# Deformation retraction onto an open subset of manifold boundary

I'd like to prove the following result, perhaps with additional assumption if needed -- I don't know whether the claim holds. Let $$M$$ be a compact connected manifold with boundary $$\partial M$$. Let $$U' \subset \partial M$$ be open in $$\partial M$$. Then there exists $$U \subset M$$ open in $$M$$ such that $$U \cap \partial M = U'$$, and $$U'$$ is a strong deformation retract of $$U$$. Any ideas welcome.

### Background

This result is true in the following specific case which I'd like to generalize. Let $$M = \overline{B_n}$$ be the closed unit n-ball at origin in $$\mathbb{R}^n$$. Then $$\partial M = S^{n - 1}$$, the unit $$(n-1)$$-sphere at origin. Let $$U' \subset S^{n - 1}$$ be open in $$S^{n - 1}$$. Let $$U = \{(1 - t)u' : t \in [0, 1/2) \text{ and } u' \in U'\}$$. Then $$U$$ is open in $$M$$ and $$U \cap \partial M = U'$$. Let $$f : U \times [0, 1] \to U$$ be such that $$f(x, t) = (1 - t)x + tx/|x|$$. Then $$f$$ is a strong deformation retraction of $$U$$ onto $$U'$$.

There is this thing called collar neighbourhood of $$\partial M$$. It's an open neighbourhood of $$\partial M$$ in $$M$$ which is homeomorphic to $$\partial M\times [0,1)$$ and the homeomorphism maps $$x\mapsto (x,0)$$ for $$x\in\partial M$$. In fact you did use a collar neighbourhood in the specific case of $$n$$-ball.
So if $$U'\subseteq\partial M$$ is open and $$\partial M$$ has a collar neighbourhood then we can restrict the collar neighbourhood to $$U'\times[0,1)$$ which is again open in $$M$$ because it is open in $$\partial M\times [0,1)$$. So we obtain an open subset of $$M$$ homeomorphic to $$U'\times[0,1)$$. And such neighbourhood easily (strongly) deformation retracts onto $$U'$$.