# Are inclusions from spaces of $C^\infty$ sections into spaces of $C^k$ sections homotopy equivalences?

[EDIT: The answer to my original question was obviously no, as user56365 pointed out. Here is what I should have asked.]

For finite-dimensional smooth manifolds $E,M$, let $E\to M$ be a smooth fibre bundle. Let $k\in\mathbb{N}=\{0,1,\dots\}$, and let $U$ be an open subset of the total space $J^kE$ of the $k$-jet bundle. For $l\in\mathbb{N}\cup\{\infty\}$, let $C^l(E)$ denote the set of $C^l$ sections in $E\to M$. For $l\geq k$, let $\mathscr{C}^l(U)$ be the set of $s\in C^l(E)$ such that the image of $j^ks$ is contained in $U$. We equip $C^l(E)$ with the compact-open $C^l$-topology, and $\mathscr{C}^l(U)$ with the subspace topology.

Smooth approximation theorems (e.g. D. Spring: Convex integration theory, p. 9) say that for every $s\in\mathscr{C}^k(U)$, there is a continuous map $h\colon[0,1]\to\mathscr{C}^k(U)$ with $h(0)=s$ and $h(1)\in\mathscr{C}^\infty(U)$.

My question is whether $h$ can be chosen depending nicely on $s$: Is the inclusion $i\colon\mathscr{C}^\infty(U)\to\mathscr{C}^k(U)$ a homotopy equivalence?

(I have read unreferenced claims that this is true in certain special cases, so the answer is probably yes. References to proofs of special cases are welcome, but I really need the general case.)

If $i$ is a homotopy equivalence, then one could ask whether a homotopy inverse $r\colon\mathscr{C}^k(U)\to\mathscr{C}^\infty(U)$ and a homotopy $H\colon[0,1]\times \mathscr{C}^k(U)\to\mathscr{C}^k(U)$ from $id$ to $i\circ r$ can be found with additional properties. For instance, the standard smooth approximation theorems say that if $s$ is already smooth on a neighbourhood of a closed subset $A\subseteq M$, then $h$ can be found such that each $h(t)$ coincides on $A$ with $s$. One would like to have an analogous property for the homotopy $H$ (which can be thought of as a $\mathscr{C}^k(U)$-parametrised family of $h$s).

• But $\mathscr{C}^\infty(U)$ is surely not closed in $\mathscr{C}^k(U)$ while (strong deformation) retracts of Hausdorff spaces are necessarily closed. – user56365 Jul 29 '14 at 14:58