My question concerns Gromov's h-principle for open diffeomorphism-invariant partial differential relations on open manifolds; see e.g. Eliashberg/Mishachev: *Introduction to the h-principle*, §6.2.A and Theorem 7.2.3. I ask whether – and if so, why – the weak homotopy equivalence in that theorem is even a homotopy equivalence.

In more detail: Let $E\to M$ be a natural smooth fibre bundle over an open manifold $M$. For some $k\in\mathbb{N}$, let $\mathcal{R}$ be a diffeomorphism-invariant open subset of the $k$-jet bundle total space $J^kE$. Let $Sec(\mathcal{R})$ be the set of $C^0$ sections in $J^kE\to M$ whose images lie in $\mathcal{R}$; we equip it with the compact-open topology. Let $Hol(\mathcal{R})$ be the subspace consisting of those elements of $Sec(\mathcal{R})$ which are $k$-jets of $C^k$ sections in $E\to M$. Gromov's theorem says that the inclusion $Hol(\mathcal{R})\to Sec(\mathcal{R})$ is a weak homotopy equivalence.

In their book, Eliashberg/Mishachev claim in a side remark on p. 62 that it is even a homotopy equivalence. Their argument is that $Sec(\mathcal{R})$ and $Hol(\mathcal{R})$ are metrisable Fréchet manifolds. It is known that metrisable manifolds modelled on locally convex topological vector spaces are dominated by CW complexes (and hence are homotopy equivalent to CW complexes). Since every weak homotopy equivalence between CW complexes is a homotopy equivalence by Whitehead's theorem, this implies that the inclusion $Hol(\mathcal{R})\to Sec(\mathcal{R})$ is a homotopy equivalence.

But why are $Sec(\mathcal{R})$ and $Hol(\mathcal{R})$ metrisable manifolds? First, Eliashberg/Mishachev claim this in a context where $\mathcal{R}$ is an arbitrary subset of $J^kE$. I doubt that in this generality the spaces are manifolds with respect to *any* reasonable topology. Second, even if we assume that $\mathcal{R}$ is open and diffeomorphism-invariant, then giving spaces of sections in $\mathcal{R}$ a nice manifold topology is difficult because $M$ is noncompact. Most importantly, we are not free to choose a topology: the topology is the compact-open one.

Does the Eliashberg/Mishachev argument work somehow? Or can you give a different proof of the claim that the inclusion is a homotopy equivalence? If so, is $Hol(\mathcal{R})$ even a strong deformation retract of $Sec(\mathcal{R})$?