Let $M$ be smooth $n$-manifold with Riemannian metric $g$. If I consider the tangent bundle $TM$, then I can see that the tangent space at a point $p\in M$ is an embedded submanifold of $TM$. Indeed, called $\pi\colon TM\longrightarrow M$ the canonical projection, and taken a chart $(U,\phi)$ of $M$, the chart $$\pi^{-1}(U)=TU\longrightarrow \phi(U)\times \mathbb{R}^n$$ gives us a diffeomorphism between the embedded submanifold $\phi(p)\times\mathbb{R^n}\subseteq \mathbb{R}^{2n}$ and $T_pM$. I would prove that the topology of $T_pM$ induced by the Riemannian metric $g$ is the same of the topology of embedded submanifold of $TM$.

In this way, if the claim above is true, I obtain that the set $V=\lbrace v\in T_pM \: : \: |v|<\epsilon\rbrace $ (for some $\epsilon >0$), where we define the exponential map $\exp_p\colon V\longrightarrow M$, is open.


closed as unclear what you're asking by Moishe Kohan, José Carlos Santos, kingW3, jvdhooft, Lord Shark the Unknown Jul 2 '17 at 13:42

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ What exactly is your question? $\endgroup$ – Math1000 Jul 2 '17 at 0:24