[proofplan] We prove both implications. First, assuming the no-antipodal-extension statement, a Tucker labeling without a complementary edge gives a piecewise affine map into $\mathbb{R}^d \setminus \{0\}$ by sending labels $\pm i$ to $\pm e_i$; after normalization this becomes a forbidden antipodal boundary extension. Conversely, assuming [Tucker's lemma](/theorems/6480), a forbidden continuous antipodal boundary extension can be sampled on finer and finer antipodal triangulations; Tucker's complementary edges then force two nearby points to have one coordinate uniformly separated with opposite signs, contradicting continuity. [/proofplan] [step:Construct a nonvanishing affine map from a Tucker labeling without complementary edges] Assume the no-antipodal-extension statement in dimension $d$. Let $K$ be a finite triangulation of $B^d$ whose boundary triangulation is invariant under the antipodal map, and let \begin{align*} \lambda: K^{(0)} \to \{\pm 1, \dots, \pm d\} \end{align*} be a Tucker labeling satisfying $\lambda(-v) = -\lambda(v)$ for every boundary vertex $v \in K^{(0)} \cap S^{d-1}$. Suppose, toward contradiction, that no edge of $K$ has endpoint labels summing to $0$. Let $e_1,\dots,e_d$ denote the standard basis of $\mathbb{R}^d$. Define the sign map $\varepsilon: \{\pm 1,\dots,\pm d\} \to \{-1,1\}$ by $\varepsilon(k)=1$ for $k>0$ and $\varepsilon(k)=-1$ for $k<0$. Define a map on vertices \begin{align*} \Phi_0: K^{(0)} \to \mathbb{R}^d \end{align*} by $\Phi_0(v)=\varepsilon(\lambda(v)) e_{|\lambda(v)|}$ for $v \in K^{(0)}$. Let \begin{align*} \Phi: B^d &\to \mathbb{R}^d \end{align*} be the unique continuous map that is affine on every simplex of $K$ and agrees with $\Phi_0$ on $K^{(0)}$. We claim that $\Phi(x) \neq 0$ for every $x \in B^d$. Indeed, let $x$ lie in a simplex $\sigma$ of $K$ with vertices $v_0,\dots,v_m$. Let $\alpha_0,\dots,\alpha_m \in [0,1]$ be the barycentric coordinates of $x$ in $\sigma$, so that \begin{align*} \sum_{j=0}^{m} \alpha_j = 1. \end{align*} Then \begin{align*} \Phi(x) = \sum_{j=0}^{m} \alpha_j \Phi_0(v_j). \end{align*} If $\Phi(x)=0$, then $0$ lies in the convex hull of the set $\{\Phi_0(v_0),\dots,\Phi_0(v_m)\}$. Since no edge of $\sigma$ is complementary, this set contains no pair $e_i,-e_i$. For each coordinate $i$, all nonzero contributions to the $i$-th coordinate therefore have the same sign. The equality $\Phi(x)=0$ forces the total barycentric weight of every vertex labeled by $+i$ or $-i$ to be $0$. This holds for every $i$, so all $\alpha_j$ are $0$, contradicting $\sum_j \alpha_j = 1$. Hence $\Phi(B^d) \subset \mathbb{R}^d \setminus \{0\}$. [guided] The goal is to turn a combinatorial labeling into a continuous map. The natural dictionary is that the label $+i$ should mean the vertex $e_i$, and the label $-i$ should mean the vertex $-e_i$ of the cross-polytope in $\mathbb{R}^d$. Let $e_1,\dots,e_d$ be the standard basis of $\mathbb{R}^d$. Define the sign map $\varepsilon: \{\pm 1,\dots,\pm d\} \to \{-1,1\}$ by $\varepsilon(k)=1$ for $k>0$ and $\varepsilon(k)=-1$ for $k<0$. Define \begin{align*} \Phi_0: K^{(0)} \to \mathbb{R}^d \end{align*} by $\Phi_0(v)=\varepsilon(\lambda(v)) e_{|\lambda(v)|}$ for $v \in K^{(0)}$. Because $K$ is a triangulation of $B^d$, every point of each simplex has barycentric coordinates. We therefore extend $\Phi_0$ affinely over every simplex and obtain a continuous piecewise affine map \begin{align*} \Phi: B^d &\to \mathbb{R}^d. \end{align*} The key point is that $\Phi$ never takes the value $0$. Fix $x \in B^d$, and let $\sigma$ be a simplex of $K$ containing $x$. Write the vertices of $\sigma$ as $v_0,\dots,v_m$, and write the barycentric coordinates of $x$ as $\alpha_0,\dots,\alpha_m \in [0,1]$. Thus \begin{align*} \sum_{j=0}^{m} \alpha_j = 1 \end{align*} and \begin{align*} \Phi(x) = \sum_{j=0}^{m} \alpha_j \Phi_0(v_j). \end{align*} Suppose $\Phi(x)=0$. Since no edge of $K$ has complementary labels, no two vertices of the simplex $\sigma$ are labeled $+i$ and $-i$ for the same $i$. Hence, in each coordinate direction, all nonzero contributions to the coordinate of $\Phi(x)$ have the same sign. A sum of nonnegative multiples of $e_i$ cannot cancel unless every coefficient in that coordinate is $0$, and the same is true for sums of nonnegative multiples of $-e_i$. Therefore every barycentric coefficient attached to every vertex of $\sigma$ must be $0$. This contradicts the barycentric identity $\sum_j \alpha_j=1$. Thus $\Phi(x)\neq 0$ for all $x \in B^d$. [/guided] [/step] [step:Normalize the affine map to contradict the no-antipodal-extension statement] Since $\Phi(B^d) \subset \mathbb{R}^d \setminus \{0\}$, define \begin{align*} H: B^d \to S^{d-1} \end{align*} by $H(x)=\Phi(x)/|\Phi(x)|$ for $x \in B^d$. The map $H$ is continuous because $\Phi$ is continuous and $|\Phi(x)|>0$ for every $x \in B^d$. For $x \in S^{d-1}$, the boundary triangulation is antipodally invariant, and the antipodal map $A: S^{d-1} \to S^{d-1}$, defined by $A(x)=-x$, is simplicial on the boundary triangulation. If $x$ lies in a boundary simplex with vertices $v_0,\dots,v_m$ and barycentric coordinates $\alpha_0,\dots,\alpha_m$, then affinity of $A$ on that simplex implies that $-x$ lies in the boundary simplex with vertices $-v_0,\dots,-v_m$ and the same barycentric coordinates. Using $\lambda(-v_j)=-\lambda(v_j)$, we get \begin{align*} \Phi(-x) = -\Phi(x). \end{align*} Therefore \begin{align*} H(-x) = -H(x) \end{align*} for every $x \in S^{d-1}$. This contradicts the no-antipodal-extension statement. Hence every Tucker labeling has a complementary edge. [/step] [step:Use a forbidden antipodal extension to label fine triangulations] Now assume Tucker's lemma in dimension $d$. Suppose, toward contradiction, that the no-antipodal-extension statement fails. Then there exists a continuous map \begin{align*} g: B^d &\to S^{d-1} \end{align*} such that $g(-x)=-g(x)$ for every $x \in S^{d-1}$. Choose a finite triangulation $K_1$ of $B^d$ whose boundary subcomplex triangulates $S^{d-1}$ and is invariant under the antipodal map, with the antipodal map simplicial on the boundary; for example, take the cone from $0$ over any finite antipodal triangulation of $S^{d-1}$. Let $K_n$ be the $(n-1)$-fold barycentric subdivision of $K_1$. Barycentric subdivision preserves antipodal invariance on the boundary, because the antipodal map sends each boundary face to the opposite boundary face and therefore sends barycentres of faces to barycentres of opposite faces. The mesh of iterated barycentric subdivisions tends to $0$, so $(K_n)_{n=1}^{\infty}$ is a sequence of finite triangulations of $B^d$ with antipodally invariant boundary and mesh tending to $0$. For each vertex $v \in K_n^{(0)}$, write \begin{align*} g(v) = (g_1(v),\dots,g_d(v)) \in S^{d-1}. \end{align*} Choose an index $i_n(v) \in \{1,\dots,d\}$ for which $|g_{i_n(v)}(v)|$ is maximal among $|g_1(v)|,\dots,|g_d(v)|$, using any fixed tie-breaking rule. Define \begin{align*} \lambda_n: K_n^{(0)} \to \{\pm 1,\dots,\pm d\} \end{align*} by setting $\lambda_n(v)=+i_n(v)$ when $g_{i_n(v)}(v) \geq 0$ and $\lambda_n(v)=-i_n(v)$ when $g_{i_n(v)}(v)<0$. Because $g(v) \in S^{d-1}$, at least one coordinate has absolute value at least \begin{align*} \frac{1}{\sqrt d}. \end{align*} Hence \begin{align*} |g_{i_n(v)}(v)| \geq \frac{1}{\sqrt d} \end{align*} for every vertex $v$. For a boundary vertex $v \in K_n^{(0)} \cap S^{d-1}$, the equality $g(-v)=-g(v)$ implies that the same tie-breaking rule chooses $i_n(-v)=i_n(v)$ and gives $\lambda_n(-v)=-\lambda_n(v)$. Thus $\lambda_n$ is a Tucker labeling of $K_n$. [guided] We now reverse the construction. Instead of starting from labels and building a map, we start from a continuous map \begin{align*} g: B^d &\to S^{d-1} \end{align*} and build labels on very fine triangulations. The reason for using finer and finer triangulations is that a complementary edge will then have endpoints that are very close, so continuity of $g$ will force their images to be close. First choose a finite triangulation $K_1$ of $B^d$ whose boundary subcomplex triangulates $S^{d-1}$ and is invariant under the antipodal map, with the antipodal map simplicial on the boundary. One concrete construction is to take the cone from $0$ over a finite antipodal triangulation of $S^{d-1}$. For each $n$, let $K_n$ be the $(n-1)$-fold barycentric subdivision of $K_1$. Barycentric subdivision preserves antipodal invariance on the boundary because opposite faces have opposite barycentres, and the mesh of iterated barycentric subdivisions tends to $0$. At a vertex $v \in K_n^{(0)}$, write \begin{align*} g(v) = (g_1(v),\dots,g_d(v)). \end{align*} Since $g(v) \in S^{d-1}$, we have \begin{align*} \sum_{j=1}^{d} |g_j(v)|^2 = 1. \end{align*} If every coordinate satisfied \begin{align*} |g_j(v)| < \frac{1}{\sqrt d}, \end{align*} then the sum of the squares would be strictly less than $1$, which is impossible. Therefore at least one coordinate has absolute value at least \begin{align*} \frac{1}{\sqrt d}. \end{align*} Choose an index $i_n(v)$ for which $|g_{i_n(v)}(v)|$ is maximal, using a fixed tie-breaking rule. Define the label by the sign of that coordinate: \begin{align*} \lambda_n: K_n^{(0)} \to \{\pm 1,\dots,\pm d\} \end{align*} by setting $\lambda_n(v)=+i_n(v)$ when $g_{i_n(v)}(v) \geq 0$ and $\lambda_n(v)=-i_n(v)$ when $g_{i_n(v)}(v)<0$. This gives the uniform estimate \begin{align*} |g_{i_n(v)}(v)| \geq \frac{1}{\sqrt d}. \end{align*} It remains to check the Tucker boundary rule. If $v \in S^{d-1}$ is a boundary vertex, then the assumed antipodality of $g$ gives $g(-v)=-g(v)$. Taking absolute values does not change under multiplication by $-1$, so the list of absolute coordinate values at $-v$ is the same as the list at $v$. Since the tie-breaking rule is fixed once and for all, it chooses the same index at $v$ and $-v$. The sign is reversed, so $\lambda_n(-v)=-\lambda_n(v)$. Thus each $\lambda_n$ is a valid Tucker labeling. [/guided] [/step] [step:Apply Tucker's lemma and pass to a fixed coordinate] By Tucker's lemma applied to $K_n$ and $\lambda_n$, for each $n$ there is an edge $\{a_n,b_n\}$ of $K_n$ and an index $j_n \in \{1,\dots,d\}$ such that \begin{align*} \lambda_n(a_n) = +j_n \end{align*} and \begin{align*} \lambda_n(b_n) = -j_n. \end{align*} Since the set $\{1,\dots,d\}$ is finite, there is an index $j \in \{1,\dots,d\}$ and an infinite subsequence, relabeled as the original sequence, for which $j_n=j$ for every $n$ in the subsequence. For this subsequence, \begin{align*} g_j(a_n) \geq \frac{1}{\sqrt d} \end{align*} and \begin{align*} g_j(b_n) \leq -\frac{1}{\sqrt d}. \end{align*} Because $\{a_n,b_n\}$ is an edge of $K_n$, the distance $|a_n-b_n|$ is bounded above by the mesh of $K_n$. Hence \begin{align*} |a_n-b_n| \to 0. \end{align*} [/step] [step:Use compactness and continuity to obtain the contradiction] The sequence $(a_n)_{n=1}^{\infty}$ lies in the compact set $B^d$, so it has a convergent subsequence. Passing to this subsequence and relabeling, let $x_* \in B^d$ satisfy \begin{align*} a_n \to x_*. \end{align*} Since $|a_n-b_n|\to 0$, we also have \begin{align*} b_n \to x_*. \end{align*} The coordinate projection \begin{align*} \pi_j: S^{d-1} \to \mathbb{R} \end{align*} defined by $\pi_j(y)=y_j$ for $y \in S^{d-1}$ is continuous, so $g_j=\pi_j \circ g$ is continuous on $B^d$. Taking limits in the two inequalities gives \begin{align*} g_j(x_*) \geq \frac{1}{\sqrt d} \end{align*} and \begin{align*} g_j(x_*) \leq -\frac{1}{\sqrt d}. \end{align*} This is impossible. Therefore no such map $g: B^d \to S^{d-1}$ exists. We have shown that the no-antipodal-extension statement implies Tucker's lemma and that Tucker's lemma implies the no-antipodal-extension statement. Hence the two statements are equivalent for every $d \geq 1$. [/step]