Sets that are diffeomorphic to $(0,1)^k$

Let $W\subset \mathbf R^{k}$ be an open set. Are there conditions on $W$ that guarantee the existence of a map $T:(0,1)^k \rightarrow W$ such that: (i) $T$ is surjective, (ii) $T$ is continuously differentiable, and (iii) $T$ has bounded derivative with everywhere nonsingular Jacobian whose determinant is bounded away from zero?

My (naive?) intuition is that $W$ must be bounded, connected and have "smooth" boundary.

Alternatively - is there a characterization of the sets in $\mathbf R^k$ that are diffeomorphic to $\mathbf R^k$?

Thank you in advance!

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For the first question, the answer is non-emptiness. $\;$ ($T$ can be chosen to affine.) $\;$ For the second question, that's equivalent to being diffeomorphic to $\mathbf{R}^k$, which seems like a nicer way to ask the question. – Ricky Demer Dec 21 '11 at 3:45
Assuming you also meant to include surjectivity, there are obvious obstructions coming from algebraic topology. – Jack Huizenga Dec 21 '11 at 3:47
Ignoring the condition in your title, Ricky Demer answered your question. If you want the map to be a diffeomorphism, then you'd need your space to be contractible among other things. Could you edit your question to make it a little more consistent? – Ryan Budney Dec 21 '11 at 4:32
Hi all: Thank you for your replies! I am sorry for the (very) poor editing. I posted in a rush and forgot two crucial conditions that hopefully make the problem more interesting: 1) The determinant of the Jacobian of $T$ should also be bounded away from zero uniformly on $(0,1)^k$. I believe this is the reason it is not the same as being diffeomorphic to $R^k$? 2) As Jack pointed out, I should have added the map $T$ needs to be surjective. – Andres Dec 21 '11 at 6:00
Do you want injectivity? If not, then $W$ could be an annulus in the plane. – Tom Goodwillie Dec 21 '11 at 6:31