If $ |\phi_1| <| \phi_2| $ and $\phi = \phi_1 + \phi_2 $ then $\mathrm {Ind} _{\phi} (0) = \mathrm {Ind} _{\phi_2} (0)$

Let $ \phi1, \phi2: [0,1] \to \mathbb {C} $ be Paths, such that $ |\phi_1 (t) | <| \phi_2 (t) | $ for all $ t \in [0,1] $. If $ 0 \leq t \leq 1 $, we set $\phi (t) = \phi_1 (t) + \phi_2 (t)$. Prove that $ 0 \notin \phi_1^* \cup \phi_2^* $ and that $ \mathrm {Ind} _{\phi} (0) = \mathrm {Ind} _{\phi_2} (0)$.

We have $ 0 \notin \phi_1^* \cup \phi_2 ^ * $ because, if $ t \in [0, 1]$, it comes: $$ 0 \leq | \phi_1 (t) | <| \phi_2 (t) |\qquad ,\qquad 0 <| \phi_2 (t) | - | \phi_1 (t) |\leq | \phi_1 (t) + \phi_2 (t) | = | \phi (t) | $$

For $ 0 \leq t, u \leq 1 $, let us set: $$ \gamma (u, t) := u \phi (t) + (1 - u) \phi_2 (t) $$ I want to show that $ \sigma $ realizes a homotopy, in $ \mathbb {C} ^ * $, but I can't show that $\phi$ and $\phi_2$ are paths with the same endpoints. Did what I wrote solve this problem? Is there any other easy way to fix this problem?

Solution 1:

Implicitly $\phi_{i}$ must be closed curves, and $\phi_{1}(t)\neq 0$. Attach a bridge between $\phi_{1}(0)$ and $\phi_{2}(0)$ and modify $\phi_{1}$ to get a closed curve that shares the same starting point as $\phi_{2}$. More explicitly, let $\varphi_{1}(t)= \begin{cases} (1-t)\phi_{2}(0)+t\phi_{1}(0), t\in [0,1] \\ \phi_{1}(t-1),t\in [1,2] \\ (3-t)\phi_{1}(0)+(t-2)\phi_{2}(0),t\in [2,3] \end{cases}$

$\varphi_{2}(t)= \begin{cases} \phi_{2}(0),t\in [0,1]\\ \phi_{2}(t-1),t\in [1,2]\\ \phi_{2}(0),t\in [2,3] \end{cases}$

Prove the proposition with $\phi_{i}$ replaced by $\varphi_{i}$.