Is $D^2 \times D^2$ a smooth manifold with boundary? I'm studying Differential Geometry, and I learned that isn't necessarily true that if $M$ and $N $ are manifolds with boundary then $M\times N $ is a manifold with boundary (An easy counter-example is $M=N=[0,1] $).
Define $D^2=\{x\in\mathbb {R}^n;\|x\|\leq 1\}$.
I would like to know if $D^2\times D^2$ is a manifold with boundary. 
Can anyone help me?
 A: Your inquiries

I would like to know if $D^2 \times D^2$ is a manifold with boundary.

and particularly

What I was really trying to do is prove that $D^2 \times D^2$ is diffeomorphic to $D^4$.

beg the question: 
What is the smooth structure you are supposed to  consider on $D^2 \times D^2$?
Your first question could be intended to mean "I would  like to know if $D^2 \times D^2$ admits some atlas which makes it a differentiable [for your purposes] manifold with boundary". If so, then it is true: indeed, pick a homeomorphism $D^2 \times D^2 \to D^4$ and transfer the structure. This imbues $D^2 \times D^2$ with a smooth manifold with boundary structure, which is trivially diffeomorphic to $D^4$. But the way you phrase your second question makes me believe your question was not intended as such.
Your second question, however, can't be interpreted so lightly. You are asking if $M_1$ and $M_2$ are diffeomorphic. Therefore, you must necessarily have a smooth structure selected. 
(In what follows, whenever I say "manifold" I mean "smooth manifold". Relevant to say that this discussion is rather mute in the $C^0$-world only.)
The point is: given $M,N$ manifolds, when we say "$M \times N$ is a manifold", we mean that the obvious way of making it a manifold works: by taking the product of charts. Likewise, if $M$ is a manifold and $N$ is a manifold with boundary, then $M \times N$ is a manifold with boundary in an obvious way. If $M$ and $N$ are manifolds with boundary, there is no direct way to make $M \times N$ a manifold with boundary. However, there is a direct way to build up a manifold with corners. So, if you are to expect a smooth structure from the notation $M \times N$ alone (where $M$ and $N$ are smooth manifolds with boundary), you should expect the structure of a manifold with corners. 

EDIT: OP asks in the comments:

Is there $F:\mathbb{R}^4 \to \mathbb{R}^4$ smooth diffeomorphism, such that $F(D^4)=D^2 \times D^2$?

Suppose this is true. Note that pre-composing $F$ with a rotation (i.e., $F \circ R$) would also be a diffeomorphism that takes $D^4$ to $D^2 \times D^2$. 
Now, let $p \in D^4$ be such that $f(p)=(-1,0,-1,0)$. By the inverse function theorem, $p$ must be in $S^3$ (for if $p$ were in the interior of $D^4$, we would have that a small ball around $f(p)$ should be in $f(D^4)$, which is not true). By the observation in the previous paragraph, we can suppose that $p=(-1,0,0,0)$.
The fact that $f(D^4) \subset D^2 \times D^2$ then implies that $Df_p$ takes $[0,+\infty) \times \mathbb{R}^3$ inside $[0,+\infty)\times \mathbb{R} \times [0,+\infty) \times \mathbb{R}$. 
But a linear transformation $A: \mathbb{R}^4 \to \mathbb{R}^4$ that takes $[0,+\infty) \times \mathbb{R}^3$ inside $[0,+\infty)\times \mathbb{R} \times [0,+\infty) \times \mathbb{R}$ can't be an isomorphism. This follows from the fact that since $A$ is linear, then $A$ would take $\mathbb{R}^4$ inside $\big([0,+\infty)\times \mathbb{R} \times [0,+\infty) \times \mathbb{R}\big) \cup  \big((-\infty,0]\times \mathbb{R} \times (-\infty,0] \times \mathbb{R}\big)$, and thus can't be an isomorphism.
Thus, $Df_p$ is not an isomorphism, and $f$ can't be a diffeomorphism.
