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EDIT: Let $X$ be an $n$-dimensional Alexandrov space with curvature bounded below. Let $f_1,\dots, f_n\colon X\to \mathbb{R}$ be $\lambda$-concave functions. Assume that at a fixed point $p$ there exist directions $\xi_1^\pm,\dots,\xi_n^\pm\in \Sigma_p$ such that $f'_i(\xi_i^+)>1,\, f'_i(\xi^-_i)<-1$ and $|f'_i(\xi_j^\pm)|<\frac{1}{100 n}$ for $i\ne j$.

Is it true that the map $$(f_1,\dots,f_n)\colon X\to \mathbb{R}^n$$ is injective in a small neighborhood of $p$?

Remark. This question is motivated by an attempt to understand the paper "Elements of Morse theory on Alexandrov spaces" by Perelman (1993). He apparently does not explain the initial step in his induction: given an admissible map $X^n\to \mathbb{R}^n$, then it is a local homeomorphism in a neighborhood of any regular point. Thus if one has a proof of injectivity in this statement, it will be great.

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The base of Perelman's induction for Main Theorem 1.4 (is this what you're talking about?) is in fact a collection of trivial statements about admissible maps $X^n \to \mathbb{R}^{n+1}$.

To move to the next step, which you're looking for, requires an application of Property 1.3(c). The complementing function $g_{n+1}$ is constantly zero.

If you want to understand issues around the Stability Theorem, I strongly recommend Kapovitch's expository article. In particular, check out Corollary 6.10. The map is actually bi-Lipschitz.

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  • $\begingroup$ Yes, I mean Thm 1.4. At the beginning of the proof of it (see paragraph 1) Perelman says that the basis of induction - the case k=n- follows directly from property 1.2 (for $k>n$ the statement of the theorem 1.4 is meaningless since there are no regular points). I do not see how the case $k=n$ follows from 1.2 since I cannot prove injectivity. This is my question. $\endgroup$
    – asv
    Commented Nov 5, 2015 at 15:41
  • $\begingroup$ I am not sure how you apply property 1.3(c). Apparently this is exactly what I want to prove. The function $g_{k+1}$ is constructed in a clever way, not just taken to be 0. If you claim that the case $k=n$ is a formality, the property that $g_{k+1}^{-1}(0)\cap g^{-1}(v)$ is a single point still has to be proven, I think. $\endgroup$
    – asv
    Commented Nov 5, 2015 at 15:54
  • $\begingroup$ The base is $k=n+1$, as he states. By property 1.2 there are no regular points, and therefore every statement about regular points is true. Some meaningless things are better than others :) $\endgroup$
    – John Harvey
    Commented Nov 5, 2015 at 22:14
  • $\begingroup$ $g_{k+1}$ is cleverly constructed, certainly, but when $k=n$ it turns out to be zero. This is guaranteed by 1.3(d), which says that where $g_{n+1}$ is not zero, you have a regular point for a function to $\mathbb{R}^{n+1}$, which we already agreed was impossible. Lemma 3.8 tells you that the function $g_{k+1}$ has a unique maximum on each fiber of $g$, in other words, that the intersection is a single point. $\endgroup$
    – John Harvey
    Commented Nov 5, 2015 at 22:15
  • $\begingroup$ You are absolutely right. I was confused by the fact that in Section 1 Perelman assumes explicitly that $k<n$. But almost all of his argument works for $k=n$. In that case it implies that $g_{n+1}=0$ as you said. Thanks a lot! $\endgroup$
    – asv
    Commented Nov 7, 2015 at 12:52

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