Proposition

Let $c$ be a differentiable curve in $\mathbb{R}^n$, that is, a differentiable function $c:[0, 1] \rightarrow \mathbb{R}^n$. Define the tangent vector $v$ of $c$ at $t$ as $c_{\ast}((e_1)_t) = ((c^1)'(t), \cdots, (c^n)'(t))_{c(t)}$. If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, show that the tangent vector to $f \circ c$ at $t$ is $f_{\ast}(v)$.

Solution

\[\begin{align*} c_{\ast}((e_1)_t) &= ((c^1)'(t), \cdots, (c^n)'(t))_{c(t)} \\ &= (Dc^1(t), \cdots, Dc^n(t))_{c(t)} \\ &= (Dc(t))_{c(t)} & \text{(Theorem 2-3-(3) [Spivak])}. \end{align*}\]

Therefore,

\[\begin{align*} (f \circ c)_{\ast}((e_1)_t) &= (((f \circ c)^1)'(t), \cdots, ((f \circ c)^m)'(t))_{(f \circ c)(t)} \\ &= (D(f \circ c)^1(t), \cdots, D(f \circ c)^m(t))_{(f \circ c)(t)} \\ &= (D(f \circ c)(t))_{f(c(t))} & \text{(Theorem 2-3-(3) [Spivak])} \\ &= (Df(c(t))Dc(t))_{f(c(t))} & \text{(Theorem 2-2[Spivak])} \\ &= f_{\ast}((Dc(t))_{c(t)}) \\ &= f_{\ast}(v). \end{align*}\]