Removing isolated points from the support

Since $D=S\setminus C$ and $T=C\cup S=C\cup D$, the upper-left open is $(X\setminus C)\cap (X\setminus D)$ and the other two opens cover $X$. Intersections of open immersions are their fiber products, and morphisms out of the two covering opens that agree on their intersection glue uniquely. Therefore the displayed scheme square is cartesian and cocartesian. All four opens are quasi-compact and quasi-separated.

Apply localization to $D\subset X$ and to the same closed set $D\subset X\setminus C$. It is closed in the latter because $D$ is a finite set of closed points and is disjoint from $C$. Open excision identifies the two supported spectra by actual restriction, compatibly with their maps to the ambient spectra. Their long exact homotopy sequences give the two exact rows and the commutative square of maps.

For disjoint finite closed subsets $A,B\subset X$, the cover $X=(X\setminus A)\cup (X\setminus B)$ and Mayer–Vietoris identify the support spectrum for $A\cup B$ with the product of the support spectra for $A$ and $B$: take the fiber over $K(X\setminus (A\cup B))$ of the cartesian restriction square and use the universal property of the pullback. The two factors map to the union-support spectrum by the corresponding maps of support fibers. Iterating, including the empty case, identifies $K_D(X)$ with the finite product of the one-point support spectra. By point dévissage each zeroth group is $\mathbb{Z}$, and the inverse generator maps to the residue-sheaf class. Open excision gives the same identification on $X\setminus C$ and preserves these generators. This proves the displayed diagram and its generator normalization.

Each motivic stage has Zariski Mayer–Vietoris on the opens used here. Indeed, stage zero has it by [F2]. If stage $q$ has it, compare its Mayer–Vietoris square with the square of the weight-$q$ graded spectrum. The latter is cartesian by derived Zariski descent in [F3]; taking fibers in the natural graded triangle shows that the stage-$(q+1)$ square is cartesian. Induction gives the assertion through stage three. The same finite-support pullback argument as in step 3.1 therefore splits the supported stage and its stage-to-$K$ map into the corresponding one-point maps.

Let $y\in D$. It lies on a three-dimensional clopen component, and the union $X_3$ of those components is pure three-dimensional. The stagewise clopen product equivalence identifies the support tower at $y$ on $X$ with that on $X_3$: the other component contributes the fiber of an identity map. On $X_3$, [F3] makes the supported graded spectra in weights $0,1,2$ contractible, since all their homotopy groups vanish. The supported graded triangles therefore make the transitions from stage three to stage zero equivalences. Stage zero is supported $K$. Thus the actual one-point stage-three map is an equivalence. By step 4.1, so is the map for the finite set $D$ on $X$; the analogous map on $X\setminus C$ follows by stagewise open excision from its Mayer–Vietoris square. Taking ${\pi }_0$ proves the two stated isomorphisms.

By [F1], $j(i(\alpha ))=\alpha {\mathrm{\mid }}_{X\setminus T}=0$. Exactness of the lower row gives an element $a$ with $i(\alpha )=g(a)$. Commutativity gives $i(f(a))=g(a)$, hence $i(\beta )=0$. By step 3.1, $f(a)$ is a finite integral combination of residue classes at points of $D$; every such point is smooth and lies on a three-dimensional component. Therefore [F4] puts $f(a)$ in $F_{\mathrm{s}\mathrm{m},\mathrm{t}\mathrm{o}\mathrm{p}}^3{K}_0(X)$ and hence in $F_{\mathrm{m}\mathrm{o}\mathrm{t}}^3{K}_0(X)$. Since this is a subgroup, $\beta$ lies in it too. Natural restriction of the image filtration then makes its vanishing in ambient $K_0(X\setminus C)$ exactly the asserted kernel condition. For pure $X$, [F4] identifies the point subgroup with $F_{\mathrm{s}\mathrm{m}}^3$. No injectivity of $f$ or $g$ is used.