# Usefulness of compact boundary proving uniqueness in gluing smooth manifolds

I have a question about the following fact which used to prove the uniqueness(upto diffeomorphism) in gluing smooth manifolds:

Let $$M$$ be a smooth manifold and let $$f_0,f_1:[0,1]\times \partial M\to M$$ be two smooth embedding that are identity map on $$\partial M\times\{0\}=\partial M$$. If $$\partial M$$ is compact then there is a diffeotopy $$H:M\times [0,1]\to M$$ relative to $$\partial M$$ from $$H_0=\text{Id}_M$$ to a diffeomorphism $$\Phi=H_1:M\to M$$ with $$\Phi\circ f_0=f_1$$.

Now, proof of this can be found in the book Differential Topology by C.T.C. Wall or Introduction to Differential Topology by Klaus Jänich and T. Bröcker.

Both books assume $$\partial M$$ to be compact to prove the above fact, is it necessary? I mean is there any sort of counterexample without the assumption $$\partial M$$ is compact?

Remark: I noticed the same assumption that $$\partial M$$ is compact in the setting of topological manifold, for example, in Collars and concordances of topological manifolds written by M.A. Armstrong.

• I guess the the assumption that $\partial M$ is compact is crucial to show you can find a tubular neighbourhood of $\partial M$ in $M$ of the form $\partial M \times [0,\varepsilon)$ defined by a riemannian metric (I mean defined by the exponential map). Nov 10, 2020 at 9:51
• Thanks for your comment. But I have proof of the collar neighborhood theorem without compactness assumption on the boundary, might be in the book Differential Topology by C.T.C. Wall. Nov 10, 2020 at 10:02
• Maybe in the proof of the statement above, there is a use of a uniform tubular neighboorhood, that is an isometric embedding $\partial M \times [0,\varepsilon) \to M$? Such a uniform tubular neighbourhood exists as soon as $\partial M$ is compact, but there exist counter examples in the non-compact case. Nov 10, 2020 at 10:05
• Good point! That might be possible. Okay, I will check it once the uniformity. Nov 10, 2020 at 10:10

Suppose $$(M,g)$$ is a riemannian manifold and suppose $$\partial M$$ has a unit normal vector field, pointing inside $$M$$, say $$\nu$$. For $$p\in \partial M$$ fixed, there is a little interval $$[0,\varepsilon(p))$$ on which the geodesic $$\gamma_p(t)=\exp_p(t\nu(p))$$ is defined. By regularity assumptions on $$(M,g)$$, one can chose $$p \mapsto \varepsilon(p)$$ as regular as $$(M,g)$$. Thus one can define a map \begin{align} E :\bigcup_{p\in \partial M}\{p\} \times [0,\varepsilon(p)) & \to M \\ (p,t) \mapsto \gamma_p(t) \end{align} Because the differential of the exponential map at $$0$$ is the identity map, by the inverse map theorem, $$E$$ is a local diffeomorphism around $$(q,0)\in\bigcup_{p\in \partial M}\{p\} \times [0,\varepsilon(p))$$: there exists an open neighbourhood $$U_q$$ of $$(q,0)$$ on which $$E$$ is a diffeomorphism onto its image. Such a neighbourhood contains an open subset $$V_q \times [0,\varepsilon'(q))$$.
If moreover, $$\partial M$$ is compact, one can chose a finite number of $$q_i \in \partial M$$ such that $$V_{q_i}\times [0,\varepsilon'(q_i))$$ covers $$\partial M$$. If $$\varepsilon = \min \varepsilon'(q_i)$$, then $$\partial M \times [0,\varepsilon)$$ is a neighbourhood on which $$E$$ is a local diffeomorphism, and we can choose $$\varepsilon$$ small enough such that it is injective. Thus, we have found a uniform tubular neighbourhood of $$\partial M$$ in $$M$$ isometrically embedded $$E\left(\partial M \times [0,\varepsilon)\right)$$.
If $$M$$ is not compact, one could not be able to chose $$\varepsilon$$ and to isometrically embed such a tubular neighbourhood. Because topology and curvature are closely related, maybe there can exist no riemannian metric on $$M$$ for which this construction can work. Right now I cannot figure out an example of such a manifold $$M$$, but one can easily construct a particular riemannian metric on a manifold with non-compact boundary such that you cannot define a uniform tubular neighbourhood. For example, the subset of $$\mathbb{R}^2$$ defined by \begin{align} M = \left\{(x,y)\in\mathbb{R}^2 ~|~ 0 \leqslant y < e^{-x^2} \right\} \end{align} with the induced euclidean metric is a submanifold with boundary $$\partial M =\mathbb{R}\times\{0\}$$, but there exists no isometric tubular neighbourhood of the form $$\mathbb{R}\times [0,\varepsilon)$$. This is a bad example, because here the metric is flat and we can easily deform it... But still, it is an example.