Smoothly extending a metric from a manifold with boundary to an attached cylinder

Let $$M$$ be a manifold with boundary $$\partial M$$. Form the manifold $$M'$$ by attaching a half-infinite cylinder $$\partial M\times[0,\infty)$$ to $$M$$ along its boundary. In other words, $$M'=M\cup_{\partial M}\partial M\times[0,\infty),$$ where we identify $$\partial M\sim\partial M\times\{0\}$$.

Let $$g$$ be a Riemannian metric on $$M$$.

Question: Does there always exist a Riemannian metric $$g'$$ on $$M'$$ such that the restriction of $$g'$$ to $$M$$ is equal to $$g$$?

Comment added later: Now that I think about it, perhaps the required property is built into the definition of smoothness of $$g$$ at the boundary, namely that it's extendible slightly beyond the boundary of the chart into some open neighborhood; one then uses a partition of unity to get a metric on all of $$M'$$.

• Just a quick thought: One big problem here seems to be whether $M'$ admits a differentiable structure - which is possibly linked to the regularity of $\partial M$. For instance if $M=\mathbb{R}_{\geq 0}\times \mathbb{R}_{\geq 0}$, i.e. the quarter space in $\mathbb{R}^2$, then the glued manifold $M'$ might look like $\partial \mathbb{R}_{\geq 0}^3$, i.e. the boundary of an octant in $\mathbb{R}^3$. Sep 25 '20 at 14:31

Take a collar neighborhood of $$\partial M$$. This is given by a flow $$\phi:\partial M \times[0,\infty)\rightarrow M$$, generated by vector field $$X$$ such that $$\phi_{*}(\partial_{t})_{(p,t)}=X_{\phi(p,t)}$$. Define $$g'$$ on $$M'$$ as follows:
$$g' = \begin{cases} g & p\in M\\ \phi_{t}^{*}g_{\phi_{t}(p)} & (p,t)\in\partial M \times[0,\infty) \end{cases}$$
Since $$\phi(p,0) = id$$, at all points in $$\partial M$$, $$g'$$ restricts to g as desired.
• I don't think that $\partial M \times [1,\infty)$ is diffeomorphic to $M$ (for example, take the closed disk $D^2$, it is not diffeomorphic to $\partial D^2 \times \mathbb{R}_+ = \mathbb{S}^1 \times \mathbb{R}_+$). But is is diffeomorphic to an open subset of $M$ and it would be sufficient I guess. Sep 25 '20 at 14:46
• The metric on $M'$ obtained in this way doesn't seem to restrict to the original metric on $M$ though. Sep 25 '20 at 15:19