automorphisms $f_\ast, g_\ast \colon {\mathcal{M}}(K)\to {\mathcal{M}}(K)$, respectively. Then $(f\circ g)_\ast = f_\ast \circ g_\ast$.

It suffices to show that $(f\circ g)_{\ast_V} = (f_{\ast} \circ g_{\ast})_V$. Consider any $(\sigma, \tau) \in {\mathcal{M}}(K)$. We have $$\begin{aligned}
(f\circ g)_{\ast_V}((\sigma,\tau)) &= (f\circ g(\sigma), f\circ g(\tau))\\
&= f_{\ast_V}(g(\sigma), g(\tau))\\
&= (f_{\ast_V} \circ g_{\ast_V})((\sigma, \tau)).\end{aligned}$$ Hence, $(f\circ g)_{\ast_V} = f_{\ast_V}\circ g_{\ast_V}$. By Lemma \[lem: comp\], we know that $f_{\ast_V} \circ g_{\ast_V} = (f_{\ast} \circ g_{\ast})_V $, so $(f\circ g)_{\ast_V} = (f_{\ast} \circ g_{\ast})_V$.

Because automorphisms of $K$ induce automorphisms of ${\mathcal{M}}(K)$, we next show that we are able to obtain not only elements of ${\mathrm{Aut}}({\mathcal{M}}(K))$, but that ${\mathrm{Aut}}(K)$ actually corresponds to a subgroup of ${\mathrm{Aut}}({\mathcal{M}}(K)).$

\[prop: autK subgroup\] The group $\text{Aut}(K)$ is isomorphic to a subgroup of $\text{Aut}(\mathcal{M}(K))$. In particular, there is an injective homomorphism $\phi\colon {\mathrm{Aut}}(K)\to {\mathrm{Aut}}({\mathcal{M}}(K)).$

Consider the function $\phi\colon \text{Aut}(K) \to \text{Aut}(\mathcal{M}(K))$ that sends each simplicial automorphism $f \in \text{Aut}(K)$ to its induced simplicial automorphism $f_{\ast}\in \text{Aut}(\mathcal{M}(K))$. We first show that $\phi$ is a homomorphism. Consider any $f,g \in \text{Aut}(K)$, and an arbitrary $\sigma = \prod_{i=0}^m(\sigma_i, \tau_i) \in \mathcal{M}(K)$ where each $(\sigma_i,\tau_i)$ is a vertex of $\sigma$. Since $f,g\in {\mathrm{Aut}}(K)$, $f\circ g\in {\mathrm{Aut}}(K)$. This induces $(f\circ g)_{\ast}\colon \mathcal{M}(K) \to \mathcal{M}(K)$. Notice that

$$\begin{aligned}
