I'm reading the beginning of the book Milnor-Stasheff "Characteristic Classes" and they take a very unusual approach to defining a manifold: they define it extrinsically, but in a potentially infinite-dimensional Euclidean space, i.e. $\mathbb{R}^A$ for some arbitrary index set $A$.
They say that a map $f: \mathbb{R}^n \supset U \to B \subset \mathbb{R}^A$ is smooth if for every $\alpha \in A$, $f_{\alpha}:U \to \mathbb{R}$ is smooth, and the partial derivative as $$ \frac{\partial f}{\partial u_{i}} = (\frac{\partial f_{\alpha}}{\partial u_{i}})_{\alpha \in A}. $$
Then they define a manifold $M \subset \mathbb{R}^A$ as a subset s.t. for every $x \in M$, there exists a local parametrization $h: U \to \mathbb{R}^A$ which is a homeomorphism onto its image $V \subset M$, which is open in $M$ and contains $x$, and s.t. $\frac{\partial h}{\partial u_{1}}(u),...,\frac{\partial h}{\partial u_{n}}$ are linearly independent.
Then they say that the vector $v \in \mathbb{R}^A$ is tangent to $M$ at $x$ if there exists a smooth curve $p:I \to M$ s.t. $p(0)=x$, $p'(0) = v$.
Now they state the following lemma and say that the proof is straightforward:
A vector $v \in \mathbb{R}^A$ is tangent to $M$ at $x$ iff for a chart $(V, h)$, $h: U \to V$, $h(u) = x$, $v$ can be expressed as a linear combination of the vectors $$ \frac{\partial h}{\partial u_{1}}(u),...,\frac{\partial h}{\partial u_{n}}(u). $$
In the courses of differential geometry where only curves and surfaces are considered, this is a standard theorem. However, I am finding a problem with the direct direction. I will outline the proof and point out the problem:
Let $p:I \to V$, $p(0)=x$, $p'(0)=v$. Then $p_{\alpha}(t) = h_{\alpha}(h^{-1} \circ p(t))$.
Denote $(h^{-1} \circ p)'(0) = (a_{1},...,a_{n})$.
Now $$p_{\alpha}'(0) = (d_{u} h_{\alpha})((h^{-1} \circ p)'(0)) = \sum_{i=1}^{n} \alpha_{i} \frac{\partial h_{\alpha}}{\partial u_{i}},$$
which shows that $$v = \sum_{i=1}^{n} \alpha_{i}\frac{\partial h}{\partial u_{i}}.$$
Question: why is $h^{-1} \circ p$ smooth? We're passing through a space $\mathbb{R}^A$ which has no differentiable structure since $A$ is just a set; can anything be done to fix this proof?