Proof of Precomposition Commutes with Pointwise Operations, and Sections of Functions Are Precompositions

In each identity of functions both sides are functions on the same set, so it suffices to show that their values agree at every point (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (i)); the value of a composite at $v$ is $(f\circ T)(v)=f(T(v))$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)). Let $v\in V$.

Proof of (i). By Positive Part, Negative Part and Absolute Value of a Real-Valued Function (i),

$$(f\circ T)^{+}(v)=\max\{f(T(v)),0\}=f^{+}(T(v)),$$

and in the same way $(f\circ T)^{-}(v)=\max\{-f(T(v)),0\}=f^{-}(T(v))$ and $|f\circ T|(v)=|f(T(v))|=|f|(T(v))$ (Positive Part, Negative Part and Absolute Value of a Real-Valued Function (ii), Positive Part, Negative Part and Absolute Value of a Real-Valued Function (iii)).

Proof of (ii). By Sums, Multiples and Products of Functions into [0,∞] (i) and Sums, Multiples and Products of Functions into [0,∞] (ii),

$$\bigl((f+g)\circ T\bigr)(v)=f(T(v))+g(T(v))=(f\circ T+g\circ T)(v),\qquad\bigl((cf)\circ T\bigr)(v)=c\,f(T(v))=\bigl(c\,(f\circ T)\bigr)(v).$$

Since $v\in T^{-1}(A)$ if and only if $T(v)\in A$ (Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii)), $\mathbf{1}_{A}(T(v))$ and $\mathbf{1}_{T^{-1}(A)}(v)$ are both $1$ or both $0$ (Indicator Function of a Subset (i)). Finally, $(h\circ T)(v)=h(T(v))=\sup_{m}f_{m}(T(v))=\sup_{m}(f_{m}\circ T)(v)$, by the hypothesis on $h$ at the point $T(v)\in W$.

Proof of (iii). By Dyadic Approximations of a Map into [0,∞] (i), $s_{m}(g)(T(v))$ and $s_{m}(g\circ T)(v)$ are given by the same expression in the number $g(T(v))=(g\circ T)(v)$, so they are equal.

Proof of (iv). For $w'\in W$, $f_{v}(w')=f(v,w')=f(i_{v}(w'))=(f\circ i_{v})(w')$ by Sections of Subsets and Functions on a Product (ii); likewise $f^{w}(v')=f(v',w)=(f\circ i^{w})(v')$ for $v'\in V$.