A Morse function can have two different critical points with the same critical value

The partial derivatives are $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-2\pi \sin (2\pi x)$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=-2\pi \sin (2\pi y)$, so the critical points are exactly the four points with $x,y\in \{0,\frac{1}{2}\}$ modulo $\mathbb{Z}$.

The Hessian is diagonal with entries $-4{\pi }^2\cos (2\pi x)$ and $-4{\pi }^2\cos (2\pi y)$. At each of the four critical points these entries are nonzero, so every critical point is nondegenerate. By [F1], the function $f$ is Morse.

The two saddle points $(0,\frac{1}{2})$ and $(\frac{1}{2},0)$ both have critical value $0$, so distinct critical points can share one critical level. This is exactly the boundary described in [L1], and by [F1] it shows that $f$ is not excellent. Therefore the displayed universal claim is false.