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This is a cross-post of this question from MSE.


Given topological manifolds $M$ and $N$ of the same dimension, let $\operatorname{Emb}(M,N)$ denote the subspace of $\operatorname{Map}(M,N)$ consisting of the (topological) embeddings $M\to N$. Here $\operatorname{Map}(M,N)$ is equipped with the compact-open topology. I am looking for a reference of the following assertion (if it is true):

Let $M$ be a topological manifold of dimension $k$. The evaluation map $$\operatorname{Emb}(\mathbb{R}^k,M)\to M$$ at the origin $0\in\mathbb{R}^k$ is a Serre fibration (or a Hurewicz fibration, preferably).

Thanks in advance.

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  • $\begingroup$ In the smooth category this is a result of Palais (vmm.math.uci.edu/PalaisPapers/…) but I've not seen it in the topological category. $\endgroup$
    – Mark Grant
    Commented May 29, 2023 at 9:05
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    $\begingroup$ @MarkGrant Ryan cites a topological version in his answer here $\endgroup$ Commented May 29, 2023 at 14:36

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It is a Serre fibration, and this is one of the rare cases where you can give an elementary argument and don't need something hard like Edwards-Kirby, Lees, or Lashof. I do not know a reference so let me outline the argument.

We construct a lift in a commutative square $$\require{AMScd} \begin{CD} I^j \times \{0\} @>{f}>> \mathrm{Emb}(\mathbb{R}^k,M) \\ @VVV @V{\mathrm{ev}_0}VV \\ I^j \times I @>{F}>> M \end{CD}$$ by finding a continuous map $H \colon I^j \times I \to \mathrm{Homeo}_c(M)$ such that $H(x,0) = \mathrm{id}_M$ and $H(x,t)(F(x,0)) = F(x,t)$. Then the desired lift will be given by $(x,t) \mapsto H(x,t) \circ f(x)$.

Subdividing the domain $I^j \times I$ of $F$ into fine enough $(j+1)$-cubes such that $F$ sends each of them into a chart of $M$ homeomorphic to $\mathbb{R}^k$, we will construct $H$ inductively over these cubes and thus may assume that $M = \mathbb{R}^k$. We can then define $H(x,t)$ as given by translation by $F(x,t)-F(x,0) \in \mathbb{R}^k$, cut off by a suitable bump function.

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  • $\begingroup$ Beautiful! I never thought of deforming $M$ (i.e., finding the map $H$ as you described). Thank you! $\endgroup$
    – Ken
    Commented Aug 11, 2023 at 3:12
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Sorry, I did not notice that your question was in $C^0$; so the following is a comment, not an answer. In the smooth ($C^\infty$) regularity class, or $C^r$ with $r\ge 1$, the restriction map is even a locally trivial fibre bundle (fibration). You can replace $R^n$ by any manifold and $0$ by any submanifold. This was (a part of) Jean Cerf's first work in 1961.

Cerf, Jean Topologie de certains espaces de plongements.(French) Bull. Soc. Math. France 89 (1961), 227–380.

His "first fibration theorem" (p. 294, section 2.2.2, corollary 2) states that for $H\subset E\subset F$ (manifolds), the restriction map $Emb(E,F)\to Emb(H,F)$ is a locally trivial fibre bundle. The theorem is actually not easy to find in the paper even if you read French perfectly, because of the formal style and of the extreme generality: the manifolds can have boundaries and corners, they can be noncompact, you can fix the embeddings on some subsets, he works with several differentiability classes, etc. Cerf was pretty young! In particular, if $E=F$ is noncompact, you can fix the germ at infinity of the embeddings $E\to E$ to be the identity; then, you get that the restriction map $Diff_c(E)\to Emb (H,E)$ is a locally trivial fibre bundle, where $Diff_c(E)$ is the group of the compactly supported diffeomorphisms of $E$...

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  • $\begingroup$ I'm afraid that my French is very poor. Can you tell me which result of the paper you are referring to? $\endgroup$
    – Ken
    Commented Jul 11, 2023 at 4:44

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