# Is the smooth mapping space a deformation-retract of the continuous one?

Let $$X,Y$$ be smooth manifolds.

Let $$F \subset G$$ denote the spaces of, respectively, smooth and continuous functions $$X\rightarrow Y$$; we give $$F,G$$ the compact-open topologies. (This should be compatible with the subspace topology on $$F$$ from $$G$$.)

Since every continuous map is homotopic to a smooth one by e.g. Whitney approxiomation, we know that $$F$$ meets every path-component of $$G$$.

I was wondering, is $$F$$ furthermore a deformation-retract of $$G$$?

• The more natural question seems to me whether the inclusion is a (weak) homotopy equivalence. Jun 24 at 21:23

No. For example if $$X=Y=\mathbb{R}$$ then the space of all smooth functions (even polynomials) is dense in $$C(X,Y)$$. And so it cannot be a retract of $$C(X,Y)$$. This idea generalizes to any smooth manifolds of positive dimension (in dimension $$0$$ smooth and continuous coincide).