Is $I \times I$ a $2$-manifold in $\mathbb{R}^2$? I am using the following definition of
k-manifold in $\mathbb{R}^n:$

The question is :
Let $I=[0,1], $is $I \times I$ a 2-manifold in $\mathbb{R}^2$? I have been searching for this problem, and I found the answer is negative because $I^2$ has the "corner" points. But I don't understand why is that true. Anyone can give a rigorous answer to this problem?
 A: HINT:
If $M$ is a manifold (with boundary) of dimension $\ge 2$, then for every $p$ in $M$ there exists a smooth map $\gamma \colon (-1,1) \to \mathbb{R}^n$ such that $\gamma ((-1,1)) \subset M$, $\gamma(0) = p$, and $\gamma'(0) \ne 0$.
Now show that you cannot have such a map for $M=[0,1]\times [0,1]$, and $p=(0,0)$ (  $\gamma$ will not stay in $I\times I$ around $0$, due to the condition  $\gamma'(0) \ne 0$).
$\bf{Added:}$ Consider a smooth map $\gamma =(\gamma_1, \gamma_2) \colon (-1,1)\to I\times I$, $\gamma(0) = (0,0)$. Since $0$ is a minimum point for both $\gamma_1$, and $\gamma_2$, we have $\gamma_1'(0)=\gamma_2'(0)=0$.
In a similar way we can show that a cube is not a manifold with boundary around any point on an edge. For if $\gamma \colon (-1,1) \to I^3$, $\gamma(0) = p=(0,0, z)$, then again $\gamma_1'(0)=\gamma_2'(0) = 0$, so the vector $\gamma'(0)$ should be along the edge on which $p$ is situated. But for any point $p$ on a $3$-dim manifold with boundary, the possible tangents to paths $\gamma$ with values in $M$ at the point $p$ has dimension at least $2$.
