Question

Notation: $\mathbf{R}^4$ is a smooth manifold with underlying topology $(\mathbb{R})^4$; ${\mathbb{R}}^4$ is the standard smooth structure.

The two things I know best about $\mathbf{R}^4$ is that it is locally diffeomorphic to $\mathbb{R} ^4$, and that it's contractible. It's easy to see that the contraction can be acheived by a smooth map ${\mathbb{R}}^4\times I\rightarrow{\mathbb{R}}^4$.

• Do I suppose correctly that the same contraction is not smooth as a map ${\mathbf{R}}^4\times I\rightarrow{\mathbf{R}}^4$?
• Do the exotic smooth structures have any smooth contractions?
• If not, are there continuous contractions $\mathbf{R}^4\times I\rightarrow\mathbf{R}^4$ within the smooth maps $\mathbf{R}^4\rightarrow\mathbf{R}^4$?

Answer 1

Community wiki answer:

(Taken From Comments)

What's a "contraction"? Is it a homotopy between the identity and a constant map? Any continuous map can be approximated by a smooth one, and nearby maps are homotopic, so it seems you can find a smooth homotopy between the identity and a constant map. The identity map and a constant map are both smooth, and it's an easy exercise from the above to show that in fact all homotopic smooth maps are smoothly homotopic (in context). It is an exercise in Milnor's Topology from a Differentiable viewpoint.