I am working on manifolds with boundary.
Assume $(E, \pi)$ is a vector bundle over a manifold with boundary $M$, for example $TM$ or $T^*M$. If $\phi : \pi^{-1}(\mathcal{U}) \to \mathcal{U} \times \mathbb{R}^m$ is a vector bundle chart of $E$ and $\varphi : \mathcal{U} \to \mathbb{R}^n$ is a chart (with boundary) of the manifold $M$, then we can (can we?) induce a manifold chart on $E$:
$$
\xi : \pi^{-1}(\mathcal{U}) \to \varphi(\mathcal{U}) \times \mathbb{R}^m : x \mapsto (\varphi(\pi(x)), t_{\pi(x)}^{-1}(x))
$$
where $t_p : \mathbb{R}^m \to \pi^{-1}(p)$ is the isomorphism between the vector space $\mathbb{R}^m$ and the fiber $\pi^{-1}(p)$ ($t_p(v) := \varphi^{-1}(p, v)$).
When $M$ is a manifold without boundary, we can then collect all the charts built in this way into an atlas and induce a manifold structure on $E$.
When $M$ is a manifold without boundary, is the construction above valid? Is $E$ equipped with this manifold structure a manifold with or without boundary?
In other words, if $M$ is a manifold with boundary, is $TM$ a manifold with or without boundary?
Best Answer
I think that the answer is yes, here there is my reasoning: $$\mathbb{R}_+^m:= \{ x \in \mathbb{R}^m\,|\,x_1 \geq 0\}$$ If $M$ is a manifold with boundary, for every boundary chart $$\varphi : U \rightarrow \mathbb{R}_+^m$$ the induced chart on $TM := \sqcup_{p \in M} T_pM$ $$\Phi : \pi^{-1}(U) \rightarrow \mathbb{R}_+^m \times \mathbb{R}^m$$ $$\Phi(X_p) = (\varphi(p),d\varphi^1|_p(X_p),..., d\varphi^m|_p(X_p) )$$ is a chart with boundary since $ \mathbb{R}_+^m \times \mathbb{R}^m = \mathbb{R}^{2m}_+$
Here $d\varphi^1|_p,..., d\varphi^m|_p$ is the dual basis of $T_p^*M$ and $\pi : TM \rightarrow M$ is the standard projection