Can a non-trivial algebraic variety carry a vector bundle whose total space is affine space? Suppose $X$ is an algebraic variety over $\mathbb{C}$, and let $Y\to X$ be an algebraic vector bundle. Suppose $Y$ is algebraically isomorphic to $\mathbb{C}^n$ for some $n$. Does it follow that $X$ is algebraically isomorphic to $\mathbb{C}^m$ for some $m$?
 A: Summing up the discussion in the comments:
As user ulrich observed, the vector bundle has to be trivial. First, since $X$ is a closed subscheme of $Y\simeq \mathbf{A}^n$ via the zero section, it has to be affine. It is also smooth. Finally, it is $\mathbf{A}^1$-contractible since $Y$ is, since $Y\to X$ induces an equivalence in $\mathbf{A}^1$-homotopy. We conclude by a theorem of Morel (see Chapter 7 in $\mathbf{A}^!$-algebraic topology over a field, here), saying that vector bundles on a smooth affine $\mathbf{A}^1$-contractible variety are trivial.
A simpler argument using Quillen-Suslin was given by Anton: $X$ is a retract of $Y \simeq \mathbf{A}^n$ via the zero section, and since every vector bundle on $Y$ is trivial, the same is true for $X$.
This turns the question into an important special case of the Cancellation Problem.

Cancellation Problem. If $X\times \mathbf{A}^m \simeq Z \times \mathbf{A}^m$, can we conclude that $X\simeq Z$?

The answer to this general problem is no (there are famous counterexamples already in dimension two, known as Danielewski surfaces). However, in the special case where $Z$ is an affine space ($\mathbf{A}^{r}$, $r+m=n$ in our case) the answer is known to be yes  for $r\leq 2$, and open for $r>2$. (In positive characteristic, the answer is no for all $r>2$:  arxiv.org/abs/1208.0483 )
This answer is "Community Wiki".  
