Let $e_{i_p}$ denote the standard basis for $T_p(\mathbb{R}^n)$. Theres a vector space isomorphism between $T_p(\mathbb{R}^n)$ and $D_p(\mathbb{R}^n)$, where $D_p$ is the set of derivations at $p$, with isomorphism $\phi$. $\phi: T_p(\mathbb{R}^n)\rightarrow D_p(\mathbb{R}^n)$ is given by $\phi(v_p)=D_{v_p}=\sum_k v^k\frac{\partial}{\partial{x}^k}|_p$. So, those partial derivatives are a basis for $D_p(\mathbb{R}^n)$.
Note, $T_p(\mathbb{R}^n)=\{p\} \times \mathbb{R}^n$
Now it is said that we may write $v_p\in T_p(\mathbb{R}^n)$ as
$v_p=\sum_iv^ie_{i_p}=\sum_iv^i \frac{\partial}{\partial{x^i}}|_p$
Isn't the last sentence a bit sloppy and not pedantic? because the second expression lives in the tangent space while the other in the space of derivations, no?
Best Answer
In general if $M$ is a smooth submanifold of $N$, then the inclusion mapping $\iota:M \to N$ will be smooth, and for each $p \in M$, the tangent mapping $T\iota_p: T_pM \to T_pN$ will be an injective linear map. As sets, very often the $T_pM$ and $T_pN$ are not subsets of each other. However, note that \begin{align} T_pM \cong \text{image}(T\iota_p) \subset T_pN. \end{align} Because of this, we can "think" of $T_pM$ as actually being a subspace of $T_pN$. Why do we make such identifications? Because it is just extremely convenient, as long as you know exactly what the isomorphism is.
Very often in math, we don't really care about "what" an object is, as opposed to "the properties of an object". For example, do you ever really think of the real numbers $\Bbb{R}$ as an equivalence class of Cauchy sequences of rationals? Oh but then how are rationals $\Bbb{Q}$ defined? Well, one possibility is to define it as a certain equivalence class of integers. How are integers defined? Well go back to natural numbers. How are natural numbers defined? As you can see, there's a lot of "going down the rabbit hole", so if you completely unwind all the definitions, you can see that it becomes EXTREMELY cumbersome to write everything out in proper notation.
Or if you come from another construction, do you ever think of real numbers as Dedekind cuts? I doubt it. All you care about for real numbers is that $(\Bbb{R}, +, \cdot, <)$, with the "usual" operations is a (complete) ordered field... basically it has all the nice properties you learn in middle/high school (plus the supremum property).
As another illustration, consider the sets $\Bbb{R}^3$, $\Bbb{R}^2 \times \Bbb{R}$ and $\Bbb{R} \times \Bbb{R}^2$. I bet most of the time, we would regard all of these sets as being the same thing, and just call it $\Bbb{R}^3$ (or whichever "version" you prefer). But set theoretically, each has a different definition, and neither set is included in the other. But clearly there are very obvious bijections between these sets.
So you see, while sometimes it is necessary, from a strictly logical perspective, to define everything step by step explicitly, in practice is is not always true that people think of the object itself as the definition states. Once you establish a logical definition, we often revert back to the intuitive way of thinking about things. Clearly most people think of the numbers $1,2,3$ etc in the same way as in kindergarten, as opposed to their set-theoretic definitions. Why? Because we usually only care about the properties of the numbers, not what they actually are.
A slightly more "sophisticated" example is to take a finite-dimensional vector space $V$ over a field $F$. Then, the map $\iota:V \to V^{**}$, defined by $[\iota(v)](f) = f(v)$, for all $v\in V$, and all $f \in V^*$ is injective, and by dimension arguments, is an isomorphism. In other words, $V$ and $V^{**}$ are isormorphic, and even canonically so. Initially, when first learning linear algebra, you might be a little uneasy with thinkning of $V$ and $V^{**}$ as being "the same thing", because their elements are completely different types of objects. Well, my claim is that the only differnce between $V$ and $V^{**}$ versus something like $\Bbb{R}^3$ and $\Bbb{R} \times \Bbb{R}^2$ is that you're more comfortable/used to the fact that the latter sets as "being the same".
So, canonical isomorphisms are meant to emphasize to us that we should really treat the two objects as being the same thing. So, I think it is fine to abuse notation slightly by suppressing the isomorphism and simply saying the elements are "equal" as opposed to the more precise statement that "one vector is the image of another vector under the isomorphism".
As another illustration, here's another definition of the tangent space (my favourite). Given a point $p$ in a smooth manifold $M$, we can define the tangent space $T_pM$ using curves as follows:
Anyway, once you prove the basics that everything is well-defined etc, do you really gain any benefit/clarity in distinguishing a curve $\gamma: I \to M$ vs the same curve but whose target space is enlarged $\gamma: I \to N$? I mean, sure set theoretically, these are different objects, but "they're obviously the same thing". I'm usually a very nitpicky person, but in this scenario, even I would admit that distinguishing these objects set-theoretically is a bit overboard and too pedantic.
Finally, once you get down to the actual computations, for example in $\Bbb{R}^3$ or $\Bbb{R}^2$ or on the spheres $S^1, S^2$ etc, you're very rarely going to use the definitions directly. Many important computations are going to be performed using a chart, and in this case, it is really extremely cumbersome to write everything down using $100\%$ correct notation. Once you learn the basic definitions, I invite you to carry out a typical calculation using correct notation vs slightly abused notation. I think you'll agree that the amount of effort needed to be entirely correct is not worth it. (Of course, you should do everything properly atleast once in your life, but after that, just do it the quickest way)