products of manifolds with boundary I am struggling a little bit with the following problem.
Let $M$ and $N$ be smooth manifolds of dimension $m$ and $n$, respectively. Show that the product $M\times N$ is a smooth manifold of dimension $m+n$. Is this also true for smooth manifolds with boundary?
I am able to do the first part (I just simply considered the product charts), but I am struggling with the part about the manifolds with boundary. My guess is that the answer is no, and most probably the space $[0,1]\times [0,1]$ should do the trick, but I am not quite able to give a rigourous proof that something goes wrong at the corners. 
 A: So, we need to proof that manifold $[0..1] \times [0..1]$ doesnt have any smooth structure. Suppose that our square $[0..1] \times [0..1]$ have smooth atlas. Let take the smooth  map $\alpha : [0..\frac{1}{2}) \times [0..\frac{1}{2}) \to [0..1) \times (0..1)$, definition of smooth manifold with boundary guarantee that such smooth map exist. If $[0..\frac{1}{2}) \times [0..\frac{1}{2})$ diffeomorphic to $[0..1) \times (0..1)$ than their boundaries is diffeomorphic too (it is most important place in proof), i.e $\{(t,0) : t \in [0..1)\} \cup \{(0,t) : t \in [0..1)\}$ diffemorphic to $(0..1)$. Define $A := \{(t,0) : t \in [0..1)\} \cup \{(0,t) : t \in [0..1)\}$, $I:= (0..1)$, $\varphi : I \to A$ - some diffeomorphism. Then $d\varphi$ is covector continuous field such that, for all points $x \in I$ we have or $d \varphi (x) = (t,0), t\neq 0$ or $d \varphi(x) = (0,t), t \neq 0$ and every case is realized at least in one point. But it is absurd.
If you want that proof in a short: it is obviously that fact "$[0..1] \times [0..1]$ smooth manifold" equivelant to "graph of $y = |x|$ diffeomorphic to $\mathbb{R}$", but if we have such diffeomorphism than differential of it must be identically $0$ in break point. Contradiction.
