The space of collars of a manifold is contractible 
Theorem: Let $M$ be a smooth manifold with boundary $\partial M$. Let $e_0,e_1 : \partial M\times [0,1]\rightarrow M$ be collars of $M$, i.e. $e_i$ are embeddings such that $e_i(x,0)=x$ for each $x\in \partial M$. Then $e_0$ and $e_1$ are isotopic through embeddings.

For some reason, it is difficult to find a clear proof of this in the literature, despite it being a fundamental result underlying the entirety of, say, cobordism theory and surgery. The only clear reference I could find was buried in Cerf's dissertation, in French. In Q&A style, I submitted my take on a proof as an answer. Are there any holes here? If so, I would appreciate any help to fix my proof.
 A: First, we can "scale" $e_0$ and $e_1$ by smooth functions $h_0,h_1:\partial M \rightarrow (0,1]$ such that the embeddings $e_i'(x,t)=e_i(x,h_i(x)t)$ have the same image in $M$. For example, we can do this by first shrinking $e_1$ until its image is properly contained in the image of $e_0$, then shrinking $e_0$ to match their images. $e_0',e_1'$ are isotopic to the original pair, so we may assume that $e_0$ and $e_1$ have the same image in $M$. 
The composition $e'=e_0^{-1} \circ e_1$ gives an automorphism of $\partial M\times [0,1]$ which is the identity on $\partial M \times \{0\}$. Define an isotopy $h:\partial M \times [0,1]\times [0,1]\rightarrow \partial M \times [0,1]$ by $h(x,t,s)=(e'(x,ts)_{\partial M},(1-s)t+se'(x,t)_{[0,1]})$, where the subscript $_{\partial M}$ denotes the projection $(x,t)\mapsto x$ onto $\partial M$ and similarly for $_{[0,1]}$. The fact that this is an isotopy follows since $e'(x,ts)_{\partial M}$ gives a flow on $\partial M$, and any two increasing self-embeddings of $[0,1]$ are isotopic.  This sets up an isotopy between $e_0^{-1}\circ e_1$ and the identity. Consequentally, $e_0$ and $e_1$ are isotopic.
