathcal{P}(T)$. For every $i\in I$, $V_{i}\subseteq T$ by The Union Set and the Power Set of a Set (ii), so by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii), for every set $s$,

$$s\in u^{-1}(V_{i})\iff(s\in S\wedge u(s)\in V_{i}). \tag{1}$$

For $s\in S$ we have $u(s)\in T$.

Proof of (i). The union $\bigcup_{i\in I}V_{i}$ is a set by The Union and Product of a Family of Sets Indexed by a Set, and the Intersection of a Family with an Inhabited Index Class, Are Sets (i), and each of its elements lies in some $V_{i}$ by Indexed Families of Sets and Their Union, Intersection and Product (ii), hence in $T$; so it is a subset of $T$. For every set $s$, by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii) and Indexed Families of Sets and Their Union, Intersection and Product (ii), $s\in u^{-1}(\bigcup_{i\in I}V_{i})$ if and only if $s\in S$ and $u(s)\in V_{i}$ for some $i\in I$; by (1) this holds if and only if $s\in u^{-1}(V_{i})$ for some $i\in I$, that is, by Indexed Families of Sets and Their Union, Intersection and Product (ii), if and only if $s\in\bigcup_{i\in I}u^{-1}(V_{i})$. So the two sets are equal by Axiom of Extensionality for Classes.

Proof of (ii). Let $I\neq\emptyset$ and fix $i_{0}\in I$. The intersection $\bigcap_{i\in I}V_{i}$ is a set by The Union and Product of a Family of Sets Indexed by a Set, and the Intersection of a Family with an Inhabited Index Class, Are Sets (ii), and each of its elements lies in $V_{i_{0}}$ by Indexed Families of Sets and Their Union, Intersection and Product (iii), hence in $T$; so it is a subset of $T$. For every set $s$, by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii) and Indexed Families of Sets and Their Union, Intersection and Product (iii), $s\in u^{-1}(\bigcap_{i\in I}V_{i})$ if and only if $s\in S$ and $u(s)\in V_{i}$ for every $i\in I$. By (1), this holds if and only if $s\in u^{-1}(V_{i})$ for every $i\in I$: one direction is immediate, and conversely $s\in u^{-1}(V_{i_{0}})$ gives $s\in S$, while $s\in u^{-1}(V_{i})$ gives $u(s)\in V_{i}$ for each $i\in I$. By Indexed Families of Sets and Their Union, Intersection and Product (iii) the latter condition says $s\in\bigcap_{i\in I}u^{-1}(V_{i})$, and the two sets are equal by Axiom of Extensionality for Classes.

Proof of (iii). Let $V\subseteq T$. Then $T\setminus V\subseteq T$ by The Boolean Operations on Classes, Disjointness, and the Universal Class (i). For every set $s$, by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii) and The Boolean Operations on Classes, Disjointness, and the Universal Class (i), $s\in u^{-1}(T\setminus V)$ if and only if $s\in S$, $u(s)\in T$ and $u(s)\notin V$. For $s\in S$, $u(s)\in T$ holds, and $u(s)\notin V$ if and only if $s\notin u^{-1}(V)$, by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii). So $s\in u^{-1}(T\setminus V)$ if and only if $s\in S$ and $s\notin u^{-1}(V)$, that is, by The Boolean Operations on Classes, Disjointness, and the Universal Class (i), if and only if $s\in S\setminus u^{-1}(V)$; the two sets are equal by Axiom of Extensionality for Classes.

Proof of (iv). $T$ is a subset of $T$, and for every set $s$, $s\in u^{-1}(T)$ if and only if $s\in S$ and $u(s)\in T$, by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii), that is, if and only if $s\in S$. So $u^{-1}(T)=S$ by Axiom of Extensionality for Classes.