[proofplan]
We prove the contrapositive. If $G$ admits a proper coloring with at most $m$ colors, then that coloring determines a graph homomorphism $G \to K_m$. Functoriality of the box complex sends this homomorphism to a $\mathbb Z/2\mathbb Z$-equivariant continuous map $B(G) \to B(K_m)$, and composing with the fixed equivariant map $B(K_m) \to S^{m-2}$ gives the forbidden map. Therefore the nonexistence of such an equivariant map forces $G$ to need at least $m+1$ colors.
[/proofplan]
[step:Turn an $m$-coloring into a graph homomorphism to $K_m$]
We prove the contrapositive. Assume that $\chi(G) \leq m$. By the definition of chromatic number, there exists a proper coloring
\begin{align*}
c: V(G) \to \{1,\dots,m\}.
\end{align*}
Let $K_m$ denote the complete graph on vertex set $\{1,\dots,m\}$. Since $c$ is proper, whenever $\{u,v\}$ is an edge of $G$, we have $c(u) \neq c(v)$. Because distinct vertices of $K_m$ are adjacent, this implies that $\{c(u),c(v)\}$ is an edge of $K_m$. Hence $c$ is a graph homomorphism
\begin{align*}
c: G \to K_m.
\end{align*}
[guided]
We begin by proving the contrapositive, because the theorem says that the absence of an equivariant map obstructs an $m$-coloring. Thus we assume instead that an $m$-coloring exists and show that it produces a $\mathbb Z/2\mathbb Z$-equivariant map into $S^{m-2}$.
The hypothesis $\chi(G) \leq m$ means precisely that there is a proper coloring
\begin{align*}
c: V(G) \to \{1,\dots,m\}.
\end{align*}
Properness means that adjacent vertices receive different colors: if $\{u,v\}$ is an edge of $G$, then $c(u) \neq c(v)$. Now view the color set $\{1,\dots,m\}$ as the vertex set of the complete graph $K_m$. In $K_m$, every pair of distinct vertices is connected by an edge. Therefore, for every edge $\{u,v\}$ of $G$, the pair $\{c(u),c(v)\}$ is an edge of $K_m$.
This is exactly the definition of a graph homomorphism from $G$ to $K_m$. Hence the coloring determines a graph homomorphism
\begin{align*}
c: G \to K_m.
\end{align*}
[/guided]
[/step]
[step:Apply functoriality of the box complex to obtain an equivariant map]
The box complex construction is functorial for graph homomorphisms. Applying it to the graph homomorphism $c: G \to K_m$ gives a continuous map
\begin{align*}
B(c): B(G) \to B(K_m).
\end{align*}
The natural $\mathbb Z/2\mathbb Z$-action on a box complex is the shore-swapping action, and functoriality preserves this action. Therefore $B(c)$ is $\mathbb Z/2\mathbb Z$-equivariant.
[/step]
[step:Compose with the complete graph model map to reach the antipodal sphere]
By the hypothesis on the chosen box complex model, there exists a continuous $\mathbb Z/2\mathbb Z$-equivariant map
\begin{align*}
q: B(K_m) \to S^{m-2},
\end{align*}
where $S^{m-2}$ carries the antipodal action. Define
\begin{align*}
F: B(G) \to S^{m-2}
\end{align*}
by
\begin{align*}
F = q \circ B(c).
\end{align*}
The [composition of continuous maps](/theorems/4960) is continuous. Moreover, if $\tau$ denotes the nontrivial element of $\mathbb Z/2\mathbb Z$, then for every point $x \in B(G)$,
\begin{align*}
F(\tau x) = q(B(c)(\tau x)) = q(\tau B(c)(x)) = \tau q(B(c)(x)) = \tau F(x).
\end{align*}
Thus $F$ is a continuous $\mathbb Z/2\mathbb Z$-equivariant map from $B(G)$ to $S^{m-2}$.
[/step]
[step:Take the contrapositive to obtain the chromatic lower bound]
We have shown that if $\chi(G) \leq m$, then there exists a continuous $\mathbb Z/2\mathbb Z$-equivariant map
\begin{align*}
B(G) \to S^{m-2}.
\end{align*}
Taking the contrapositive, if no such equivariant map exists, then $\chi(G) > m$. Since $\chi(G)$ is an integer, this is equivalent to
\begin{align*}
\chi(G) \geq m+1.
\end{align*}
This proves the theorem.
[/step]
