and $(f(\sigma), f(\tau))\in V({\mathcal{M}}(K)).$

\[prop: f\_\* simplicial\] Let $f\colon K \to K$ be a simplicial automorphism. Then the induced map $f_{\ast}$ is simplicial.

Suppose for the sake of contradiction that $f_{\ast}\colon \mathcal{M}(K)\to \mathcal{M}(K)$ is not a simplicial map. Then there is some $\alpha = v_{i_0}v_{i_1}\cdots v_{i_m}\in \mathcal{M}(K)$ such that $f_{\ast} (\alpha) = f_{\ast_V}(v_{i_0}) f_{\ast_V}(v_{i_1})\cdots  f_{\ast_V}(v_{i_m})\not\in \mathcal{M}(K)$. This implies that the induced function $f_\ast(\alpha)$ either does not induce a gradient vector field (i.e. two vertices in $f_\ast(\alpha)$ are not compatible), or it induces a discrete vector field that contains a nontrivial closed $V$-path.\
**Case 1:** Suppose for the sake of contradiction that there are at least two distinct simplices, say $v_{i_j} = (\sigma_j, \tau_j)$ and $v_{i_k} = (\sigma_k, \tau_k)$, which are incompatible. This implies that exactly one of the following holds:

1.  $f(\sigma_i) = f(\sigma_j)$

2.  $f(\tau_i) = f(\tau_j)$

3.  $f(\tau_j) = f(\sigma_k)$

4.  $f(\sigma_j) = f(\tau_k)$.

Suppose $(1)$ is true. Since $f$ is an automorphism, this implies $\sigma_i = \sigma_j$. However, since $\tau_i \neq \tau_j$, we cannot simultaneously have both $(\sigma_i, \tau_i)$ and $(\sigma_i, \tau_j)$, so this would imply $\alpha \not\in \mathcal{M}(K)$, a contradiction. The cases in which $(2)$, $(3)$, or $(4)$ are true lead to similar contradictions, again due 