\phi(f\circ g)(\sigma)  &=(f\circ g)_{\ast}(\sigma) \\
&=  \prod_{i=0}^m (f\circ g)_{\ast_V}((\sigma_i, \tau_i))  \\
&= \prod_{i=0}^m((f\circ g)(\sigma_i), (f\circ g)(\tau_i)).\end{aligned}$$

On the other hand, we have

$$\begin{aligned}
(\phi(f) \circ \phi(g))(\sigma) &= (f_\ast \circ g_\ast)(\sigma) \\
&= f_\ast\left(\prod_{i=0}^mg_{\ast_V}((\sigma_i, \tau_i)) \right) \\
&= f_\ast\left(\prod_{i=0}^m(g(\sigma_i), g(\tau_i))\right) \\
&= \prod_{i=0}^m((f\circ g)(\sigma_i), (f\circ g)(\tau_i))\end{aligned}$$

Thus $\phi(f \circ g) = \phi(f)\circ \phi(g)$.\
We now need to show that $\phi$ is injective. Consider any $f \in {\mathrm{Ker}}(\phi)$. We claim that $f = {\mathrm{id}}_K$. Again, consider an arbitrary $\sigma = \prod_{i=0}^m(\sigma_i, \tau_i) \in \mathcal{M}(K)$. Notice $f$ induces a simplicial map $f_\ast \colon {\mathcal{M}}(K) \to {\mathcal{M}}(K)$. Since $f \in {\mathrm{Ker}}(\phi)$, we have $$\begin{aligned}
\phi(f)(\sigma) &= \sigma\\
f_\ast (\sigma) &= \sigma\\
\prod_{i=0}^m f_{\ast_V}((\sigma_i,\tau_i)) &= \prod_{i=0}^m (\sigma_i,\tau_i)\\
\prod_{i=0}^m (f(\sigma_i),f(\tau_i)) &= \prod_{i=0}^m (\sigma_i,\tau_i).\end{aligned}$$ As this holds for any choice of $\sigma$, we conclude that $f = {\mathrm{id}}_K$. Since ${\mathrm{Ker}}(\phi)$ is trivial, it follows that $\phi$ is injective.

Proposition \[prop: autK subgroup\] guarantees that ${\mathrm{Aut}}(K)\leq {\mathrm{Aut}}({\mathcal{M}}(K))$ by inducing an automorphism of ${\mathcal{M}}(K)$ from an automorphism of $K$. Thus if every automorphism of ${\mathcal{M}}(K)$ is induced by an automorphism of $K$, then ${\mathrm{Aut}}({\mathcal{M}}(K))={\mathrm{Aut}}(K).$ We will first show that this is indeed the case for $K\neq C_n, {\partial\Delta}^n$.

\[prop: aut(M(K))=aut(K)\] If $K$ is a simplicial complex other than $C_n$ 