[proofplan] We prove the contrapositive estimate attached to an arbitrary proper coloring. A proper $m$-coloring of $G$ is the same as a graph homomorphism $G \to K_m$, and functoriality of the box complex turns this into a $\mathbb{Z}/2$-equivariant simplicial map $B(G) \to B(K_m)$. The box complex of the complete graph $K_m$ has $\mathbb{Z}/2$-index $m-2$, so monotonicity of the $\mathbb{Z}/2$-index gives $\operatorname{ind}_{\mathbb{Z}/2}(B(G)) \leq m-2$. Minimizing over all proper colorings gives the stated lower bound for $\chi(G)$. [/proofplan]

[step:Turn a proper coloring into a graph homomorphism to a complete graph] Let $m \in \mathbb{N}$ and suppose that $G$ admits a proper $m$-coloring. Write $V(G)$ for the vertex set of $G$ and write $K_m$ for the complete graph on the color set $\{1,\dots,m\}$. A proper $m$-coloring is a map $c: V(G) \to \{1,\dots,m\}$ such that whenever $\{u,v\}$ is an edge of $G$, one has $c(u) \neq c(v)$. Since two distinct vertices of $K_m$ are adjacent, this condition says precisely that $c$ is a graph homomorphism $c: G \to K_m$. [/step]

[step:Use functoriality of the box complex to construct an equivariant simplicial map]We use the standard poset model of the box complex. Its vertices are ordered pairs $(A,B)$ of nonempty subsets $A,B \subset V(G)$ such that every vertex of $A$ is adjacent in $G$ to every vertex of $B$, ordered by coordinatewise inclusion. If no such pair exists, then $B(G)$ is the empty simplicial complex, and every map below is interpreted as the unique map from the empty complex when its domain is empty. The $\mathbb{Z}/2$-action is the involution $\tau_G: B(G) \to B(G)$ defined by $\tau_G(A,B)=(B,A)$. If $B(G)$ is nonempty, define a map on poset elements $B(c): B(G) \to B(K_m)$ by $B(c)(A,B)=(c(A),c(B))$, where $c(A) := \{c(a) : a \in A\}$ and $c(B) := \{c(b) : b \in B\}$. This map is well-defined. Indeed, if every $a \in A$ is adjacent to every $b \in B$ in $G$, then $c(a) \neq c(b)$ for every such pair because $c$ is a proper coloring. Hence every vertex of $c(A)$ is adjacent in $K_m$ to every vertex of $c(B)$, so $(c(A),c(B))$ is a poset element of $B(K_m)$. The map is order-preserving because $A \subset A'$ and $B \subset B'$ imply $c(A) \subset c(A')$ and $c(B) \subset c(B')$. Therefore it induces a simplicial map on the order complexes. Finally, $B(c)$ is $\mathbb{Z}/2$-equivariant. If $B(G)$ is empty, this is vacuous. Otherwise, for every $(A,B) \in B(G)$, one has $B(c)(\tau_G(A,B)) = B(c)(B,A) = (c(B),c(A)) = \tau_{K_m}(c(A),c(B)) = \tau_{K_m}(B(c)(A,B))$.[/step]

[guided]The goal of this step is to convert combinatorial data, namely a coloring, into topological data, namely an equivariant map between box complexes. We use the standard poset model of $B(G)$: an element is a pair $(A,B)$ of nonempty subsets of $V(G)$ such that all edges between $A$ and $B$ are present in $G$. If there is no such pair, then $B(G)$ is the empty simplicial complex. The involution is the map $\tau_G: B(G) \to B(G)$ defined by $\tau_G(A,B)=(B,A)$. If $B(G)$ is nonempty, define $B(c): B(G) \to B(K_m)$ by $B(c)(A,B)=(c(A),c(B))$, where $c(A) := \{c(a) : a \in A\}$ and $c(B) := \{c(b) : b \in B\}$. If $B(G)$ is empty, $B(c)$ denotes the unique map from the empty complex to $B(K_m)$. We must verify that this formula really lands in $B(K_m)$. Let $(A,B) \in B(G)$. By definition of the box complex, every $a \in A$ is adjacent in $G$ to every $b \in B$. Since $c$ is a proper coloring, adjacent vertices receive distinct colors, so $c(a) \neq c(b)$ for all $a \in A$ and $b \in B$. In the complete graph $K_m$, distinct colors are adjacent. Thus every vertex of $c(A)$ is adjacent to every vertex of $c(B)$ in $K_m$, which proves that $(c(A),c(B))$ is an element of $B(K_m)$. The map is also compatible with the simplicial structure. If $(A,B) \leq (A',B')$ in the poset, then $A \subset A'$ and $B \subset B'$, hence $c(A) \subset c(A')$ and $c(B) \subset c(B')$. Thus $B(c)$ is order-preserving, so it induces a simplicial map on the corresponding order complexes. Finally, the map respects the $\mathbb{Z}/2$-actions. If $B(G)$ is empty, equivariance is vacuous because there are no points to check. Otherwise, for every $(A,B) \in B(G)$, one has $B(c)(\tau_G(A,B)) = B(c)(B,A) = (c(B),c(A)) = \tau_{K_m}(c(A),c(B)) = \tau_{K_m}(B(c)(A,B))$. This proves that $B(c)$ is a $\mathbb{Z}/2$-equivariant simplicial map.[/guided]

[step:Compare indices using the complete graph box complex]We use two standard facts about the $\mathbb{Z}/2$-index. First, if $X$ and $Y$ are free $\mathbb{Z}/2$-simplicial complexes and there is a $\mathbb{Z}/2$-equivariant simplicial map $f: X \to Y$, then $\operatorname{ind}_{\mathbb{Z}/2}(X) \leq \operatorname{ind}_{\mathbb{Z}/2}(Y)$. The map constructed above is $\mathbb{Z}/2$-equivariant, so this gives \begin{align*} \operatorname{ind}_{\mathbb{Z}/2}(B(G)) \leq \operatorname{ind}_{\mathbb{Z}/2}(B(K_m)). \end{align*} Second, for $m \geq 2$, the standard computation of the box complex of a complete graph gives a $\mathbb{Z}/2$-equivariant homotopy equivalence between $B(K_m)$ and the sphere $S^{m-2} := \{x \in \mathbb{R}^{m-1} : |x|=1\}$ with the antipodal action $x \mapsto -x$, and hence $\operatorname{ind}_{\mathbb{Z}/2}(B(K_m)) = m-2$. For $m=1$, $K_1$ has no edge, so $B(K_1)$ is empty and the same formula holds under the empty-complex convention $\operatorname{ind}_{\mathbb{Z}/2}(\varnothing)=-1$. Combining these statements yields \begin{align*} \operatorname{ind}_{\mathbb{Z}/2}(B(G)) \leq m-2. \end{align*}[/step]

[guided]The index comparison uses two separate ingredients, and we verify their hypotheses explicitly. The first ingredient is monotonicity of the $\mathbb{Z}/2$-index: an equivariant map $f: X \to Y$ between free $\mathbb{Z}/2$-simplicial complexes forces $\operatorname{ind}_{\mathbb{Z}/2}(X) \leq \operatorname{ind}_{\mathbb{Z}/2}(Y)$, because any equivariant map from $Y$ to an antipodal sphere can be composed with $f$ to produce an equivariant map from $X$ to the same sphere. In the previous step we constructed precisely such an equivariant simplicial map $B(c): B(G) \to B(K_m)$, with the empty-domain case interpreted as the unique map from the empty complex. Therefore \begin{align*} \operatorname{ind}_{\mathbb{Z}/2}(B(G)) \leq \operatorname{ind}_{\mathbb{Z}/2}(B(K_m)). \end{align*} The second ingredient is the complete-graph computation. For $m \geq 2$, the box complex $B(K_m)$ is $\mathbb{Z}/2$-equivariantly homotopy equivalent to the sphere $S^{m-2} := \{x \in \mathbb{R}^{m-1} : |x|=1\}$, where $\mathbb{Z}/2$ acts by the antipodal involution $x \mapsto -x$. The defining property of the index then gives $\operatorname{ind}_{\mathbb{Z}/2}(B(K_m)) = m-2$. The exceptional case $m=1$ is consistent with the same formula: $K_1$ has no edge, hence $B(K_1)$ is the empty complex, and by the empty-complex convention $\operatorname{ind}_{\mathbb{Z}/2}(\varnothing)=-1=m-2$. Substituting this complete-graph value into the monotonicity inequality gives \begin{align*} \operatorname{ind}_{\mathbb{Z}/2}(B(G)) \leq m-2. \end{align*}[/guided]

[step:Minimize over all proper colorings] The preceding argument applies to every $m \in \mathbb{N}$ for which $G$ has a proper $m$-coloring. Hence every such $m$ satisfies \begin{align*} m \geq \operatorname{ind}_{\mathbb{Z}/2}(B(G)) + 2. \end{align*} Taking the minimum over all admissible $m$ gives \begin{align*} \chi(G) \geq \operatorname{ind}_{\mathbb{Z}/2}(B(G)) + 2. \end{align*} This is the desired chromatic lower bound. [/step]