[proofplan] We use the configuration space-test map method. If no $r$ pairwise disjoint faces have images with a common point, then $f$ produces an equivariant map from the $r$-fold deleted join of $\Delta^N$ to a representation sphere built from the standard sum-zero permutation representation. Since $r=p^k$, we restrict the action to the transitive elementary abelian $p$-subgroup $G=(\mathbb{Z}/p\mathbb{Z})^k$. The standard prime-power deleted-join [Borsuk-Ulam theorem](/theorems/6462) forbids such a $G$-equivariant map after its connectivity, dimension, fixed-point, and action hypotheses are verified. [/proofplan]

[step:Assume that no $r$ disjoint faces have images with a common point] Let $K$ denote the geometric simplex $\Delta^N$. Suppose, toward a contradiction, that for every ordered $r$-tuple $(\sigma_1,\dots,\sigma_r)$ of pairwise disjoint faces of $K$ one has \begin{align*} f(\sigma_1) \cap \cdots \cap f(\sigma_r) = \varnothing. \end{align*} Let $X$ denote the $r$-fold deleted join $K^{*r}_{\Delta}$. 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 continuously on $X$ by permuting the $r$ labels. [/step]

[step:Construct the deleted-join test map]Let $\mathbb{R}^r$ be the real permutation representation of $S_r$, 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*} Define the real $S_r$-representation \begin{align*} V = W_r^{\oplus(d+1)}. \end{align*} Let \begin{align*} \Phi: X &\to \mathbb{R}^r \oplus (\mathbb{R}^d)^r \end{align*} be the continuous map given 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*} Let $L$ 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*} of $\mathbb{R}^r \oplus (\mathbb{R}^d)^r$, and let \begin{align*} \rho: \mathbb{R}^r \oplus (\mathbb{R}^d)^r &\to L^\perp \end{align*} be [orthogonal projection](/theorems/437) with respect to the standard Euclidean [inner product](/page/Inner%20Product). The $S_r$-representation $L^\perp$ is $S_r$-equivariantly isomorphic to $V$, because the orthogonal complement of the constant line in each copy of $\mathbb{R}^r$ is $W_r$. For every $x \in X$, $\rho(\Phi(x)) \ne 0$. Indeed, write \begin{align*} x = \lambda_1 x_1 + \cdots + \lambda_r x_r. \end{align*} If $\rho(\Phi(x))=0$, then $\Phi(x) \in L$. The first component is constant, and since $\sum_{i=1}^r \lambda_i=1$, this gives $\lambda_i=1/r$ for every $i$. The second component then gives \begin{align*} \frac{1}{r}f(x_1)=\cdots=\frac{1}{r}f(x_r), \end{align*} hence $f(x_1)=\cdots=f(x_r)$. Since all $\lambda_i$ are positive, the points $x_i$ lie in pairwise disjoint faces $\sigma_i$ of $K$, contradicting the standing assumption. Let $S(V)$ denote the unit sphere of $V$. Therefore normalization defines a continuous map \begin{align*} \Psi: X &\to S(V) \end{align*} by \begin{align*} \Psi(x)=\frac{\rho(\Phi(x))}{|\rho(\Phi(x))|}. \end{align*} Every construction commutes with permutation of the $r$ labels, so $\Psi$ is $S_r$-equivariant.[/step]

[guided]The test map is designed so that its zeros encode exactly the forbidden Tverberg intersection. 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 join coordinates vary continuously on the join and $f:K\to\mathbb{R}^d$ is continuous. 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 situation where all weights are equal and all weighted image points are equal. Let \begin{align*} \rho: \mathbb{R}^r \oplus (\mathbb{R}^d)^r &\to L^\perp \end{align*} be orthogonal projection. Since the orthogonal complement of the constant vectors in $\mathbb{R}^r$ is \begin{align*} W_r = \left\{(a_1,\dots,a_r) \in \mathbb{R}^r : \sum_{i=1}^r a_i = 0\right\}, \end{align*} the space $L^\perp$ is $S_r$-equivariantly isomorphic to $W_r^{\oplus(d+1)}$, which we denote by $V$. We verify that $\rho(\Phi(x))$ never vanishes. Take \begin{align*} x = \lambda_1 x_1 + \cdots + \lambda_r x_r \in X. \end{align*} If $\rho(\Phi(x))=0$, then $\Phi(x)\in L$. The first coordinate of $\Phi(x)$ is $(\lambda_1,\dots,\lambda_r)$, so membership in the diagonal line forces all $\lambda_i$ to be equal. Since their sum is $1$, each equals $1/r$. The second coordinate then gives \begin{align*} \frac{1}{r}f(x_1)=\cdots=\frac{1}{r}f(x_r), \end{align*} and hence $f(x_1)=\cdots=f(x_r)$. Because every $\lambda_i$ is positive, the deleted-join condition says that the points $x_i$ lie in pairwise disjoint faces $\sigma_i$ of $K$. Thus the common value of the $f(x_i)$ lies in \begin{align*} f(\sigma_1) \cap \cdots \cap f(\sigma_r), \end{align*} contradicting the assumption. Thus $\rho(\Phi(x))\ne 0$ for every $x\in X$, and normalization gives \begin{align*} \Psi: X &\to S(V) \end{align*} by \begin{align*} \Psi(x)=\frac{\rho(\Phi(x))}{|\rho(\Phi(x))|}. \end{align*} The maps $\Phi$, $\rho$, and the identification $L^\perp\cong V$ are compatible with permutations of the labels, so $\Psi$ is $S_r$-equivariant.[/guided]

[step:Restrict the test map to the elementary abelian $p$-subgroup] Let $G$ be the additive group of the [vector space](/page/Vector%20Space) $(\mathbb{Z}/p\mathbb{Z})^k$. Since $|G|=p^k=r$, choose a bijection between $G$ and the label set $\{1,\dots,r\}$. Left translation gives a transitive action of $G$ on this label set, and hence an embedding $G \le S_r$. Restricting the $S_r$-action to $G$, the map $\Psi$ is a continuous $G$-equivariant map \begin{align*} \Psi: X &\to S(V). \end{align*} The action of $G$ on $X$ is free. To see this, write a point of the deleted join as \begin{align*} x=\lambda_h x_h *_{h\in G}, \end{align*} where the active labels are those $h$ with $\lambda_h>0$, and where active points $x_h$ lie in pairwise disjoint faces of $\Delta^N$. Suppose that $g\in G$ fixes $x$. Equality in join coordinates implies that the active label set is invariant under left translation by $g$, and that \begin{align*} \lambda_{gh}=\lambda_h,\qquad x_{gh}=x_h \end{align*} for every active label $h$. If $g\neq 0$, then $gh\neq h$. Thus two distinct active labels $h$ and $gh$ would have the same point $x_h=x_{gh}$ in two disjoint faces of $\Delta^N$, which is impossible because disjoint faces have disjoint geometric realizations. Hence no nonzero $g$ fixes a point of $X$. The representation $V$ has no nonzero $G$-fixed vector. Indeed, a $G$-fixed vector in $\mathbb{R}^r$ is constant on the transitive $G$-orbit, and the condition defining $W_r$ forces that constant to be $0$; hence $(W_r)^G=\{0\}$ and $V^G=\{0\}$. [/step]

[step:Invoke the prime-power deleted-join Borsuk-Ulam theorem with all hypotheses verified] We use the following standard external theorem, often called the prime-power deleted-join Borsuk-Ulam theorem or the Volovikov-Dold obstruction theorem for the deleted join. Let $G=(\mathbb{Z}/p\mathbb{Z})^k$ act transitively on $r=p^k$ labels by left translation. Let $X=(\Delta^N)^{*r}_{\Delta}$, where \begin{align*} N=(r-1)(d+1). \end{align*} Then $X$ is $(N-1)$-connected, the induced $G$-action on $X$ is free, and for every real $G$-representation $V$ of dimension $N$ with $V^G=\{0\}$ there is no continuous $G$-equivariant map \begin{align*} X &\to S(V). \end{align*} Here $S(V)$ is the unit sphere of $V$, so $\dim S(V)=N-1$. We verify the hypotheses. The group $G=(\mathbb{Z}/p\mathbb{Z})^k$ is elementary abelian of order $r=p^k$ and acts transitively on the $r$ labels by left translation. The space used above is exactly the deleted join $X=(\Delta^N)^{*r}_{\Delta}$. The representation used above is \begin{align*} V=W_r^{\oplus(d+1)}. \end{align*} Since $\dim W_r=r-1$, we have \begin{align*} \dim V=(d+1)(r-1)=N. \end{align*} The preceding step proved that $V^G=\{0\}$ and that the $G$-action on $X$ is free. Thus the theorem applies and forbids the continuous $G$-equivariant map $\Psi:X\to S(V)$ constructed above. This contradiction shows that the assumption made at the start of the proof is false. [/step]

[step:Extract the Tverberg partition] Since the contradiction assumption is false, 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*} Forgetting the order gives the required collection of $r$ pairwise disjoint faces. This proves the theorem. [/step]