In my course we have just been introduced to and will only be dealing with regularly embedded submanifolds.
Let $M = [0,1]\times [0,1] \subset \mathbb{R}^2$.
I don't think it's a submanifold. If it is, then is it a smooth submanifold? My thinking is that if we try and take open subsets $U$ and $V$ in $\mathbb{R}$ such that a corner point (say $(1,1)$) is in $U\times V$, then we can't find a continuous function $\phi : U \to V $ such that $M \cap (U \times V) =$ graph$(\phi)$ as it will not be well defined at $1 \in U$ and as $U$ is an open set in $\mathbb{R}$ that contains $1$, it contains $1+\delta$ (for some small $\delta > 0$) but $\phi (1+\delta)$ can't take a value in $V$ in order for $M \cap (U \times V) =$ graph$(\phi)$ to hold.
Is my thinking correct? How would this change if M were not embedded in $\mathbb{R}^2$?
Best Answer
Your thinking is correct. The unit square is not a manifold for the reason you gave in your question, but it is a topological manifold with boundary. Keeping smoothness out of the picture for a moment, a topological manifold with boundary is a second countable Hausdorff space where every point has a neighborhood homeomorphic to either $\mathbb{R}^n$, or the half-space
$$ \mathbb{H}^n \;\; =\;\; \{(x_1, x_2, \ldots, x_n) \in \mathbb{R}^n \; | \; x_n \geq 0\}. $$
You can see that the edges of the square lie in a neighborhood that is homeomorphic to one of these half-spaces and the edge point lies on the edge of $\mathbb{H}^n$. Similarly, a corner point lies on the boundary of one of these half-spaces since it has a neighborhood that can be continuously deformed into a half space (think of unfolding the two edges so that they flatten out). Generally if you have an $n$-manifold with boundary $M$ it can be decomposed as $M = Int(M) \cup \partial M$ where the interior is an $n$-manifold (in the traditional sense) and $\partial M$ is an $(n-1)$-dimensional manifold with $Int(M)\cap \partial M = \emptyset$.
If we now consider smoothness again, then the closed unit square is not a differentiable manifold strictly because of the corners. While there may be a homeomorphism taking a corner neighborhood to the half-space, there isn't a diffeomorphism that accomplishes that.