[proofplan] We encode the labelling as a simplicial map from $K$ into the boundary complex of the $m$-dimensional crosspolytope, with the absence of complementary edges exactly ensuring that every simplex has a legal image. For $d=0$ the assertion is immediate from the single vertex of $B^0$, so the main argument treats $d\geq 1$. We then double the finite triangulated ball along its antipodal boundary to obtain a finite antipodal triangulation of $S^d$, extend the labelling antipodally to the double, and apply the separately established sphere form of Ky Fan's parity lemma. The resulting alternating $d$-simplex in the double corresponds to an alternating $d$-simplex in the original ball, and such a simplex uses $d+1$ distinct absolute labels, forcing $m \geq d+1$. [/proofplan] [step:Handle the zero dimensional case] If $d=0$, then $B^0$ consists of one point and $K$ has exactly one $0$-simplex, say $\{v\}$. Since $m\geq 1$ by hypothesis, the label $\lambda(v)$ exists and the simplex $\{v\}$ is alternating by the definition of an alternating $0$-simplex. Also $m\geq 1=d+1$. Thus the theorem holds when $d=0$. [/step] [step:Build the signed label complex and the induced simplicial map] Assume for the rest of the proof that $d\geq 1$. Let $C_m$ denote the boundary complex of the $m$-dimensional crosspolytope. Its vertex set is \begin{align*} V(C_m)=\{+1,-1,\dots,+m,-m\}, \end{align*} and a finite subset $S \subset V(C_m)$ is a simplex of $C_m$ exactly when $S$ contains no pair $\{+i,-i\}$. Define a vertex map $L_0: V(K) \to V(C_m)$ by sending each vertex $v \in V(K)$ to the signed label $\lambda(v) \in V(C_m)$. Because $K$ has no complementary edge, no simplex of $K$ contains two vertices labelled $+i$ and $-i$ for the same $i$. Indeed, if a simplex contained such two vertices, then the edge joining them would be a complementary edge. Hence the image under $L_0$ of every simplex of $K$ is a simplex of $C_m$. Therefore $L_0$ extends uniquely to a simplicial map \begin{align*} L: K \to C_m. \end{align*} Moreover, on the boundary subcomplex $\partial K$, the map $L$ is antipodal in the sense that \begin{align*} L(-v)=-L(v) \end{align*} for every boundary vertex $v$. [guided] The target complex is chosen so that its simplices record exactly the label sets that do not contain an opposite pair. Let $C_m$ be the boundary complex of the $m$-dimensional crosspolytope, with vertices labelled by the signed integers $\pm 1,\dots,\pm m$. A set of signed labels spans a simplex of $C_m$ precisely when it does not contain both $+i$ and $-i$ for any $i$. We define the vertex map $L_0: V(K) \to V(C_m)$ by the rule that $L_0(v)=\lambda(v)$ for every vertex $v \in V(K)$. To extend $L_0$ to a simplicial map, we must verify that the image of every simplex of $K$ is a simplex of $C_m$. Suppose a simplex of $K$ had two vertices labelled $+i$ and $-i$. Since every two vertices of a simplex are joined by an edge, this would give an edge $\{u,v\}$ with $\lambda(u)+\lambda(v)=0$, contradicting the hypothesis that there is no complementary edge. Thus every simplex maps to a simplex of $C_m$, and so $L_0$ extends to a simplicial map \begin{align*} L: K \to C_m. \end{align*} Finally, the boundary condition on $\lambda$ says exactly that this simplicial map respects the antipodal involution on the boundary: for every boundary vertex $v$, \begin{align*} L(-v)=\lambda(-v)=-\lambda(v)=-L(v). \end{align*} This equivariance on $\partial K$ is the source of the odd parity count below. [/guided] [/step] [step:Double the triangulated ball to an antipodal triangulated sphere] Let $K_+$ be a copy of the finite complex $K$, and let $K_-$ be a second copy of $K$. Define the doubled complex $D(K)$ by gluing the boundary subcomplex of $K_+$ to the boundary subcomplex of $K_-$ through the boundary antipodal map: the boundary vertex $v$ in $K_+$ is identified with the boundary vertex $-v$ in $K_-$. Since $K$ triangulates $B^d$ and the gluing is along the entire boundary sphere $\partial B^d$, the geometric realization of $D(K)$ is a triangulation of $S^d$. Define the involution $\tau: D(K) \to D(K)$ as follows. On an interior vertex $v$ of $K_+$, let $\tau(v)$ be the corresponding vertex in $K_-$, and conversely. On a boundary vertex, this agrees with the given antipodal map because of the gluing rule. Thus $D(K)$ is an antipodal triangulation of $S^d$. Define a labelling $\Lambda: V(D(K)) \to \{\pm 1,\dots,\pm m\}$ by setting $\Lambda(v)=\lambda(v)$ on $K_+$ and $\Lambda(v_-)=-\lambda(v)$ on the corresponding vertex $v_-$ of $K_-$. This is well-defined on the glued boundary: if the boundary vertex $v$ in $K_+$ is identified with the boundary vertex $-v$ in $K_-$, then the two formulae give $\lambda(v)$ and $-\lambda(-v)$, which are equal because $\lambda(-v)=-\lambda(v)$. The doubled labelling is antipodal, since by construction $\Lambda(\tau(x))=-\Lambda(x)$ for every vertex $x\in V(D(K))$. It also has no complementary edge. Indeed, every edge of $D(K)$ lies either in one copy of $K$ or in the glued boundary subcomplex, and each such edge corresponds to an edge of $K$; the original hypothesis excludes opposite labels on all of those edges. [guided] The purpose of the double is to replace the ball problem by the already established sphere problem. Let $K_+$ and $K_-$ be two copies of the finite triangulation $K$. We glue them along their boundary, but not by the identity map: a boundary vertex $v$ in $K_+$ is glued to the boundary vertex $-v$ in $K_-$. This gluing produces a triangulated sphere $D(K)$, because it is the usual double of a triangulated $d$-ball along its boundary sphere. Now define the antipodal involution $\tau: D(K)\to D(K)$. On interior vertices it exchanges the two copies. On boundary vertices it becomes the given antipodal map, precisely because the boundary gluing identified $v$ in one copy with $-v$ in the other. Therefore $D(K)$ is a finite antipodal triangulation of $S^d$. We next extend the labelling. On the positive copy $K_+$, put $\Lambda(v)=\lambda(v)$. On the negative copy $K_-$, if $v_-$ corresponds to $v\in V(K)$, put $\Lambda(v_-)=-\lambda(v)$. This definition is compatible with the boundary identifications: the vertex $v$ in $K_+$ is glued to the vertex $-v$ in $K_-$, and the two labels are \begin{align*} \lambda(v) \end{align*} and \begin{align*} -\lambda(-v)=\lambda(v), \end{align*} using the boundary condition $\lambda(-v)=-\lambda(v)$. Hence $\Lambda$ is a well-defined labelling on $D(K)$. Finally, $\Lambda$ has the two hypotheses needed for the sphere form of Ky Fan's lemma. It is antipodal because the two copies carry opposite labels. It has no complementary edge because every edge in the double comes from an edge of the original triangulation $K$, either inside one copy or on the glued boundary, and the original labelling has no complementary edge on $K$. [/guided] [/step] [step:Apply the sphere form of Ky Fan's lemma to the doubled complex] We use the separately established sphere form of Ky Fan's parity lemma: if $T$ is a finite antipodal triangulation of $S^d$ and $\mu:V(T)\to\{\pm1,\dots,\pm m\}$ is an antipodal labelling with no complementary edge, then the number of alternating $d$-simplices in $T$ modulo antipodal pairing is odd. In particular, $T$ contains at least one alternating $d$-simplex. Apply this result to the finite antipodal triangulation $D(K)$ and the labelling $\Lambda$ constructed above. The hypotheses have been verified in the previous step: $D(K)$ is a finite antipodal triangulation of $S^d$, the labelling $\Lambda$ is antipodal, and $\Lambda$ has no complementary edge. Hence $D(K)$ contains an alternating $d$-simplex $\Sigma$. The equatorial subcomplex coming from $\partial K$ has dimension $d-1$, so the $d$-simplex $\Sigma$ lies entirely in one of the two copies of $K$. If $\Sigma$ lies in $K_+$, then it is already an alternating $d$-simplex of $K$ with respect to $\lambda$. If $\Sigma$ lies in $K_-$, let $\sigma$ be the corresponding $d$-simplex in $K_+$. The labels on $\Sigma$ are the negatives of the labels on $\sigma$, and multiplying all labels by $-1$ preserves the property that their absolute values are strictly increasing after reordering and their signs alternate. Thus $\sigma$ is an alternating $d$-simplex of $K$. [guided] We now use the sphere theorem in its precise form. The separately established sphere form of Ky Fan's parity lemma says: for a finite antipodal triangulation $T$ of $S^d$ and an antipodal labelling $\mu:V(T)\to\{\pm1,\dots,\pm m\}$ with no complementary edge, the number of alternating $d$-simplices modulo antipodal pairing is odd. An odd count is nonzero, so such a triangulation contains at least one alternating $d$-simplex. The previous step verified every hypothesis for $T=D(K)$ and $\mu=\Lambda$. The complex $D(K)$ is finite because it is made from two copies of the finite triangulation $K$. It triangulates $S^d$ because it is the double of a triangulated $d$-ball along its boundary. The involution is antipodal, the labelling satisfies $\Lambda(\tau(x))=-\Lambda(x)$ on every vertex, and no edge has opposite labels. Therefore the sphere theorem gives an alternating $d$-simplex $\Sigma$ in $D(K)$. It remains only to bring this simplex back to the original ball. No $d$-simplex is contained in the glued boundary, because that boundary has dimension $d-1$. Hence $\Sigma$ is contained in one copy of $K$. If it is contained in $K_+$, it is an alternating simplex of the original triangulation. If it is contained in $K_-$, take the corresponding simplex $\sigma$ in $K_+$. The label list on the copy in $K_-$ is obtained from the label list on $\sigma$ by multiplying every entry by $-1$. This operation does not change the absolute labels and only flips all signs simultaneously, so an alternating sign pattern remains alternating. Therefore the original triangulation $K$ contains an alternating $d$-simplex. [/guided] [/step] [step:Conclude the dimension bound] The complex $C_m$ has dimension $m-1$, because a simplex of $C_m$ can contain at most one of $+i$ and $-i$ for each $i \in \{1,\dots,m\}$. An alternating $d$-simplex uses $d+1$ distinct absolute labels. Hence the existence of an alternating $d$-simplex implies that \begin{align*} d+1 \leq m. \end{align*} Thus $m \geq d+1$, and the alternating $d$-simplex obtained above proves the theorem. [/step]