Homotopy-fiber weights reduce to componentwise ratios

Replace $f$ by the mapping-path fibration $p_f$ in [F2]. For a component $x$ mapping to $y$, let $P_x\subseteq {\pi }_0(F_y)$ be the components mapping to $x$. A point $z\in P_x$ bases the fiber over $y$; $p_f$ is a Hurewicz, hence Serre, fibration, and $j_f$ identifies its total-space homotopy groups with those of $X$, so [F3] applies. Its pointed-set tail identifies $P_x$ as one orbit of ${\pi }_1(Y,y)$, with stabilizer the image of ${\pi }_1(X,x)\to {\pi }_1(Y,y)$; hence $\mathrm{\mid }{P}_x\mathrm{\mid }=\mathrm{\mid }{\pi }_1(Y,y)\mathrm{\mid }\mathrm{/}\mathrm{\mid }\mathrm{im}({\pi }_1(X,x)\to {\pi }_1(Y,y))\mathrm{\mid }$. For $z\in P_x$, the group tail includes ${\pi }_2(Y,y)\to {\pi }_1(F_y,z)\to {\pi }_1(X,x)\to {\pi }_1(Y,y)$, so ${\pi }_1(F_y,z)$ need not be the kernel of the last map. All groups are finite, and exactness makes the fiber groups vanish in high degrees. Expanding each finite group order as the product of its kernel and image orders through this entire tail, and using the displayed orbit count at its pointed-set end, gives $\mathrm{\mid }{P}_x\mathrm{\mid }=r_f(x,y){\prod }_{j>0}\mathrm{\mid }{\pi }_j(F_y,z){\mathrm{\mid }}^{(-1{)}^{j+1}}\mathrm{.}$ The action transports components within $P_x$, so their higher homotopy groups have the same orders.

The total weight of the fiber components mapping to $x$ is $\mathrm{\mid }{P}_x\mathrm{\mid }$ times the weight of one of them, since their homotopy-group orders agree by step 1.1. Multiplying its formula by ${\prod }_{j>0}\mathrm{\mid }{\pi }_j(F_y,z){\mathrm{\mid }}^{(-1{)}^j}$ cancels the fiber factors and gives $r_f(x,y)$.

Substitute the componentwise identity of step 2.1 into the defining fiber sum for $f_{!}$ and group terms by their image component $x$. Each $P_x$ is finite by step 1.1, and only finitely many $x$ have $\alpha (x)\mathrm{\ne }0$, so this gives the stated ratio formula and the finiteness of $f_{!}(\alpha )(y)$.