Finite relative homotopy lifting across a weak equivalence

The ordinary cylinder formulas and embeddings in [F1] are valid without separation assumptions. On $L\times I$ define a homotopy starting at $kv{\mathrm{\mid }}_L$ by $B(l,s)=\{\begin{array}{ll}kT(l,2s),& 0\le s\le 1\mathrm{/}2,\\ [u(l),2s-1],& 1\mathrm{/}2\le s\le 1.\end{array}$ At $s=1\mathrm{/}2$ the two values are $kfu(l)=[u(l),0]$, so finite closed pasting gives continuity. At $s=1$, its value is $ju(l)$. By [F3], extend $B$ from $L$ to a homotopy $V:K\times I\to M$ starting at $kv$. Put $b=V(-,1)$, so $b{\mathrm{\mid }}_L=ju$. Projection by $r$ on $L$ gives the precise formula $rB(l,s)=T(l,\min (2s,1))$.

We compress this $b$ into $j(P)$ rel $L$ using only finitely many source-cell choices. Write $D_d=L\cup K^d$, with $D_{-1}=L$. Suppose a current map $b_{d-1}:K\to M$ equals $ju$ on $L$ and takes $D_{d-1}$ into $j(P)$. For a relative $d$-cell, its characteristic disk followed by $b_{d-1}$ has boundary in $j(P)$. If $d\ge 1$, mark a fixed boundary point and use its actual image $j(p)$ as basepoint. The relative class is null by [F1], so [F2] gives a compression into $j(P)$ fixing all boundary points. If $d=0$, the component-surjectivity of $j$ in [F1] gives a path from the image of that vertex into $j(P)$. There are only finitely many relative cells in this dimension, so [F4] supplies their finitely many compression witnesses.

Glue these disk homotopies to the stationary homotopy on $D_{d-1}$. They agree on every attaching identification, because disk boundaries were fixed. By [F4], $D_d$ is the quotient of $D_{d-1}$ and the finitely many characteristic $d$-disks by their boundary identifications, and the product of this quotient with $I$ is again quotient. The compatible continuous homotopies on those pieces therefore descend to a continuous homotopy on $D_d\times I$, even with an infinite-dimensional $L$. Extend it to $K$ by [F3] for $(K,D_d)$. Its endpoint $b_d$ sends $D_d$ into $j(P)$ and retains $ju$ on $L$. Starting from $b_{-1}=b$, perform these stages through the maximum dimension of the finite set of relative cells. The HEP is specified without choices, and the remaining witness selections are a finite sequence. Concatenation gives a continuous $C:K\times I\to M$ from $b$ to a map into $j(P)$, fixed on $L$. If there are no relative cells, take $C$ constant. Its endpoint factors continuously through the embedded subspace $j(P)$ by [F1]; denote the resulting map by $w:K\to P$. It satisfies $w{\mathrm{\mid }}_L=u$.

Concatenate $V$ and $C$ on the two half-intervals and compose with $r$: $J(x,t)=\{\begin{array}{ll}rV(x,2t),& 0\le t\le 1\mathrm{/}2,\\ rC(x,2t-1),& 1\mathrm{/}2\le t\le 1.\end{array}$ The seam is $rb(x)$, the initial value is $v(x)$ and the final value is $fw(x)$. On $L$, step 1.1 gives $J(l,t)=T(l,\min (4t,1))$ for the first half; on the second half $C$ is constantly $ju$, and its projection is $fu(l)=T(l,1)$. Hence the displayed formula holds for all $t$. In particular any stationary $T$ track stays stationary, and strict commuting data yield the rel-$L$ conclusion.

If $P$ is empty, weak equivalence forces $Y$ empty. Existence of $v$ then forces $K$ and $L$ empty, and the unique maps satisfy the result. Empty $L$ otherwise imposes no boundary condition; zero relative cells give $w=u$ and the same reparametrized $T$. A zero-cell uses a path, and a one-cell compression fixes its two possibly distinct endpoints by [F2]. No choice is made on all of $L$: its homotopy is prescribed as data, and only the finitely many cells outside it request witnesses. The formula in step 4.1 checks $t=0,1\mathrm{/}4,1\mathrm{/}2,1$, so the plateau in $\lambda$ is intentional and no claim of extending the original time parametrization is made. This proves every assertion choice-free.