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I'm learning differential topology from Lee's Introduction to Manifolds and was reading about the tangent space $T_pM$ of a smooth manifold $M$, of dimension say, $n$ at some point $p \in M$. Now, I do understand that $T_pM$ is an $n$ dimensional vector space, but I'm having trouble understanding what does an open set in $T_pM$ look like. I can't visualize what it is. Any help towards that end is appreciated.

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  • $\begingroup$ Start thinking about dimensions 1 and 2. $\endgroup$ – Anubhav Mukherjee Oct 19 '17 at 22:24
  • $\begingroup$ Why does $T_{p}M need to have a topology? The tangent bundle has a natural topology, but the tangent space at each point is just a vector space. $\endgroup$ – Aritro Pathak Oct 19 '17 at 22:26
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Since $T_pM$ is finite dimensional, it is isomorphic to $\mathbb{R}^n$ for some $n$. It is endowed with the standard topology.

Remark: Of course the isomorphism $T_pM\cong\mathbb{R}^n$ is not canonical, but any such isomorphism gives the same topology, so it doesn't really matter in this situation.

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  • $\begingroup$ Alright. So you're saying that I should try and think of any derivation $v \in T_pM$ as its coordinate vector? $\endgroup$ – DpS Oct 19 '17 at 22:26
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    $\begingroup$ No! Coordinates are not canonical, you should always try to work without them when it's possible. I'm saying that you have a canonical topology on the tangent space (which behaves well with respect to coordinate choices). Another point of view is that $TM$ is canonically a manifold, and thus a topological space. Now take the subspace topology on $T_pM\subset TM$. (Of course, it will be the same topology I defined above.) $\endgroup$ – Daniel Robert-Nicoud Oct 20 '17 at 5:45

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