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.
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?
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.
