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**xxp9** Suppose the pre-images of the two maxima are isolated this would be true and the proof involves using the Morse theory I think.

The idea is, define $\displaystyle D^a = f^{-1}(-\infty, a]$, then $\displaystyle D^0$ is nothing but $\displaystyle \partial{D}$, suppose $\displaystyle x \le y$ are the two local maximal values , then with a small positive number $\displaystyle \varepsilon$, $\displaystyle D^{x-\varepsilon}$ has at least two holes around $\displaystyle f^{-1}(x)$ and $\displaystyle f^{-1}(y)$, for that $\displaystyle \forall p \in f^{-1}(x)\cup f^{-1}(y), p \notin D^{x-\varepsilon} $, and since the critical points are isolated, by choosing $\displaystyle \varepsilon$ sufficiently small, those holes will not connect each other to merge to a single hole. So the (n-1)th homology group of $\displaystyle D^{x-\varepsilon}$ has at least dimension 2.

However since D is connected $\displaystyle D^0$ has only one hole inside, represented by a homology group of dimension 1.

According to Morse theory, between 0 and x, there must be other critical value z, so that the homotopic type of $\displaystyle D^{z+\varepsilon}$ is got by gluing one or more (n-1) cells to $\displaystyle D^{z-\varepsilon}$, to increase the dimension of the (n-1)th homology group.

Then the theory says that the critical point z must have an index of (n-1), so it has an indefinite Hessian, with a signature of (1, -1, -1, ..., -1).