Suppose $G\subset \Bbb C$ is open, $0\notin G$, and some closed curve in $G$ has non-zero index about the origin. Does it follow that some closed curve has index $1$ about the origin?
(To avoid an XY problem: All I really need to know is that if the index is always even then it is always $0$.)
Seems clear, but as sometimes happens in topology I have no idea how to prove it.
My work so far: Oh gimme a break.
Context: complex analysis.
On
Argue by induction on the number of self intersections, $n$, of the polygonal path. If $n=0$ then apply the Jordan curve theorem. If $n=k+1$ write the path as the sum of $2$ paths with self intersection numbers $0$ and $m<k+1$.
I have been trying for a long time to find a proof that does not use some version of the Jordan curve theorem.
This is more like a sketch of how a solution might go, rather than an actual solution. There may be some insurmountable error within.
You can replace your original path with a polygonal closed path with the same winding number about $0$. You can also assume that all the edges of this path have different slopes. There will probably be some nontrivial self-intersections. If so break the path into a "sum" of simple closed paths. One of these must have nonzero winding number. Now appeal to the (polygonal) Jordan curve theorem to see that the winding number is this curve is $\pm1$.