[proofplan] We argue by contradiction. From a Tucker labeling with no complementary edge, we build a continuous piecewise-affine map from the triangulated ball to $\mathbb{R}^d$ by sending label $+i$ to $e_i$ and label $-i$ to $-e_i$. The absence of complementary edges implies that no simplex has image containing $0$, so the map can be radially normalized to a continuous map into $S^{d-1}$. On the boundary, the Tucker antipodality condition makes this normalized map antipodal, contradicting the Borsuk-Ulam no-antipodal-extension obstruction. [/proofplan]

[step:Build the affine map determined by the signed coordinate labels] Let $e_1,\dots,e_d \in \mathbb{R}^d$ denote the standard coordinate vectors. Define the signed-coordinate map \begin{align*} \phi: \{\pm 1,\dots,\pm d\} \to \mathbb{R}^d \end{align*} by $\phi(i) = e_i$ and $\phi(-i) = -e_i$ for each $i \in \{1,\dots,d\}$. Assume, for contradiction, that no edge of $K$ is complementary. Define \begin{align*} F: |K| \to \mathbb{R}^d \end{align*} to be the unique map that is affine on each simplex of $K$ and satisfies \begin{align*} F(v) = \phi(\lambda(v)) \end{align*} for every vertex $v \in V(K)$. Since affine definitions agree on common faces, $F$ is a well-defined continuous piecewise-affine map on $|K| = B^d$. [/step]

[step:Show that the affine map never takes the value $0$]We prove that $0 \notin F(|K|)$. Let $\sigma$ be a simplex of $K$, and let $v_0,\dots,v_m$ be the vertices of $\sigma$. Since every pair of distinct vertices in a simplex spans an edge, the assumption that there is no complementary edge implies that there is no index $i \in \{1,\dots,d\}$ for which both labels $i$ and $-i$ occur among $\lambda(v_0),\dots,\lambda(v_m)$. Suppose $x \in \sigma$. Write $x$ in barycentric coordinates as \begin{align*} x = \sum_{j=0}^{m} a_j v_j \end{align*} where $a_j \geq 0$ for each $j$ and \begin{align*} \sum_{j=0}^{m} a_j = 1. \end{align*} By affine linearity of $F$ on $\sigma$, \begin{align*} F(x) = \sum_{j=0}^{m} a_j \phi(\lambda(v_j)). \end{align*} For each $i \in \{1,\dots,d\}$, define the sign $\varepsilon_i \in \{-1,0,1\}$ as follows: $\varepsilon_i = 1$ if some vertex of $\sigma$ has label $i$, $\varepsilon_i = -1$ if some vertex of $\sigma$ has label $-i$, and $\varepsilon_i = 0$ if neither label occurs. This is well-defined because $i$ and $-i$ cannot both occur on $\sigma$. Define \begin{align*} A_i = \sum_{\{j : |\lambda(v_j)| = i\}} a_j. \end{align*} Then $A_i \geq 0$ for each $i$, and \begin{align*} F(x) = \sum_{i=1}^{d} \varepsilon_i A_i e_i. \end{align*} If $F(x) = 0$, then every coordinate of the last expression is zero, so $\varepsilon_i A_i = 0$ for each $i$. Whenever $A_i > 0$, some vertex has label $i$ or $-i$, hence $\varepsilon_i \neq 0$, a contradiction. Therefore $A_i = 0$ for every $i$, and then \begin{align*} 1 = \sum_{j=0}^{m} a_j = \sum_{i=1}^{d} A_i = 0, \end{align*} which is impossible. Thus $F(x) \neq 0$ for every $x \in \sigma$. Since $\sigma$ was arbitrary, $0 \notin F(|K|)$.[/step]

[guided]The key point is that the only way a convex combination of signed coordinate vectors can equal $0$ is by using opposite signed vectors in some coordinate direction. We now make that statement precise on one simplex. Let $\sigma$ be a simplex of $K$, and let $v_0,\dots,v_m$ be its vertices. Because $\sigma$ is a simplex, every pair $\{v_j,v_k\}$ with $j \neq k$ is an edge of $K$. Our contradiction assumption says that no edge has labels $i$ and $-i$ for any $i \in \{1,\dots,d\}$. Therefore, on this one simplex, the labels cannot contain both $i$ and $-i$ for the same coordinate index $i$. Take any point $x \in \sigma$. By the definition of barycentric coordinates, there are [real numbers](/page/Real%20Numbers) $a_0,\dots,a_m$ such that $a_j \geq 0$ for each $j$, \begin{align*} \sum_{j=0}^{m} a_j = 1, \end{align*} and \begin{align*} x = \sum_{j=0}^{m} a_j v_j. \end{align*} Since $F$ is affine on $\sigma$, it respects barycentric combinations, so \begin{align*} F(x) = \sum_{j=0}^{m} a_j F(v_j) = \sum_{j=0}^{m} a_j \phi(\lambda(v_j)). \end{align*} Now group the terms by coordinate direction. For each $i \in \{1,\dots,d\}$, define $\varepsilon_i \in \{-1,0,1\}$ by setting $\varepsilon_i = 1$ if the label $i$ occurs on $\sigma$, setting $\varepsilon_i = -1$ if the label $-i$ occurs on $\sigma$, and setting $\varepsilon_i = 0$ if neither label occurs. This definition is unambiguous because $i$ and $-i$ cannot both occur on $\sigma$. Also define the non-negative coefficient \begin{align*} A_i = \sum_{\{j : |\lambda(v_j)| = i\}} a_j. \end{align*} Then the contribution in the $i$-th coordinate direction is exactly $\varepsilon_i A_i e_i$, and hence \begin{align*} F(x) = \sum_{i=1}^{d} \varepsilon_i A_i e_i. \end{align*} Assume that $F(x) = 0$. Since $e_1,\dots,e_d$ are the standard basis vectors of $\mathbb{R}^d$, equality to $0$ forces each coordinate coefficient to vanish: \begin{align*} \varepsilon_i A_i = 0 \end{align*} for every $i \in \{1,\dots,d\}$. If $A_i > 0$, then at least one vertex of $\sigma$ has label $i$ or $-i$, so by construction $\varepsilon_i$ is either $1$ or $-1$. Thus $\varepsilon_i A_i \neq 0$, a contradiction. Therefore $A_i = 0$ for every $i$. But the sets of indices $\{j : |\lambda(v_j)| = i\}$ partition $\{0,\dots,m\}$ as $i$ ranges from $1$ to $d$. Hence \begin{align*} \sum_{i=1}^{d} A_i = \sum_{j=0}^{m} a_j = 1. \end{align*} This contradicts $A_i = 0$ for all $i$. Therefore $F(x) \neq 0$ for every $x \in \sigma$. Since the simplex $\sigma$ was arbitrary, the map $F$ never takes the value $0$ on $|K|$.[/guided]

[step:Normalize the map to obtain a sphere-valued extension over the ball] Since $F(x) \neq 0$ for every $x \in B^d$, define \begin{align*} H: B^d \to S^{d-1} \end{align*} by \begin{align*} H(x) = \frac{F(x)}{|F(x)|}. \end{align*} The Euclidean norm map $y \mapsto |y|$ is continuous on $\mathbb{R}^d$, and $F$ avoids $0$, so $H$ is continuous. Thus $H$ is a continuous extension over the ball of its boundary restriction \begin{align*} G: S^{d-1} \to S^{d-1} \end{align*} defined by $G(x) = H(x)$ for $x \in S^{d-1}$. [/step]

[step:Use the Tucker boundary condition to make the boundary map antipodal] We show that $G(-x) = -G(x)$ for every $x \in S^{d-1}$. Let $x \in S^{d-1}$, and let $\sigma$ be a boundary simplex of $K$ containing $x$. Write the barycentric representation of $x$ in $\sigma$ as \begin{align*} x = \sum_{j=0}^{m} a_j v_j, \end{align*} where $v_0,\dots,v_m$ are the vertices of $\sigma$, $a_j \geq 0$, and \begin{align*} \sum_{j=0}^{m} a_j = 1. \end{align*} Because $K$ is an antipodal triangulation, $-\sigma$ is a boundary simplex with vertices $-v_0,\dots,-v_m$, and \begin{align*} -x = \sum_{j=0}^{m} a_j(-v_j). \end{align*} Using affine linearity of $F$ on boundary simplices and the Tucker condition $\lambda(-v_j) = -\lambda(v_j)$, we get \begin{align*} F(-x) = \sum_{j=0}^{m} a_j \phi(\lambda(-v_j)). \end{align*} Since $\phi(-\ell) = -\phi(\ell)$ for every $\ell \in \{\pm1,\dots,\pm d\}$, \begin{align*} F(-x) = -\sum_{j=0}^{m} a_j \phi(\lambda(v_j)) = -F(x). \end{align*} Therefore \begin{align*} G(-x) = \frac{F(-x)}{|F(-x)|} = \frac{-F(x)}{|F(x)|} = -G(x). \end{align*} Thus the boundary map $G: S^{d-1} \to S^{d-1}$ is antipodal. [/step]

[step:Contradict the Borsuk-Ulam obstruction] The map $H: B^d \to S^{d-1}$ is continuous, extends $G$, and its boundary restriction $G$ is antipodal. This contradicts the no-antipodal-extension form of the [Borsuk-Ulam theorem](/theorems/6462): there is no continuous map $H: B^d \to S^{d-1}$ whose boundary restriction satisfies $H(-x) = -H(x)$ for every $x \in S^{d-1}$ (citing a result not yet in the wiki: Borsuk-Ulam no-antipodal-[extension theorem](/theorems/59)). The contradiction came from assuming that no complementary edge exists. Hence there must be an edge $\{v,w\}$ of $K$ and an index $i \in \{1,\dots,d\}$ such that the labels of its endpoints are $i$ and $-i$. This is the desired complementary edge. [/step]