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This question was previously posted on MSE.

Let $M, N$ be smooth connected manifolds (without boundary), where $M$ is a compact manifold, so we can put a topology in the space $\mathcal C^\infty(M, N)$ using $\mathcal{C}^1$ Whitney Topology.

Now, consider $S\subset M$ a compact submanifold of $M$ with boundary such that $\text{dim}S=\text{dim}M$, using the same process we can put a topology in $\mathcal C^\infty(S,N)$ using the $\mathcal{C}^1$ Whitney Topology. There is a natural continous projection of $\mathcal C^\infty(M, N)$ on $\mathcal C^\infty(S,N)$, definided by

\begin{align*} \pi: \mathcal C^\infty(M, N) &\to \mathcal C^\infty(S,N)\\ f&\mapsto \left.f\right|_{S}. \end{align*}

My Question: Is $\pi$ an open map or at least a quotient map?


Some comments

$\mathcal{C}^1$-Whitney Topology is also called $\mathcal{C}^1$-strong topology.

As noticed for the user Adam Chalumeau, on the book "Morris W. Hirsh Differential Topology" there is the following exercise

[Exercise 16, page 41]: Let $M, N$ be $\mathcal{C}^r$ manifolds. Let $V⊂M$ be an open set then

  • The restriction map $$δ:\mathcal{C}^r(M,N)→\mathcal{C}^r(V,N)$$ $$δ(f)=f|_V$$ is continuous for the weak topology, but not always for the strong.

  • $δ$ is open for the strong topologies, but not always for the weak".

Since our $M$ is compact weak topology = strong topology. However, I don't know how to solve this exercise let alone adapt such proof to the case that I want.

Does anyone know anything about this problem?

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Let me try to give at least some partial answers (and apologies that I am leaving the realm of topological spaces to give you these answers, but I am a differential geometer, so you have been warned):

  1. In general the restriction map will fail to be a quotient map, as it will fail to be surjective (and quotient maps are by definition surjective mappings). Extending a $C^\infty$-mapping from a compact submanifold with boundary to a larger manifold is not always possible due to topological reasons. As a simple example take a smooth map $f \colon \mathbb{S}^2\supseteq C \rightarrow \mathbb{S}^1$, where the $\mathbb{S}^i$ are the unit spheres in $\mathbb{R}^{i+1}$ and $C$ is an 'equatorial belt' (which is a compact submanifold with smooth boundary). Then $f$ cannot extend to $\mathbb{S}^2$ if has non-zero winding number (cf. Lee, Introduction to smooth manifolds, Corollary 6.27).

  2. It might be helpful to note that the $C^k$-Whitney topologies on all your spaces coincide with the compact open $C^k$-topology (not sure whether you meant this topology with 'weak topology', just wanted to be sure we are on the same page). If you had endowed your spaces of smooth functions with the $C^\infty$-topology (instead of the $C^1$-topology), then the answer to your question is positive: The continuous projection $\pi$ is open. This follows from a recent paper of mine (and D. Roberts): Extending Whitney’s extension theorem: nonlinear function spaces Theorem B establishes that $\pi \colon C^\infty (M,N) \rightarrow C^\infty (S,N)$ is a submersion (but in general NOT surjective, see 1.) between infinite-dimensional manifolds (the meaning of this is that it is in local charts a projection, thus an open map by the usual proof). Note that in the cited paper, the compact submanifold $S$ is allowed to have a much more irregular boundary then the smooth boundary you want to consider.

  3. To adapt what was said in 2. to your case with the much weaker $C^1$-topology, I guess there would be two (relatively straight forward) ways to achieve this: (a)Try to copy the argument in 2.: The function space techniques which allow one to drag the Whitney extension theorem from chart domains to a compact manifold are (apart from the Whitney extension theorem) all amenable to change from $C^\infty$ to $C^k$ ($k$ finite), i.e. if you change the topology on the spaces, you obtain smooth maps with respect to the weaker topologies (to see this, compare with the description of the manifold charts etc. for $C^k (M,N)$ in Appendix A of Lie groupoids of mappings taking values in a Lie groupoid). Hence you can just copy the proof steps and either see that the Whitney map (from the extension theorem) is also continuous in the $C^1$-topology (Check this!) or you use a corresponding extension theorem for $C^1$-maps due to Fefferman (and check that it yields $C^\infty$-mappings if it gets a $C^\infty$-map as input [I have no idea if this will work]). The arguments then yield a local continuous inverse for $\pi$ and this should imply that $\pi$ is open. (b) Ignore all the infinite-dimensional manifold stuff and try to construct something in the topology. As a subbase ($\{f\in C^\infty (M,N) \mid Tf (L) \subseteq O \subseteq TN\}, L \subseteq TM$ compact and $O$ open) of the compact open $C^1$-topology is well known you can check openness of $\pi$ on the BASE generated by the subbase. Since $S$ is compact we can identify $TS$ with a closed subset of $TM$, whence $L \cap TS$ is compact in the subspace topology. Now you probably want to distinguish cases of whether $L \cap TS =\emptyset$ or not. Note however, that it is not quite clear to me how to identify the image of the (sub)basic neighborhoods under $\pi$. Since we are dealing here with a mapping which is in general non surjective, one probably needs at this stage an extension argument to see that the image of such a neighborhood is an open neighborhood of each of its elements.

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    $\begingroup$ I also thought of referencing that recent paper of yours ;-) but if the OP wants submanifolds-with-boundary of positive codimension (for instance a non-closed simple path in a surface, or an embedding of a surface with boundary inside some random manifold of higher dimension), then its results don't apply. $\endgroup$ – David Roberts May 27 at 11:09
  • $\begingroup$ Hei David, sorry edited a reference to you in (did not mean to diminish your part in the paper and was a bit strapped for time this morning so I was not too careful). However, I thought that as dim $S$ = dim $M$ we are dealing with the situation of our paper (i.e. an open subset with boundary). Though it is true that for lower dimensional submanifolds we did not write up something, I think that as long as one knows what submanifold charts should be (which is not clear for the rough boundary we consider, but should be clear for smooth boundary), our approach holds up $\endgroup$ – Alexander Schmeding May 27 at 14:17
  • $\begingroup$ @AlexanderSchmeding thx for the answer. I will read your paper and try to apply your amazing suggestion. If it leads me to the desired result I will accept your question, I was really very lost, your comment gave me a way to follow. $\endgroup$ – Matheus Manzatto May 27 at 17:39
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    $\begingroup$ @AlexanderSchmeding ah, I missed the dimension criterion! As you were (I wasn't meaning to pull you up on the paper, just felt odd recommending my own work when I didn't think it applied!) $\endgroup$ – David Roberts May 27 at 21:50

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