[proofplan] We argue by contradiction using the configuration space-test map scheme. If no $r$ pairwise disjoint faces have intersecting images, the map $f$ induces an equivariant map from the $r$-fold deleted join of $\Delta^N$ to the unit sphere of a zero-sum permutation representation introduced below. Since $r$ is a prime power, write $r=p^k$ for a prime $p$ and an integer $k\geq 1$, then restrict the symmetric-[group action](/page/Group%20Action) to a transitive elementary abelian $p$-subgroup $G \subset S_r$. The deleted-join [Borsuk-Ulam theorem](/theorems/6462) for this $G$ forbids the resulting $G$-equivariant map, giving the required contradiction. [/proofplan]

[step:Assume that all images of pairwise disjoint faces have empty common intersection] Let $K = \Delta^N$, where $N = (r - 1)(d + 1)$. Suppose, toward a contradiction, that for every ordered $r$-tuple $(\sigma_1, \dots, \sigma_r)$ of pairwise disjoint faces of $K$, \begin{align*} f(\sigma_1) \cap \cdots \cap f(\sigma_r) = \varnothing. \end{align*} Let $X = K^{*r}_{\Delta}$ denote the $r$-fold deleted join of $K$. Thus $X$ consists of join points \begin{align*} \lambda_1 x_1 + \cdots + \lambda_r x_r \end{align*} where $\lambda_i \ge 0$, $\sum_{i=1}^r \lambda_i = 1$, each $x_i \in K$, and the minimal faces of $K$ containing those $x_i$ with $\lambda_i > 0$ are pairwise disjoint. The symmetric group $S_r$ acts on $X$ by permuting the $r$ join labels. [/step]

[step:Build the equivariant test map from the deleted join]Define the real permutation representation \begin{align*} \mathbb{R}^r = \{(a_1, \dots, a_r) : a_i \in \mathbb{R}\} \end{align*} with $S_r$ acting by permuting coordinates, and define \begin{align*} W_r = \left\{(a_1, \dots, a_r) \in \mathbb{R}^r : \sum_{i=1}^r a_i = 0\right\}. \end{align*} Let \begin{align*} \Phi: X \to \mathbb{R}^r \oplus (\mathbb{R}^d)^r \end{align*} be the continuous map \begin{align*} \Phi(\lambda_1 x_1 + \cdots + \lambda_r x_r) = \left((\lambda_1, \dots, \lambda_r),(\lambda_1 f(x_1), \dots, \lambda_r f(x_r))\right). \end{align*} Let $L \subset \mathbb{R}^r \oplus (\mathbb{R}^d)^r$ be the diagonal linear subspace \begin{align*} L = \{((a, \dots, a),(z, \dots, z)) : a \in \mathbb{R}, z \in \mathbb{R}^d\}. \end{align*} Let \begin{align*} \rho: \mathbb{R}^r \oplus (\mathbb{R}^d)^r \to L^\perp \end{align*} be the [orthogonal projection](/theorems/437) onto the Euclidean orthogonal complement of $L$. The representation $L^\perp$ is naturally $S_r$-equivariantly isomorphic to $W_r^{\oplus(d+1)}$. For every $x \in X$, $\rho(\Phi(x)) \ne 0$. Indeed, if $x = \lambda_1 x_1 + \cdots + \lambda_r x_r$ and $\rho(\Phi(x)) = 0$, then $\Phi(x) \in L$. Since $\sum_i \lambda_i = 1$, membership in $L$ forces $\lambda_i = 1/r$ for every $i$. The second component then forces \begin{align*} \frac{1}{r} f(x_1) = \cdots = \frac{1}{r} f(x_r), \end{align*} hence $f(x_1) = \cdots = f(x_r)$. The points $x_i$ lie in pairwise disjoint faces of $K$, contradicting the standing assumption. Therefore the formula \begin{align*} \Psi: X \to S(W_r^{\oplus(d+1)}) \end{align*} given by \begin{align*} \Psi(x) = \frac{\rho(\Phi(x))}{|\rho(\Phi(x))|} \end{align*} defines a continuous $S_r$-equivariant map.[/step]

[guided]The purpose of the deleted join is to remember $r$ points chosen from pairwise disjoint faces, together with barycentric weights. A point of $X = K^{*r}_{\Delta}$ has the form \begin{align*} x = \lambda_1 x_1 + \cdots + \lambda_r x_r, \end{align*} where $\lambda_i \ge 0$, $\sum_i \lambda_i = 1$, and the active points $x_i$ with $\lambda_i > 0$ lie in pairwise disjoint faces of $K$. We now encode both the weights and the images under $f$. Define \begin{align*} \Phi: X \to \mathbb{R}^r \oplus (\mathbb{R}^d)^r \end{align*} by \begin{align*} \Phi(\lambda_1 x_1 + \cdots + \lambda_r x_r) = \left((\lambda_1, \dots, \lambda_r),(\lambda_1 f(x_1), \dots, \lambda_r f(x_r))\right). \end{align*} This map is continuous because it is induced from the continuous map $f: K \to \mathbb{R}^d$ and the join coordinates. The diagonal subspace \begin{align*} L = \{((a, \dots, a),(z, \dots, z)) : a \in \mathbb{R}, z \in \mathbb{R}^d\} \end{align*} records the forbidden situation: all weights equal and all weighted images equal. Let \begin{align*} \rho: \mathbb{R}^r \oplus (\mathbb{R}^d)^r \to L^\perp \end{align*} be orthogonal projection. The complement $L^\perp$ is exactly the part measuring deviation from the diagonal. Since the orthogonal complement of the constant vectors in $\mathbb{R}^r$ is $W_r$, we have an $S_r$-equivariant identification \begin{align*} L^\perp \cong W_r^{\oplus(d+1)}. \end{align*} We must check that $\rho(\Phi(x))$ never vanishes. If $\rho(\Phi(x)) = 0$, then $\Phi(x) \in L$. The first coordinate of $\Phi(x)$ is $(\lambda_1,\dots,\lambda_r)$, and because $\sum_i \lambda_i = 1$, membership in the constant-coordinate line forces \begin{align*} \lambda_1 = \cdots = \lambda_r = \frac{1}{r}. \end{align*} The second coordinate then gives \begin{align*} \frac{1}{r} f(x_1) = \cdots = \frac{1}{r} f(x_r), \end{align*} so $f(x_1)=\cdots=f(x_r)$. Since all $\lambda_i$ are positive, each $x_i$ belongs to one of the pairwise disjoint active faces of $K$. This produces a point in \begin{align*} f(\sigma_1) \cap \cdots \cap f(\sigma_r), \end{align*} contrary to the contradiction hypothesis. Thus $\rho(\Phi(x)) \ne 0$ for every $x \in X$, so normalization is valid. We obtain \begin{align*} \Psi: X \to S(W_r^{\oplus(d+1)}) \end{align*} by \begin{align*} \Psi(x) = \frac{\rho(\Phi(x))}{|\rho(\Phi(x))|}. \end{align*} All constructions commute with permutation of the $r$ labels, so $\Psi$ is $S_r$-equivariant.[/guided]

[step:Restrict the test map to a transitive elementary abelian $p$-subgroup] Since $r$ is a prime power, choose a prime $p$ and an integer $k\geq 1$ such that $r = p^k$. Let $G$ be the additive group of the [vector space](/page/Vector%20Space) $(\mathbb{Z}/p\mathbb{Z})^k$. Choose a bijection between $G$ and the label set $\{1,\dots,r\}$. The left-translation action of $G$ on itself gives a transitive embedding $G \subset S_r$. Restricting the $S_r$-action to this subgroup, the map $\Psi$ is a continuous $G$-equivariant map \begin{align*} \Psi: X \to S(W_r^{\oplus(d+1)}). \end{align*} The $G$-representation $W_r$ has no nonzero fixed vector: a $G$-fixed vector in $\mathbb{R}^r$ is constant on the transitive $G$-orbit, and the condition that its coordinate sum is zero forces that constant to be $0$. Hence $W_r^{\oplus(d+1)}$ is fixed-point-free as a $G$-representation. [/step]

[step:Apply the prime-power deleted-join Borsuk-Ulam theorem] We now use the prime-power deleted-join Borsuk-Ulam theorem: for $r=p^k$, $N=(r-1)(d+1)$, $K=\Delta^N$, and the transitive elementary abelian $p$-subgroup $G \subset S_r$, there is no continuous $G$-equivariant map \begin{align*} K^{*r}_{\Delta} \to S(W_r^{\oplus(d+1)}). \end{align*} (citing a result not yet in the wiki: prime-power deleted-join Borsuk-Ulam theorem) The hypotheses of this theorem are satisfied: $K$ is the $N$-simplex with $N=(r-1)(d+1)$, $G=(\mathbb{Z}/p\mathbb{Z})^k$ acts transitively on the $r$ labels, and $W_r^{\oplus(d+1)}$ is a fixed-point-free real $G$-representation. The map $\Psi$ constructed above is exactly the forbidden $G$-equivariant map. This contradiction proves that the assumption of no common intersection for images of pairwise disjoint faces is false. [/step]

[step:Conclude the existence of the Tverberg partition] Therefore there exists an ordered $r$-tuple $(\sigma_1,\dots,\sigma_r)$ of pairwise disjoint faces of $\Delta^N$ such that \begin{align*} f(\sigma_1) \cap \cdots \cap f(\sigma_r) \ne \varnothing. \end{align*} This is precisely the desired Tverberg partition for the continuous map $f: \Delta^{(r-1)(d+1)} \to \mathbb{R}^d$. [/step]