to the fact that $f$ is a simplicial automorphism.\
**Case 2:** Suppose for the sake of contradiction that $f_\ast(\alpha)$ contains a nontrivial closed V-path. Then, $f_\ast(\alpha)$ contains some vertices $$f_\ast(\sigma_0, \tau_0), f_\ast(\sigma_1, \tau_1), f_\ast(\sigma_2, \tau_2), \dots , f_\ast(\sigma_{k}, \tau_{k})$$ $$= (f(\sigma_0), f(\tau_0)), (f(\sigma_1), f(\tau_1)), (f(\sigma_2), f(\tau_2)), \dots , (f(\sigma_{k-1}), f(\tau_{k-1}))$$ such that $f(\sigma_0), f(\tau_0), f(\sigma_1), f(\tau_1), \dots , f(\sigma_{k-1}), f(\tau_{k-1}), f(\sigma_0)$ is a closed V-path, in which $f(\sigma_i) < f(\tau_{i-1})$ for $0 \leq i \leq k-1$ (where the indices are taken mod $k$). Since $f$ is an automorphism, it follows that $\sigma_0, \tau_0, \sigma_1, \tau_1, \dots , \sigma_{k-1}, \tau_{k-1}, \sigma_0$ must also be a nontrivial closed V-path. Since $(\sigma_0, \tau_0), (\sigma_1, \tau_1), \dots , (\sigma_{k-1},\tau_{k-1})$ are vertices in $\alpha$, it follows that $\alpha \not\in {\mathcal{M}}(K)$, a contradiction. Therefore, $f_\ast(\alpha) \in {\mathcal{M}}(K)$, so $f_{\ast}$ must be a simplicial map.

We give an example to show that if the simplicial map $f\colon K \to K$ is not an automorphism, then $f_*$ is not necessarily a simplicial map. Indeed, consider the simplicial complex $K = \Delta^2$ with vertex set $\{a,b,c\}$. Define $f_V\colon V(K)\to V(K)$ by $f_V(v) = a$ for all vertices $v$. It is easy to verify that $f$ is a simplicial map. 