This is a crosspost of [this MSE question](https://math.stackexchange.com/q/2619713/223002).


Given a locally Euclidean (locally homeomorphic to some Euclidean space) topological subspace $X\subset\mathbb R^n$ and $p\in X$, let $\mathrm{T}\_pX$ denote the *tangent set* of $X$ at $p$, namely the set of derivatives of curves in $X$ based at $p$ and differentiable at zero.


**Questions.** 


1. Are the following conditions equivalent? What are the implications
between them?
2. Does the second condition imply $\pi\_V$ is a local homeomorphism about $p$?
3. Does the first condition naturally furnish a local homeomorphism at $p$ between a neighborhood of $p$ in $\mathrm T\_pX$ and a neighborhood of $p$ in $X$?


(For 3, the only natural candidate I see here is orthogonal projection. I don't know how to prove it's locally injective. The conditions of the [everywhere differentiable inverse function theorem](https://terrytao.wordpress.com/2011/09/12/the-inverse-function-theorem-for-everywhere-differentiable-maps/) may be false as exemplified e.g by $f(x)=(x^5\cos(\frac 1x))^{\frac13}$, so that won't work.)




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1. $\mathrm{T}\_pX$ is a linear subspace of $\mathbb R^n$ of dimension $\dim\_pX$.
2. There's a linear subspace $V\subset\mathbb R^n$ of dimension $\dim\_pX$ for which we have the following equality, where the limit is taken in a translated neighborhood in $X$ of $p$. $$\lim\_{h\to 0} \frac{\pi\_{V^\perp}(h)}{\|h\|}=0$$ (Here $\pi\_{V^\perp}$ is the orthogonal projection onto the orthogonal complement of $V\subset\mathbb R^n$.)



Is the space of tangents actually the tangent space?
I don't think the two conditions are equivalent. Take for example


$$X:=\{ (x,y)\in \mathbb{R}^2\mid x\geq0,y=x^2 \}\cup\{ (x,y)\in \mathbb{R}^2\mid x\geq0,y=2x^2 \}$$


(or the cusp $X:=\{(x,y)\in\mathbb{R}^2\mid y^2=x^3\}\;$).


In this case, $T\_p X=0$ at $p=(0,0)\in\mathbb{R}^2$. And $V=\{(x,y)\mid y=0\}$.


Also, in this case the second condition is true but $\pi$ is not a local diffeo around the origin.




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**Added:** Maybe I have a counterexample that shows that 1. does not imply 2.


Basically: take a continuous function $f:\mathbb{R}^2\to\mathbb{R}$ that has directional derivatives (existing and) equal to zero along every direction through the origin $(0,0)$ but its restriction along the $y=x^2$ parabola in $\mathbb{R}^2$ goes like $|x|$, i.e. $f(x,x^2)=|x|$. Set $X:=$ the graph of $f$ in $\mathbb{R}^2$. Then, if I'm not mistaken, $T\_p X=\{z=0\}$. Also, the only possible $V$ could only be $V=\{z=0\}$ too, but $\pi(h)/||h||$ would not tend to $0$ as $h$ approaches the origin along the "approaching" subset $\Sigma=\{(x,x^2,|x|)\}\subset X$.


Now I'll try to formalize this. Set


$$\Sigma:=\{(x,x^2,|x|)\in\mathbb{R}^3\mid x\in\mathbb{R}\}\subset \mathbb{R}^3.$$


For every $x\in\mathbb{R}$,


* let $\ell'\_x$ be the straight segment in $\mathbb{R}^3$ joining $(x,\frac{1}{2}x^2,0)$ to $(x,x^2,|x|)$
* let $\ell''\_x$ be the straight segment in $\mathbb{R}^3$ joining $(x,2x^2,0)$ to $(x,x^2,|x|)$.


Set 


$$S:=\{(x,y)\in\mathbb{R}^2\mid \frac{1}{2}x^2\leq y \leq 2x^2\}\subset\mathbb{R}^2$$


and $D:=\mathbb{R}^2\smallsetminus S$, so $D\times\{0\}\subset\mathbb{R}^3$.


Finally


$$X:=(D\times\{0\})\cup \bigcup\_{x\in\mathbb{R}}(\ell'\_x\cup\ell''\_x)\subset\mathbb{R}^3.$$


If I got this right, $X$ should be the graph of a function $f:\mathbb{R}^2\to\mathbb{R}$ with the following properties:


* $f$ is continuous.
* For every direction $v\in\mathbb{R}^2$, the directional derivative $\partial\_v f(0)$ at the origin exists and is zero.
* $T\_p X=\{z=0\}$ for $p=(0,0,0)$.
* $\Sigma\subset X$.


If a linear subspace $V\subseteq\mathbb{R}^3$ as in condition 2. of the OP exists, then, looking at the straight lines through the origin, I think it would be forced to be $V=\{z=0\}=T\_p X$. But the condition $\pi(h)/||h||\to 0$ for $h\to 0$ is not met for $h\in\Sigma$ since 


$$\pi(x,x^2,|x|)=|x|,\quad h=(x,x^2,|x|)\in\Sigma$$


hence it's not met for $h\in X$. 


