# Tangent bundle topology

Suppose $M \subset \mathrm{R}^n$ is a smooth manifold. Does the typical definition of the topology of the tangent bundle $TM$ of $M$ by charts coincide with the topology of $TM$ regarded as a subspace of $\mathrm{R}^{2n}$?

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How do you regard $TM$ as a subspace of $\mathbb{R}^{m+n}$? – Jason DeVito Jun 10 '12 at 0:42
As the set of points $(p,v) \in \mathrm{R}^{m+n}$ such that $p \in \mathrm{R}^n$ and $v$ is tangent to $M$ at $p$. The topology would then be defined by the open sets $U \cap TM$, where $U \subset R^{m+n}$ is open. – Daniel Jun 10 '12 at 0:46
I'm still not understanding. Here's a simpler question: If $M = S^1\subseteq \mathbb{R}^2$, what does $TM\subseteq\mathbb{R}^3$ look like? (I can see a way to embed $TS^1$ into $\mathbb{R}^3$, but it uses the fact that $S^1$ is parallelizable.) – Jason DeVito Jun 10 '12 at 0:49
Wouldn't you want it to sit inside of $\mathbf R^{2n}$ then? – Dylan Moreland Jun 10 '12 at 0:49
Sorry, it really is $\mathrm{R}^{2n}$, as Dylan pointed out – Daniel Jun 10 '12 at 0:52