Perfect K theory localization and open excision

The localization sequence is [F1] applied to $j:U\hookrightarrow X$ and its full kernel ${\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{f}}_T(X)$. This identifies the category-defined $K_T(X)$ with the actual fiber, without assuming that identification in its definition or comparing with a different K-theory model.

For a two-open cover $X=U\cup V$, quasi-separatedness makes $U\cap V$ quasi-compact, so [F1] applies both to $U\subseteq X$ and $U\cap V\subseteq V$. Put $T=X\setminus U$, a closed subset of $X$ contained in $V$. Their fiber terms are $K_T(X)$ and $K_T(V)$, and [F2] identifies the induced map between these fibers with an equivalence. The coherent restriction square therefore has equivalent horizontal fibers. To check it is cartesian, put $P=K(U){\times }_{K(U\cap V)}K(V)$. The induced map $K(X)\to P$ lies over $K(U)$; its fiber map over that base is exactly $K_T(X)\to K_T(V)$ because pullbacks commute. Thus its map on fibers is an equivalence and its map on bases is the identity. The long exact sequences of spectrum homotopy groups give isomorphisms ${\pi }_{i}K(X)\to {\pi }_{i}P$ for all integers $i$, hence an equivalence of spectra. This proves Mayer–Vietoris. Repeated binary covering decompositions give the finite covering form.

For the stated support excision take $W$ in [F2]. Its support equivalence is induced by actual restriction and is compatible with inclusions. Because $X\setminus T$ is quasi-compact, so is $W\setminus T=W\cap (X\setminus T)$ by quasi-separatedness; hence both support terms are also the fibers in step 1.1 with their natural maps. This proves the entire interface, with every original qcqs, closed-support and quasi-compact-open condition retained.