Edge states in the "half BHZ" model Consider the "half BHZ" Hamiltonian
$${\cal H}=\sum_{\mathbf{k}}\left(A\sin(k_{x})\sigma_{x}+A\sin(k_{y})\sigma_{y}+{\cal M}(\mathbf{k})\sigma_{z}\right)c_{\mathbf{k}}^{\dagger}c_{\mathbf{k}}$$ where $M(\mathbf{k})=M-2B\left(2-\cos(k_{x})-\cos(k_{y})\right)$
For simplicity, take A=1 and B > 0. M < 0 and M > 8B give a trivial phase, while 0 < M < 4B and 4B < M < 8B give topological phases with Chern number $\pm1$ and the corresponding edge states. The existence of the edge states can be easily verified by numerically diagonalizing the Hamiltonian for periodic boundary conditions in, say, the x direction and vanishing boundary conditions in the y direction. They can also be found analytically by expanding the Hamiltonian around the TRIM points $(k_x, k_y) = (0, 0), (0, \pi), (\pi, 0), (\pi, \pi)$ to second order and looking for solutions localized near a boundary. The trouble is, the expansion around $(0,0)$ seems to imply that the only condition for the existence of a localized state at $k_x=E=0$ is MB > 0. But we know that for M > 4B there's no such state, so what happens to it? I think it's supposed to somehow pair-annihilate with the $(0, \pi)$ edge state when M becomes bigger than 4B, but according to the expansion around $(0, \pi)$ the $(0, \pi)$ edge state vanishes on its own. Is there some other explanation for the disappearance of the $(0, 0)$ edge state? Or does it annihilate with the $(0, \pi)$ edge state after all, but there's no way to see it from the local Hamiltonians?
 A: You are correct. However, I would rephrase what you said to a technically more accurate form: the band inversions at $\mathbf{k} = (0, \; 0)$ and $\mathbf{k} = (0, \; \pi)$ annihilate each other at $M = 4B$. I am strictly following the formalism of Fu and Kane (since the BHZ model has inversion symmetry):

Liang Fu and Charles L. Kane. “Topological insulators with inversion symmetry.” Physical Review B 76, no. 4 (2007): 045302. (arXiv)

You can define a quantity called the time-reversal polarization at the Time-Reversal Invariant Momentum (TRIM) values ($\Lambda_{a}$) on the edge (surface) Brillouin Zone (BZ) of a 2D (3D) topological insulator (Eq. (2.6)): $$\pi_{a} = \delta_{a1}\delta_{a2}$$ where $\delta_{i}$ is defined in Eq. (3.10) in terms of the band parities $\xi_{2n}$ $$\delta_{i}=\prod_{m=1}^{N}\xi_{2m}(\Gamma_{i})$$ The $\pi$’s are basically the projection of the $\delta$’s in the bulk BZ onto the surface BZ (see Fig. 1 (b) and (d)) at the TRIMs.
In the BHZ model, we only have two $\pi$’s: at $k_{x} = 0$ and $k_{x} = \pi$. When $M$ rises above $0$ the gap closes at $\mathbf{k} = (0, \; 0)$ and the bands invert (only at $\mathbf{k} = (0, \; 0)$) and we get $\delta_{(0, \; 0)} = -1$ and $\pi_{k_{x} = 0} = -1$ (the rest are $+1$). Eq. (2.10) picks up an extra $(-1)$ and $\nu$ goes from $0$ to $1 \; {\rm mod} \; 2$; i.e. the system becomes topologically nontrivial. When $M$ rises above $4B$ the gap closes at two points: $\mathbf{k} = (\pi, \; 0)$ and $\mathbf{k} = (0, \; \pi)$, and we get $\delta_{(\pi, \; 0)} = -1$ and $\delta_{(0, \; \pi)} = -1$. Then we get $\pi_{k_{x} = 0} = (-1)^2 = +1$ and $\pi_{k_{x} = \pi} = -1$. Thus the intersection of edge states moves from $k_{x} = 0$ to $k_{x} = \pi$ when $M$ rises above $4B$.
So to answer your question: yes, there is no way to see changes in edge states from the local Hamiltonians. Recall that the phase transition at $M = 4B$, which is accompanied by the closing of a bulk gap, results in the rearrangement of the entire electronic system. The Fu and Kane analysis involves analysis (above) of the entire bulk bandstructure can capture edge state shift.
