Reversal negates and concatenation adds winding numbers

The map $t\mapsto a+b-t$ is a decreasing continuous bijection of $[a,b]$ onto itself, so by [L3] and [L4] the reversal ${\gamma }^{-}$ is a path of the same length as $\gamma$, hence rectifiable, and its trace is $\gamma ([a,b])={\gamma }^{\ast }$; it is closed because ${\gamma }^{-}(a)=\gamma (b)=\gamma (a)={\gamma }^{-}(b)$ by [L3].

By [L3] the concatenation $\alpha \ast \beta$ is continuous on $[0,1]$, its restrictions to $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ are monotone reparametrizations of $\alpha$ and $\beta$, so both are rectifiable by [L4] and $\alpha \ast \beta$ is rectifiable by [L5]; its trace is ${\alpha }^{\ast }\cup {\beta }^{\ast }$ by the two-piece formula.

If $\alpha$ and $\beta$ are closed then $(\alpha \ast \beta )(0)=\alpha (0)=\alpha (1)=\beta (0)=\beta (1)=(\alpha \ast \beta )(1)$ by [L3], so $\alpha \ast \beta$ is closed.

With $p\mathrm{\notin }{\gamma }^{\ast }$ the function $z\mapsto 1\mathrm{/}(z-p)$ is continuous on ${\gamma }^{\ast }=({\gamma }^{-}{)}^{\ast }$, so all the integrals below exist by [L6]; applying the reversal identity of [L2] to it and dividing by $2\pi i$ gives $n({\gamma }^{-},p)=-n(\gamma ,p)$ through [L1].

With $p\mathrm{\notin }{\alpha }^{\ast }\cup {\beta }^{\ast }$ the function $z\mapsto 1\mathrm{/}(z-p)$ is continuous on that union, so the concatenation identity of [L2] applies to it and gives the displayed additive formula, all three integrals existing by [L6].

If $\alpha$ and $\beta$ are closed, step 1.3 makes $\alpha \ast \beta$ closed, so [L1] turns step 2.2 into $n(\alpha \ast \beta ,p)=n(\alpha ,p)+n(\beta ,p)$; all three values are integers by [L7], consistently with the identity.