Complex path length is invariant under monotone surjective reparametrization

The composite $\gamma \circ \phi$ is continuous. Indeed, fix $s_0\in [c,d]$ and $\epsilon >0$. By continuity of $\gamma$ at $\phi (s_0)$ choose $\delta >0$ such that $u\in [a,b]$ and $\mathrm{\mid }u-\phi (s_0)\mathrm{\mid }<\delta$ imply $d_{\mathbb{C}}(\gamma (u),\gamma (\phi (s_0)))<\epsilon$. By continuity of $\phi$ at $s_0$ choose $\eta >0$ such that $s\in [c,d]$ and $\mathrm{\mid }s-s_0\mathrm{\mid }<\eta$ imply $\mathrm{\mid }\phi (s)-\phi (s_0)\mathrm{\mid }<\delta$. The implication in [F5] then proves continuity of the composite at $s_0$.

If $c=d$, surjectivity forces $a=b$, so both lengths are zero by [F1]. If $a=b$ while $c<d$, then $\gamma$ and $\gamma \circ \phi$ are constant paths and both lengths are zero by [F1]. Thus the equality covers all singleton cases.

Put $\mathrm{\Gamma }=\mathrm{\Phi }\circ \gamma$. By the isometry in [F2], $\mathrm{\Gamma }$ is continuous and $\mathrm{\Phi }\circ (\gamma \circ \phi )=\mathrm{\Gamma }\circ \phi$. The Euclidean reparametrization theorem [F3] gives $L_{{\mathbb{R}}^2}(\mathrm{\Gamma }\circ \phi )=L_{{\mathbb{R}}^2}(\mathrm{\Gamma })$.

Apply the complex-to-coordinate length equality [F2] to $\gamma$ and, using step 1.1, to $\gamma \circ \phi$. Together with step 2.1 this yields $L_{\mathbb{C}}(\gamma \circ \phi )=L_{\mathbb{C}}(\gamma )$.

By [F4], the nondecreasing case preserves order and the nonincreasing case reverses it. If $a<b$, a nondecreasing surjection must attain $a$ at $c$ and $b$ at $d$; a nonincreasing surjection must attain $b$ at $c$ and $a$ at $d$. If $a=b$, the map is constant and there is no nontrivial orientation.