Let $f:A(r,1) \rightarrow \mathbb{C}$ be a holomorphic, injective and non-zero function. $f$ maps the annulus $A(r,1)$ for some $0<r<1$ onto some domain which is bounded by $|z| = 1$ and a closed curve which lies in the unit circle such that $0 \not \in f(A(r,1))$.
I want to show that $\operatorname{Ind}_\Gamma(0) = 1$ for $\Gamma = f \circ \gamma$ and $\gamma(t) = \rho e^{it}$ for some $r < \rho < 1, \ t \in [-\pi,\pi]$. Obviously $\operatorname{Ind}_\gamma(0) = 1$, but how do I convince myself that also $\operatorname{Ind}_\Gamma(0) = 1$?
Do you have any suggestions?
Consider the index number as a function of $\rho$. Show that it is continuous. But because it is discrete, this means it must be constant. Either as $\rho \to 1$ or as $\rho \to r$, $\Gamma$ approaches the unit circle. Because $f$ is injective, it can only go around this circle once. hence the index must be $1$.
Edit: or $-1$, such as is the case for $f(z)= \frac1{2z}$ on $A(1/2,1)$. This is the best you can do with the information given. The index would definitely be $1$ if you required that $f$ is the identity on the unit circle. It might also be only $1$ if you just required that $f$ carry the unit circle to itself - though I'd have to think on that one some more.