Do Carmo defines a piecewise differential curve in the following manner:
A piecewise differentiable curve is a continuous mapping $c:[a,b]\to M$ of a closed interval $[a,b]\subset \mathbb{R}$ into $M$ satisfying the following condition: there exists a partition
$$a=t_0<t_1<\dots <t_{k-1}=b$$
of $[a,b]$ such that the restrictions $c\rvert_{[t_i, t_{i+1}]},i=0,\dots,k-1,$ are differentiable. We say that $c$ joins the points $c(a)$ and $c(b)$. $c(t_i)$ is called a vertex of $c$, and the angle formed by $\lim_{t\to t_i^+} c'(t)$ with $\lim_{t\to t_i^-} c'(t)$ is called the vertex angle at $c(t_i)$; here $\lim_{t\to t_i^+}$ ($\lim_{t\to t_i^-})$ signifies that $t$ approaches $t_i$ trough values above (below) that of $t_i$.
As far as I know, do Carmo hasn't defined what is meant by differentiability of a curve $c: [t_i, t_{i+1}]\to M$ on a closed set. If I understand the answer given by John B in On the definition of piecewise differentiable curves correctly, it means that $c$, as a curve on $(t_i, t_{i+1})$, is differentiable and that the one-sided derivatives exist at the endpoints $t_i$ and $t_{i+1}$. However, how would one go about calculating these derivatives or defining their existence? We have
$$c'(t) f = (f \circ c)'(t)=\frac{d}{dt} (f \circ c)(t) = \lim_{h\to 0} \frac{f(c(t+h))-f(c(t))}{h}\quad \forall f\in C^{\infty}(M), t\in (t_i,t_{i+1}).$$
Does that mean that $\lim_{t\to t_i^+} c'(t)$ exists if
$$\lim_{h\to 0^+} \frac{f(c(t_i^++h))-f(c(t_i^+))}{h}$$
exists for all $f\in C^{\infty}(M)$? This doesn't seem correct to me since a definition without the use of $f$ would be preferable.
John B also mentioned that the differentiability defined by the existence of the one-sided derivatives is equivalent to the fact that there exists a differentiable extension of $c$ which is defined on a larger open interval. Could someone provide me with a reference for this fact?
Best Answer
Everything here is basically generalising the $\mathbb{R}$ or $\mathbb{R}^n$ version to manifolds using charts. One-sided derivatives are defined by, for example, the equivalence class of curves version:
Definition: The one-sided derivative (from the right) of a curve $c\colon[0,\ell)\to M^n$ at $0$, $c'_+(0)\in T_{c(0)}M$ is the equivalence class $[\gamma]$ of curves $\gamma\colon (-\epsilon,\epsilon)\to M$, $\gamma(0)=c(0)$ that has $(\varphi\circ\gamma)'(0)=(\varphi\circ c)'_+(0)$ for any (hence all) charts $\varphi$ at $c(0)$. Similarly one-sided derivative from the left $c'_-$
For extendability, again we only need to do the case $\mathbb{R}$-valued functions and extend to $\mathbb{R}^n$ on each coordinate and to $M$ by charts.
Recall: For any differentiable $f\colon[a,b]\to\mathbb{R}$ with one-sided derivatives $f'_+\colon[a,b)\to\mathbb{R}$ and $f'_-\colon(a,b]\to\mathbb{R}$, you can extend to a differentiable $\tilde{f}\colon\mathbb{R}\to\mathbb{R}$ by $$ \tilde{f}(t)= \begin{cases} f(a)+(t-a)f'_+(a) & t<a\\ f(t) & t\in[a,b]\\ f(b)+(t-b)f'_-(b) & t>b \end{cases} $$ Moreover, if $f'_{\pm}$ are continuous then $\tilde{f}$ is $C^1$ (and you can extend this to higher-order derivatives suitable degree-$k$ polynomials instead of linears). This allows us to define piecewise $C^k$ function to $\mathbb{R}$. For $C^\infty$ you will need to use some bump functions (exercise).