Supported perfect complexes are invariant under an open neighbourhood

Let $F\in {\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{f}}_T(V)$. Exactness and the stalk formula in [F1] show that the cohomology of $v_{!}F$ equals that of $F$ on $V$ and is zero off $V$, so its support is contained in $T$. On the open $V$ it is perfect because $v^{\ast }{v}_{!}F=F$; on the open $X\setminus T$ it is acyclic, hence equivalent to the zero perfect object. These opens cover $X$ because $T\subseteq V$. Thus local perfection in [F2] makes $v_{!}F$ perfect on $X$. This is why closed ambient support, rather than preservation of arbitrary quasi-coherent sheaves by $v_{!}$, suffices.

The degreewise unit identifies $v^{\ast }{v}_{!}$ with the identity. For $E\in {\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{f}}_T(X)$ the actual counit $v_{!}{v}^{\ast }E\to E$ is a quasi-isomorphism: at points of $V$ it is an isomorphism on complex stalks, and outside $V$ the source is zero while the target is acyclic since those points lie outside $T$. Hence this natural transformation becomes an equivalence in the supplied infinity-localization. Both functors are exact by [F1], and step 1.1 verifies their restrictions have precisely the claimed perfect/support domain. They therefore give mutually inverse exact functors of the actual enhanced full subcategories.

Apply the functor in [F2] to this exact equivalence. The square formed by the two support inclusions and the global/support restriction functors commutes already on complexes, and the common-family realization of the supplied interface represents the same restriction. Thus the K equivalence is compatible with the claimed maps. Extension by zero is used solely as an inverse on these support categories, without requiring it to preserve the common K-flat presentation or to be a global scheme pullback.