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Let $M$ be a smooth $n$-manifold without boundary, and let $B$ be the open unit ball in $\mathbb{R}^n$. I am trying to understand the space $\text{Emb}(B,M)$ of smooth embeddings of $B$ into $M$. Let $\pi : FM \rightarrow M$ be the bundle of frames on $M$, so for $p \in M$ the fiber $\pi^{-1}(p)$ is the set of all ordered bases for the tangent space of $M$ at $p$. We can define a map $\psi : \text{Emb}(B,M) \rightarrow FM$ as follows. Let $\vec{e}_1,\ldots,\vec{e}_n$ be some fixed basis for the tangent space to $B$ at the origin. Consider an embedding $i : B \hookrightarrow M$. We then define $\psi(i)$ to be the point $(i_{\ast}(\vec{e}_1),\ldots,i_{\ast}(\vec{e}_n))$ in the fiber of $FM$ over $i(0)$.

I believe that it is the case that $\psi$ is a homotopy equivalence. In fact, I have read various articles that make this claim with the following sketch of a proof. First, you prove that for all points $\theta \in FM$ the fiber $\psi^{-1}(\theta)$ is contractible, and then you deduce that $\psi$ is a homotopy equivalence. Here are my questions.

  1. The fact that the fiber $\psi^{-1}(\theta)$ is contractible is supposed to be a souped up version of the uniqueness up to isotopy of tubular neighborhoods. That uniqueness theorem definitely says that $\psi^{-1}(\theta)$ is connected -- does anyone know a reference (preferably in English) that proves that it is contractible?

  2. How does knowing that the fibers of $\psi$ are contractible prove that $\psi$ is a homotopy equivalence? Is it maybe a fiber bundle?

Alternate proofs that $\psi$ is a homotopy equivalence that do not follow the above outline are also welcome.

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For 2, see page 318 of ``Topologie de certains espaces de plongements" by J. Cerf. –  Oscar Randal-Williams Oct 28 '13 at 22:18
Here are a couple of references with a few details of a proof for the result you seek. 1. See theorem V.4.5 in my thesis at arxiv.org/abs/1210.7909 which actually deals with the case of multiple balls (be careful that the target space is mistyped for the case of more than one ball). This gives a very brief proof sketch that $\psi$ is locally trivial and a homotopy equivalence. 2. See also proposition 6.4 in arxiv.org/abs/1307.0322 which deals with the case of a single ball. –  Ricardo Andrade Oct 28 '13 at 22:28
Regarding your second question: In particular, Oscar's reference and the ones I give above show that the map $\psi$ is locally trivial, i.e. a fibre bundle. So the contractibility of the fibres implies that the map is a homotopy equivalence. This implication uses the paracompactness of the base space, and follows from results of Dold and tom Dieck: see theorem 13.3.3 of tom Dieck's book "Algebraic topology". –  Ricardo Andrade Oct 28 '13 at 22:29
@RicardoAndrade : Thank you very much for the references. They are very helpful. And if you do have the time, any further details would be very welcome. –  Lars Oct 29 '13 at 1:09
@OscarRandal-Williams : Thanks for the reference! –  Lars Oct 29 '13 at 1:10

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