Let $\gamma_0$ and $\gamma_1$ be piecewise differentiable closed curves in an open set $U$ in the plane $\mathbb{C}$. (Of course, you may consider a more general setting if relevant.)
Suppose that there is a continuous deformation (i.e. homotopy) from one curve to the other: that is, there is a family of continuous closed curves $\gamma_s$ ($0 \le s \le 1$) in $U$ such that the two-variable function $\psi(t,s) = \gamma_s(t)$ is continuous.
Does this imply that there is a homotopy where each curve $\gamma_s$ is not merely continuous, but piecewise differentiable?
(The analogous result for regular curves is not true, since regular homotopy must preserve the turning number / total curvature / index / whatever you want to call it. But it seems like there should be no such issue in the piecewise differentiable setting, where the deformation is free to introduce cusps.)
Yes, if $H:I\times I\to U$ is a homotopy from $\gamma_0$ to $\gamma_1$, we can approximate it by a homotopy $\tilde H:I\times I\to U$ such that $t\mapsto\tilde H(s, t)$ is a differentiable curve, for all $s\in I$. For instance, you can do this by using the Weierstrass approximation theorem. Now, if $\epsilon>0$ is such that the $\epsilon$-neighborhood of the image $H(I\times I)$ is contained in $U$, and $\sup_{(s,t)}\|\tilde H(s,t)-H(s,t)\|<\epsilon$, then the initial and terminal curves $\tilde H(0,t)$ and $\tilde H(1,t)$ are within $\epsilon$ of the original curves $\gamma_0$ and $\gamma_1$, so you can just perform a linear homotopy to get from $\gamma_0(t)$ to $\tilde H(0,t)$ and then from $\tilde H(1,t)$ to $\gamma_1(t)$. Gluing these homotopies together gives you a homotopy from $\gamma_0(t)$ to $\tilde H(0,t)$ to $\tilde H(1,t)$ to $\gamma_1(t)$.
So, this is actually stronger in the fact that you can have a homotopy such that every intermediate curve is differentiable everywhere except that the curves at it's beginning and end are only piecewise differentiable.