$M\setminus \mathrm{Cut}(L)$ deforms to $L$ Let $M$ be a connected complete Riemannian manifold, $p\in M$ and $\mathrm{Cu}(p)$ denotes the cut locus of a point. This is a standard result that $M\setminus\mathrm{Cu}(p)$ deforms to $p$.
Now if $L$ is a compact submanifold of $M$ and $\mathrm{C}u(L)$ denotes the cut locus of $L$ then is it true that $M\setminus \mathrm{Cu}(L)$ deforms to $L$. I checked some of the examples and it worked. But I am unable to prove this fact. Any reference or proof will be appreciated.
Edit

*

*We say $q\in \mathrm{Cu}(L)$ if any distance minimal geodesic joining $L$ to $q$ is no longer distance minimal beyond $q$.

*By deformation, I mean to find  $$H:M\setminus \mathrm{Cu}(L)\times [0,1]\to M\setminus \mathrm{Cu}(L)$$ such that $H(x,0)= x,~ H(q,1)\in L$ and $H(q,t)=q$  (if $q\in L$).

Thanks!
 A: I am assuming the submanifold has no boundary.  In this case it is true that $M\setminus cut(L)$ deformation retracts to $L$.  I'm not sure what happens if there is boundary.
Let $\nu L$ denote the normal bundle of $L$.  For each $v\in \nu L$, let $t_v\in [0,\infty]$ denote the time where the geodesic $\exp(tv)$ stops minimizing distance to $L$.
We define $U\subseteq \nu L$ by $U = \{tv \in \nu L: \|v\| = 1\text{ and } t<t_v\}$.
Then, according to this paper, $\exp|_{U}:U\rightarrow M\setminus Cut(L)$ is a diffeomorphism.  Just to make notation a bit nicer, I'll use $\rho$ to denote $\exp|_{U}$.
Now, define $H:M\setminus C(L)\times [0,1]\rightarrow M\setminus C(L)$ by $H(m,t) = \exp((1-t)\rho^{-1}(m))$.
This is a composition of continuous functions, so is continuous.  Further, $H(m,0) = \exp(\rho^{-1}(m)) = m$.  Further, $H(m,1) = \exp(0\cdot \rho^{-1}(m)) \in L$.  Lastly, for any $\ell\in L$, $\rho^{-1}(\ell)$ is in the zero section of $\nu L$, so $\exp((1-t) \rho^{-1}(\ell)) = \ell$, independent of $t$.
