defining $C^\infty$ structure on finite-dim vector space, homeomeomorphism to tangent bundle, such that independent of choice of bases If $V$ is a finite dimensional vector space over $\mathbb{R}$, how would I go about defining a $C^\infty$ structure on $V$ and a homeomorphism from $V \times V$ to $TV$ which is independent of bases?
My thoughts so far. Using notation from Spivak's Comprehensive Introduction to Dif. Geo. Vol. 1, as in the case of $\mathbb{R}^n$, for $v, w \in V$ we will denote by $v_w \in V_w$ the vector corresponding to $(w, v)$. Let $\mathcal{V}$ be a finite dimensional vector space over $\mathbb{R}$ with $\dim(\mathcal{V}) = n$. Let $\mathcal{A}$ be the collection of charts on $\mathcal{V}$ consisting of all pairs of the form $(\mathcal{V}, \varphi)$ for $\varphi: \mathcal{V} \to \mathbb{R}^n$ a vector space isomorphism.
It's clear that every linear isomorphism between finite dimensional vector spaces over $\mathbb{R}$ is a homeomorphism, so that $\mathcal{A}$ truly is a collection of charts.
I'm not sure what to do next.
 A: It appears you are using the notation of Spivak, so I will too. $($See Theorem 3.1 and Problem 3.8 for notation and background.$)$
You have a good start on the problem. The next thing to do is to check that $\mathcal{A}$ is a $C^\infty$ atlas on ${V}$. Given two members $({V}, \varphi)$ and $({V}, \psi)$ of $\mathcal{A}$, we have that $\varphi, \psi$ are vector space isomorphisms, so $$\varphi \circ \psi^{-1}: \mathbb{R}^n \to \mathbb{R}^n,\text{ }\psi \circ \varphi^{-1}: \mathbb{R}^n \to \mathbb{R}^n$$ are also vector space isomorphisms. We observe that given any isomorphism $\Phi: \mathbb{R}^n \to \mathbb{R}^n$, $\Phi$ can be represented by some matrix $A$, so that $\Phi(x) = Ax$. This shows that the component functions $\Phi^i(x)$ are polynomials in $x_1, \dots, x_n$, whence they are smooth. Thus $\Phi$ is smooth, so in fact any isomorphism from $\mathbb{R}^n$ onto $\mathbb{R}^n$ is a $C^\infty$ map. It follows that $\varphi \circ \psi^{-1}$ and $\psi \circ \varphi^{-1}$ are $C^\infty$, and we conclude that $\mathcal{A}$ is a $C^\infty$-atlas on ${V}$.
Next, we show that there is a natural diffeomorphism from $V \times V \to TV$. Let $$t: \mathbb{R}^n \times \mathbb{R}^n \to T\mathbb{R}^n,\text{ }(p, v) \mapsto [(\mathbb{R}^n, \text{Id}, v)]_p$$ be a diffeomorphism $($for more information on this, see Theorem 3.1 and Problem 3.8 of Spivak$)$. Given a linear isomorphism $\varphi: V \to \mathbb{R}^n$, we have $\varphi \in \mathcal{A}$, so $\varphi$ is a diffeomorphism and hence $\varphi_*: TV \to T\mathbb{R}^n$ is also a diffeomorphism. It follows that $$\varphi_*^{-1} \circ t \circ (\varphi \times \varphi): V \times V \to TV$$ is a diffeomorphism. It is given by $$(p, q) \overset{\varphi \times \varphi}{\longmapsto} (\varphi(p), \varphi(q)) \overset{t}{\longmapsto} [(\mathbb{R}^n, \text{Id}, \varphi(q))]_{\varphi(p)} \overset{\varphi_*^{-1}}{\longmapsto} [(U, \varphi, \varphi(q))]_p.$$To check that this is independent of the choice of $\varphi$, we need to check that for any other linear isomorphism $\psi: {V} \to \mathbb{R}^n$, we have$$({V}, \varphi, \varphi(q))_p \underset{p}{\sim} ({V}, \psi, \psi(q))_p.$$Notice that $$D(\psi \circ \varphi^{-1})_p = \psi \circ \varphi^{-1}(\varphi(q)) = \psi(q),$$where the first equality due to $\psi \circ \varphi^{-1}$ being linear. Hence, our diffeomorphism is independent of the choice of $\varphi$.
