Every Whitehead class is realized by a finite CW homotopy equivalence

Choose $A=(a_{ij})\in G{L}_m(R)$ with $R=\mathbb{Z}[\pi ]$ representing $\eta$, allowing $m=1$ and $A=I$ for $\eta =0$. Choose an even integer $n\ge \max (4,\dim X+1)$. Attach $m$ $n$-cells by constant maps at the chosen base vertex to form $X_n=X\vee {\bigvee }_{j=1}^m{S}_j^n$. By [F6], ${\pi }_1{X}_n=\pi$ and no lower relative homotopy is introduced.

Each $a_{ij}$ is a finite integral sum of elements of $\pi$. The $j$th wedge sphere, preceded by a based whisker representing a group element and repeated by signed pinch maps, realizes its coefficient in ${\pi }_n(X_n)$. Since $n\ge 2$, this is an abelian group and finite sums of these maps are represented by based maps $S^n\to X_n$. For each column $j$ choose one such map $f_j$ whose relative coordinates are $(a_{1j},\dots ,a_{mj}{)}^{\mathsf{T}}$; only finitely many explicit choices are needed. Attach $m$ $(n+1)$-cells along the $f_j$ to form finite $Y$. The right-module column convention and [F2] make the relative cellular differential exactly $A$.

By [F6] the map $i:X\hookrightarrow Y$ induces a component bijection and an isomorphism on ${\pi }_1$. The inverse image $\tilde{X}$ of $X$ in the universal cover $\tilde{Y}$ is the connected universal cover of $X$: the ${\pi }_1$-isomorphism makes the restricted cover connected, and a loop in it maps to a null loop in $Y$ and hence is null in $X$. Both $\tilde{X}$ and $\tilde{Y}$ are simply connected. Their relative cellular chain complex is $0\to R^m\stackrel{A}{\to }{R}^m\to 0$ in degrees $n+1,n$, so it has zero homology in every degree because $A$ is invertible. By [F4], $H_k(\tilde{Y},\tilde{X};\mathbb{Z})=0$ for every $k$.

Suppose some ${\pi }_k(\tilde{Y},\tilde{X})$ were nonzero and take the least such $k$. Since the pair has no relative cells below $n\ge 4$, it is at least $3$-connected and $k\ge n\ge 4$; its base $\tilde{X}$ is simply connected. The choice-free comparison [F3] gives ${\pi }_k(\tilde{Y},\tilde{X})\cong H_k(\tilde{Y},\tilde{X};\mathbb{Z})=0$, a contradiction. Hence all relative homotopy groups vanish. Coverings induce isomorphisms on higher homotopy groups by lifting spheres and homotopies; together with the ${\pi }_1$-isomorphism of step 3.1, $i$ is a weak homotopy equivalence.

Both $X$ and $Y$ are finite, so the finite choice-free clause [F5] makes $i$ a homotopy equivalence. Its relative complex has upper degree $q=n+1$, and $n$ is even, so $(-1{)}^{q+1}=(-1{)}^{n+2}=+1$. Therefore [F7] gives $i_{\ast }^{-1}\tau (i)=[A]=\eta$. For finitely many connected components, repeat the construction separately on each component and take their finite disjoint union; no component transport or infinite choice is involved. ∎