So it is well known that if $y$ is a boundary point for a $k$-manifold $M$ in $\Bbb R^n$ then a vector $\vec v$ of $\Bbb R^n$ is tangent to $M$ at $y$ if and only if there exist a parametrized curve $\gamma$ defined in a open neighborhood of $0$ that carries the right or left part of this neighborhood into $M$ such that $$ y=\gamma (0)\,\,\,\text{and}\,\,\,\vec v=\gamma'(0) $$
So I ask if this result is still true for corner points of a $k$-Manifold of class $C^r$ for any $r\ge 1$ with corners that is defined to be a $k$-manifold having some local patch defined in an open set of $\Bbb R^{k-h}\times[0,+\infty)^{h}$ for any $h\le k$.
To follow some reference about well know definitions and elementary results about manifolds: I put here them to show the formalism I use about manifolds; moereover I point out that theese reference (from the text Analysis on Manifolds by James Munkres) are for manifolds with boudary but reading them it is not hard to see that the definition and results can be easily (is this false?) extended to a manifolds with corners. So if you like click here to see it, otherwise (provided you like it!) see my proof attempt.
MY PROOF ATTEMPT
So let be $\gamma$ a parametrized curve defined in a open neighborhood of $0$ that carries the right (or left) part of this neighborhood into $M$ so that $\gamma(0)$ is a corner point for $M$ and we shall prove that that its derivative $\gamma'$ at $\gamma (0)$ is a tangent vector to $M$. So let be $\alpha:U\rightarrow V$ a coordinate patch about $\gamma(0)$ and thus putting $$ y=\alpha(x):=\gamma(0)\,\,\,\text{and}\,\,\,\vec v:=\frac{d}{dt}\gamma(t)|_{t=0}\,\,\,\text{and thus}\,\,\,\vec u:=D\alpha^{-1}(y)\cdot\vec v $$ we observe that $$ \alpha_*(x;\vec u)=(y;D\alpha(x)\cdot\vec u)=(y;D\alpha(x)\cdot D\alpha^{-1}(y)\cdot\vec v)=(y;D(\alpha\circ\alpha^-1)(y)\cdot\vec v)=(y;\vec v) $$ and so we conclude that $\vec v$ is a tangent vector to $M$ at $y$.
So let be $y$ a corner point for a $k$-manifold $M$ in $\Bbb R^n$ and we shall prove there exist a parametrized curve having the properties described above. So is $y$ is a corner point for $M$ coordinate patch $\alpha:U\rightarrow V$ defined in a set $U$ that is open in $\Bbb R^{k-h}\times[0,+\infty)^{h}$ for any $h\le k$ but not in $\Bbb R^k$ so that $$ y=\alpha(x) $$ for some $x\in\text{bd}\Big(\Bbb R^{k-h}\times[0,+\infty)^{h}\Big)$. Now if $(y,\vec v)$ is a tangent vector to $M$ at $y$ then there exist $\vec u\in\Bbb R^k$ such that $\vec v=D\alpha(x)\cdot\vec u$ and moreover since $U$ is open in $\Bbb R^{k-h}\times[0,+\infty)^{h}$ there exist an open cube $C(x,r)$ centred at $x$ and of radius $r>0$ such that. $$ C(x;r)\cap \Bbb R^{k-h}\times[0,+\infty)^{h}\subseteq U $$ and so first we suppose that $h=1$. Well in this case observing that $x+t\vec u\in C(x,r)$ iff $x_i+tu_i\in(x_i-r,x_i+r)$ and so iff $tu_i\in(-r,r)$ let be $$ \epsilon:=\min\Big\{\frac{r}{|u_i|}:i=1,...,k\Big\} $$ so that $$ x+t\vec u\in C(x;r)\cap\Bbb R^{k-h}\times[0,+\infty)^{h} $$ for any $t\in[0,\epsilon)$ or $t\in (-\epsilon,0]$ if $u_k\ge 0$ or $u_k\le 0$ respectively. So observing that $$ D\alpha(x+t\vec u)|_{t=0}=D\alpha(x+0\cdot\vec u)\cdot\vec u=\vec v $$ we conclude the result holds for $h=1$. So unfortunately if $h>1$ then $$ x+t\vec u\notin C(x;r)\cap\Bbb R^{k-h}\times[0,+\infty)^{h} $$ because it could be that $x_{i_1}=x_{i_2}=0$ for $i_1,i_2=1,...,k$ and $u_{i_1}<0$ and $u_{i_2}>0$ so I do not be able to implement the above arguments. So could someone help me, please?
