Take an exotic $\mathbb{R}^4$ i.e. $V = (\mathbb{R}^4,d)$ such that $V$ is not diffeomorphic to $\mathbb{R}^4$ with standard metric.


Is it true (obvious?) that any two smooth maps $f\_1, f\_2: S^k \to V$ are equivalent via **smooth** homotopy?


edited: (for any k)



smooth homotopy on exotic R^4
Yes, since smooth maps which are (continuously) homotopic are always smoothly homotopic. See Kosinski's "Differential Manifolds", Theorem III.2.5 and Corollary III.2.6.


