# Homotopy equivalence vs weak homotopy equivalence in Gromov's h-principle

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 $Sec(\mathcal{R})$?

• If $M$ is compact, then $Sec(\mathcal{R})$ is an open subset of $Sec(\mathcal{J^kE})$, hence a manifold with the compact-open topology. Of course, the same cannot be said when $M$ is non-compact, unless the fibers of $\mathbb{R}$ differ from those of $J^kE$ only over a compact set $K\subset M$. On the other hand, we can write $Sec(\mathcal{R}) = \bigcap_K Sec(\mathcal{R}_K)$, where $K$ ranges over a countable exhaustion of $M$ by compacts, if one exists, and $\mathcal{R}_K$ agrees with $\mathcal{R}$ over $K$ and with $J^kE$ outside it. Could that be enough for a CW structure? – Igor Khavkine Jul 29 '14 at 3:49
• Hi Igor, I see a priori no reason why countable intersections of "good" spaces in our context should be good (where being good means at least that weak homotopy equivalence implies homotopy equivalence). I would rather suspect that in general they are not (whatever "good" means precisely). Of course, even if they are not good, that will not tell us whether our inclusions (which are quite special weak homotopy equivalences) are always homotopy equivalences. – Marc Nardmann Jul 30 '14 at 0:00
• It was just a thought, so you may be right. It seems that the CW type of mapping spaces is an old and well studied question. In particular, it seems that $C(X,Y)$ is of CW type when both $X$ and $Y$ are of CW type, with $X$ finite dimensional and $Y$ has only finitely many non-vanishing homotopy groups (not sure how to check that last condition, unfortunately): arxiv.org/abs/0708.2838 – Igor Khavkine Jul 30 '14 at 14:43
• That is a useful reference, thank you. It mentions also several necessary conditions for $C(X,Y)$ to be of CW type, for instance Example 3 on p. 25. Let us take a trivial bundle $E=M\times F\to M$ and, say, $\mathcal{R}=J^kE$. Then $Sec(\mathcal{R})$ is homotopy equivalent to $C(M,F)$. Suitable manifolds $M,F$ are up to homotopy equivalence of the form of Example 3, e.g. with $T=\mathbb{R}$, all $m_\lambda=1$, $dim M=2$ and $Y=S^2$. Then $C(M,F)$ is not of CW type. I.e., the Eliashberg/Mishachev argument fails in general. Of course, the inclusion might still always be a homotopy equivalence. – Marc Nardmann Jul 31 '14 at 5:45