Spivak's definition of tangent vectors

Question

Let me copy some notation (I borrowed it from "Calculus on Manifolds" by Spivak) from my previous question:

Given a differentiable map $f:\mathbb R^n\to \mathbb R^m$, define $f_\ast: \mathbb R_p^n\to \mathbb R_{f(p)}^m$ by $$f_\ast(v_p)=(Df(p)(v))_{f(p)}.$$ Note that the notation $v_p$ stands for the vector $v$ at $p$ (so the RHS is a vector at $f(p)$).

Let $c$ be a differentiable curve in $\mathbb R^n$, i.e., a differentiable function $c:[0,1]\to \mathbb R^n$. Define the tangent vector $v$ of $c$ at $t$ as $c_\ast((e_1)_t)=((c^1)'(t),\dots,(c^n)'(t))_{c(t)}$.

There are a couple of things which bother me. Below I assume $c=(c^1,\dots,c^n)$, which I believe Spivak does.

• Spivak defines $f_\ast$ when $f$ is a map on the whole $\mathbb R^n$, not on its subset. But he talks about $c_\ast$ where $c$ is defined on a (closed) subset of $\mathbb R^1$. Should I assume that the definition of $f_\ast$ is the same if $f$ is defined on an arbitrary subset of $\mathbb R^n$?
• Why does the equality $c_\ast((e_1)_t)=((c^1)'(t),\dots,(c^n)'(t))_{c(t)}$ hold? If I assume the my previous point is true, then by definition, $c_\ast((e_1)_t)=((dc)_te_1)_{f(t)}$, where the RHS is the differential of $c$ at $t$ applied to the vector $e_1$, and the whole thing is considered as a vector at $f(t)$. But $(dc)_t$ is the linear map whose matrix is $((c^1)'(t),\dots,(c^n)'(t))$, and if one applies it to $e_1$, one gets $(c^1)'(t)$. But Spivak claims this equals $((c^1)'(t),\dots,(c^n)'(t))_{c(t)}$.

Answer 1

1. The $f_*$ is defined whenever the domain $M$ of $f$ is an $n$-dimensional submanifold of $\Bbb R^n$, with or without boundary, and $f$ is smooth. The definition is the same: given $f:M\to\Bbb R^m$, for every point $p\in M$, we define $f_{*p}:\Bbb R^n{}_p\to\Bbb R^m{}_{f(p)}$,$$f_{*p}(v_p)=(Df(p)(v))_{f(p)}.$$ Note that the domain of $f_{*p}$ is a whole vector space $\Bbb R^n{}_p$, even if $M$ is not the whole space $\Bbb R^n$. You can't have the domain of $f$ to be arbitrary, because you need $f$ to be smooth, and you can talk about smoothness only when the domain is nice enough.

2. The matrix representation of the linear map $(dc)_t$ is not $((c^1)'(t),\dots,(c^n)'(t))$ but is its transpose. So when you apply that matrix to $(e_1)_t$, you get the whole column of $((c^1)'(t),\dots,(c^n)'(t))^T$.