High relative cells do not change lower homotopy

By [F6] the domain is compact, and [F2] supplies a finite subcomplex $T\subseteq X$ containing its image. If $T\subseteq A$, the constant homotopy suffices. Otherwise take a cell $e$ of largest dimension among the finitely many cells of $T$ outside $A$. Its dimension $k$ is at least $n>i$. The complement $W=T\setminus e$ is a subcomplex: a cell in $A$ has its entire closure in $A$ and cannot meet $e$; any other remaining cell has dimension at most $k$, and its boundary consists of strictly lower-dimensional cells, so cannot meet the distinct $k$-cell $e$. Thus $T$ is obtained from $W$ by this one $k$-cell, even if $T\cap A$ has cells of dimension greater than $k$.

Apply [F3] to deform $u$ into $W$. Since $W$ contains $T\cap A$, this deformation fixes every point of the original $u^{-1}(A)$. Repeat with the current map and the smaller finite subcomplex $W$, each time choosing a cell of maximal dimension outside $A$. The number of such cells strictly decreases, so finitely many applications end in $T\cap A$. At every stage the original points of $u^{-1}(A)$ still have their original values in $A$, hence are fixed by every subsequent deformation. Concatenating the finite list gives the claimed homotopy. Its choices form only a finite sequence [F6]; no sequence over all maps or cells of $X$ is selected. This proof also works for $i=0$.

A Euclidean disk and cube are homeomorphic as pairs: on the centered cube the radial map $v\mapsto v\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_2$, with zero sent to zero, has inverse $w\mapsto w\mathrm{\parallel }w{\mathrm{\parallel }}_2\mathrm{/}\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus step 2.1 applies to each disk map of dimension less than $n$, fixing its boundary when that boundary maps into $A$. By [F1] the pair is $(n-1)$-connected. In dimension zero it gives a path from any point of $X$ into $A$, proving surjectivity on components. If $n\ge 2$, a path in $X$ between two points of $A$ has dimension one less than $n$; compress it by step 2.1 fixing both endpoints to obtain a path in $A$. Hence two $A$ components cannot merge in $X$, proving component injectivity.

Let $a\in A$ and $1\le i<n$. A based $i$-cube in $X$ has its boundary in $A$, so step 2.1 compresses it into $A$ while fixing that boundary at $a$. The resulting based class maps to the original class; thus the inclusion is surjective on ${\pi }_i$. If also $i+1<n$ and a based cube in $A$ represents an element of the kernel, take its based nullhomotopy $I^i\times I\to X$. The entire boundary of this $(i+1)$-cube lies in $A$: the bottom is the given cube, the top is constant, and the side boundary is constantly $a$. Step 2.1 compresses this map into $A$ fixing that whole boundary, giving a based nullhomotopy in $A$. Hence the induced homomorphism has trivial kernel and is injective. This works for the nonabelian degree-one group as well and uses no vertex restriction on $a$.

If there are no relative cells the compression is constant. If $A$ is empty, the no-low-cell hypothesis forces $X$ empty: a nonempty CW complex contains a zero-cell, since descending through the nonempty boundary image of any positive-dimensional characteristic disk eventually reaches dimension zero. Thus there is no map of a nonempty cube into $X$ in this case, and the component map is the bijection of empty sets. The case $n=1$ asserts only the zero-disk compression and component surjectivity, with no positive-degree surjection at the undefined relative degree zero. At $i=n-1\ge 1$ surjectivity was proved, but injectivity would require a dimension-$n$ compression, which was not assumed or claimed. Equal endpoints, constant maps and nonregular attaching maps retain their fixed inverse-image data in step 2.1. Steps 3.1 and 3.2 prove all the consequences, and every selection was finite.