**Definition.** Let $X\subset \mathbb R^n$ be a locally Euclidean subset. Say it has a tangent space at $p\in X$ if there exists a linear subspace $V\leq \mathbb R^n$ satisfying the following conditions.


1. $\dim V=\dim\_pX$.
2. There exists a neighborhood $U\subset X$ of $p$ in $X$ for which
$$\lim\_{\substack{h\to 0\\\text{in
}U-p}}\frac{\|\pi\_{V^\perp}(h)\|}{\|h\|}=0.$$ Here $U-p$ is the
neighborhood of the origin in $X-p$ obtained by translation of $U$
by $(-p)\in X\subset\mathbb R^n$.
3. The orthogonal projection of $X$ onto $V$ is locally injective at $p$.


**Question.** Is $V\leq \mathbb R^n$ unique?


The usual argument for the uniqueness of the derivative of a map on an *open* subset of Euclidean spaces relies on (more or less) convexity of open balls, so is not applicable here.


I am hoping the local injectivity at $p$ of the orthogonal projection will eliminate singular cases by preventing $X$ from "wrapping around $V$" about $p$ as in the cubic cusp (a sort of ramification). Without this condition the other two do not suffice. The example I was given is the surface of revolution of a cusp, e.g $\{(x,y,\sqrt[4]{x^2+y^2})\mid x,y\in \mathbb R\}\subset\mathbb R^3$.



Uniqueness of tangent space given local injectivity of orthogonal projection onto it
By invariance of domain, the condition $\dim V=\dim\_pX$ implies that the local injectivity is locally a bijection to a neighborhood of $p.$ So there are open sets $V\_0\subseteq V$ and $X\_0\subseteq X$ with $V\_0$ containing $0,$ and a function $g:V\_0\to X\_0$ such that $\pi\_V(g(v)-p)=v$ and $g(0)=p.$ By condition 2 we have $g(v)=p+v+o(\|v\|)$ with implicit constants depending only on $X,V$ and $p.$


Assume there's another $V'$ satisfying the same conditions. For each $v\in V$ we have $\lim\_{\lambda\to 0^+}\frac{\|\pi\_{V'^\perp}(g(\lambda v)-p)\|}{\|g(\lambda v)-p\|}=0.$ Using $g(\lambda v)=p+\lambda v+o(\lambda\|v\|)$ we get $\pi\_{V'^\perp}(v)=0,$ so $v\in V'.$ Since $\dim V=\dim V'$ we must have $V=V'.$


