I tried to understand in what way an orientation on $M$ induces an orientation on $\partial M$. Note that I am learning in the context of the extrinsic definition of manifolds, i.e. everything is assumed to be $\subset ℝ^n$.
I stumbled upon the following sentence:
A basis $(w_1, {}_\cdots, w_{k-1})$ of $T_p(\partial M)$ is said to be positively oriented if $(v, w_1, {}_\cdots, w_{k-1})$ being a basis of $T_p(M)$ is positively oriented
Where $v$ is said to be “pointing outwards”, which has been defined beforehand.
However, I thought: “How are $T_p(M)$ and $T_p(\partial M)$ even different for $P\in \partial M$? doesn't $T_p(M)$ ‘collapse’ by one dimension when looking at the boundary? Let's look at the definitions.”
Definition (Tangent space)
Let $M\subset ℝ^n$ be a k-manifold, $p\in M$. $T_p(M) := \{ v\inℝ^n \vert\, ∃ε>0\:∃γ \in C^1((-ε,ε),\,M) : γ(0)=p\wedge γ^\prime(0)=v \}$
However, when $p\in \partial M$, for every such map $γ$ we find a chart $Φ: \mathcal T(ℝ^k_-)\ni T→M$ (choosing $ε$ s.t. $γ[(-ε,ε)]\subset W$) s.t. $Φ^{-1}\circ γ$ is a continuously differentiable curve in the half-space with $(Φ^{-1}\circ γ)\,(0) ∈\partial ℝ^k_-$. Thus, the first component of $v=(Φ^{-1}\circ γ)$ is zero and thus $\dim T_p(M) = k-1$.
What is wrong with my argumentation? Or are the definitions of the tangent space insufficient?
I will happily give more of the definitions we used if needed.