In my course we were given several proofs of Cauchy's theorem, each at various points in the course, each version stronger than the previous. I'd like to learn a proof of the theorem, so naturally went to the last and most general one we had, which also happens to have the shortest proof, at least in appearance.
The key statement is the following:
Let $\phi,\psi:[a,b]\to U$ be homotopic piecewise $C^1$ closed paths in a domain $U$, then there exists a chain $\phi=\nu_0,\,\nu_1,\dots,\nu_n=\psi$ such that $\nu_i$ piecewise $C^1$ and each $\nu_{i+1}$ is obtained from $\nu_i$ by elementary deformation.
Having $F:[0,1]\times [a,b]\to U$ be the homotopy, the proof of the statement boils down to constructing: $$x_j=a+(b-a)\tfrac{j}{n},\quad \nu_i(t)=F\big(\tfrac{i}{n},t\big),\quad B_{ij}=B\big(F(\tfrac{i}{n},x_j),\epsilon\big)$$ where everything has been chosen such that if $s\in (\tfrac{i-1}{n},\tfrac{i}{n}),\;t\in [x_{j-1},x_j]$ then $F(s,t)\in B_{ij}$.
Then as a corollary it has $\int_{\phi}\,f=\int_{\psi}\,f$ and so the theorem follows from simple connectedness. My questions are:
- does this proof rely on the weaker version for convex domains? I can see that $B_{ij}$ is convex; is the idea that each $\nu_i$ deforms continuously to $\nu_{i+1}$ within convex sets?
- if it does rely on the weaker theorem, is it possible to modify this proof so that it works as a stand-alone proof?
- last, this is not as important, but what exactly is happening here? What makes the transformations elementary and how can I "visualise" them?