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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.

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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

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