Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
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 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 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.