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The notes I am reading cover differential geometry at only a very basic level (say around Do Carmo's Differential Geometry of Curves and Surfaces), so I apologize in advance if this question would be considered incredibly misguided for anyone with a substantial knowledge of the subject.

Here is the definition of tangent space which I have:

Let $M \subset \mathbb{R}^n$ be open. Then $TM := M \times \mathbb{R}^n$ is called the tangent bundle of $M$. For $p \in M$, $T_pM := \{ p \} \times \mathbb{R}^n$ is called the tangent space of $M$ at $p$.
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If now $f: M \subset \mathbb{R}^n \to \mathbb{R}^m$ smooth, then we consider the differential of $f$ to be a map $df: TM \to T\mathbb{R}^m,$ and $d_p f:T_p M \to T_{f(p)}\mathbb{R}^m$.
$\dots$
Let $S \subset \mathbb{R}^3$ be an embedded surface, $q \in S$ be a point of $S$, and $f: U \to \mathbb{R}^3$ be a parametrization of $S$ such that $f(p)=q$. The tangent space of $S$ at $q$ is defined to be $T_q S := (d_p f)(T_p U) \subset T_{f(p)} \mathbb{R}^3$ and the tangent bundle $TS \subset T\mathbb{R}^3$ of $S$ is $TS := \bigcup\limits_{q \in S} T_q S$.

My question:

Why the notation $T_pU$? The $U$ here does not provide any useful information about the nature of the object. All it says is that $p \in U$, which seems almost redundant.

The most important aspect of the definition is that $T_p U \simeq \mathbb{R}^n$, but this is impossible to deduce from the notation $T_pU$. I find it very difficult to understand any discussion of tangent spaces using this notation unless ample context about the ambient spaces is given nearby, which it usually isn't.

Why would anyone care more about $p \in U$ than the fact that $T_p U \simeq \mathbb{R}^n$? What am I missing?

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    $\begingroup$ The spaces $T_pM$ are usually identified with the space of derivations on $p$, that is with the vector space $\{ D \in \mathcal L(C^\infty(M),\mathbb R) \mid D(f\cdot g) = D(f) g(p)+ f(p) D(g) \, \forall f,g\in C^\infty(M) \}$. Perhaps here it is clear, that while the tangent spaces at different points may be isomorphic, they are essentially different spaces. $\endgroup$ – s.harp Aug 2 '16 at 18:43
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    $\begingroup$ No, derivations act on functions on $M$, they are not functions on $M$. They are linear maps from $C^\infty$ to $\mathbb R$ that satisfy the derivation property (the product rule in the pervious comment). $\endgroup$ – s.harp Aug 2 '16 at 18:47
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    $\begingroup$ If you wanted to refer to a tangent space, you'd have to say what space it's tangent to and at what point, otherwise what tangent space are you even talking about? As for why bother with tangent spaces at all if they're all just isomorphic to $\Bbb R^n$ as vector spaces, it's because their elements are tangent vectors which are the basis of doing calculus and geometry in the space. If we're studying a nice submanifold of $\Bbb R^m$, all tangent vectors at all points can be identified with vectors in $\Bbb R^m$, greatly simplifying matters, but this can't be done with abstract manifolds. $\endgroup$ – arctic tern Aug 2 '16 at 18:49
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    $\begingroup$ To connect the comment by @arctictern and by me: You can look at differentiable paths from $\gamma: (-\epsilon,\epsilon)\to M, \gamma(0)=p$ and construct from such a path a derivation: $[\gamma](f):=\frac{d}{dt} f(\gamma(t))\lvert_{t=0}$. The product rule follows from it holding for functions from $\mathbb R\to\mathbb R$ and the map is linear in $f$, so this is a derivation. You can also see that it essentially describes $\dot\gamma(0)$, ie the "tangent" of the path. A further result is that every derivation on a manifold comes from such a path. $\endgroup$ – s.harp Aug 2 '16 at 18:53
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    $\begingroup$ Even with submanifolds of $\Bbb R^m$, there is no canonical isomorphism $T_pM \to \Bbb R^n$. You cannot canonically identify the differential $df$ with some matrix without picking a basis for $T_pM$ and $T_{f(p)}N$, and if you're allowed to pick an arbitrary basis, you can get an arbitrary matrix of fixed rank. In addition, when you assemble the tangent spaces into the tangent bundle, the bundle is usually not trivial. $\endgroup$ – user98602 Aug 2 '16 at 20:39
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The issue here is that in an open subset $U\subset\mathbb R^n$, there is no need to distinguish between the tangent spaces at different points. All tangent spaces can be naturally identified with the space $\mathbb R^n$. However, if you are trying to extend the concepts of analysis to more general objects, then it is necessary to distinguish between tangent spaces at different points. Depending on the generality you are working in, they are either (for submanifolds in $\mathbb R^N$) different subspaces of some fixed $\mathbb R^N$ or (for abstract manifolds) different vector spaces, which all have the same dimension but otherwise are unrelated unless you make additional choices (say of a chart).

In preparation for the general concepts, one usually tries to reconsider the case of an open subset $U\subset\mathbb R^n$ first. Doing this, you can see that indeed already in this case, you can say that the tangent space at $p\in U$ is not really $\mathbb R^n$ but a copy of $\mathbb R^n$ which is "attached" at $p$. Correspondingly, the derivative of $f:U\to\mathbb R^m$ is the best linear approximation to the map $v\mapsto f(p+v)-f(p)$ (which means that it maps the space "attached at $p$" to the space "attached at $f(p)$". But as I said above, if you want to stay with open subsets, there is no real need to observe this. If you move to some concept of manifold, this shows that before you can differentiate, you need to construct a vector space which "approximates the manifold at $p$", and there are various ways to to this (the most general one being derivations as mentioned in the comments to your question).

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