I believe there should be a simple topological proof of the Nullhomotopical Cauchy Integral Formula based only on the Cauchy Integral Formula over a Circle, but I can't quite finish the argument and would appreciate some help. (Hopefully I am on the right track.)

Nullhomotopical Cauchy Integral Formula: Let $U\subseteq\mathbb{C}$ be an open and path-connected subset, let $z_0\in U$, and let $\gamma\subseteq U$ be some, say, smooth loop such that $z_0\notin\gamma$ and $\gamma\simeq\rm{pt.}$ in $U$, where by abuse of notation $\gamma$ stands for both the curve and its support. If $f:U\to\mathbb{C}$ is a holomorphic function, then: $$ f(z_0)\operatorname{ind}(\gamma,z_0) = \frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z-z_0}\mathrm{d}z $$ Attempted proof: I know that, being holomorphic, the 1-form $$ \omega:=\frac{f(z)}{z-z_0}\mathrm{d}z $$ is $\mathrm{d}$-closed, hence (its integral is) homotopy-invariant. Since $\gamma\subseteq U\setminus\{z_0\}$ and $[\gamma]=0$ in $\pi_1(U)$, I believe that $[\gamma]$ should induce a well-defined class $[\gamma]'$ in $\pi_1(D\setminus\{z_0\})$, where $D$ is a sufficiently small open disk around $z_0$ contained in $U$, but I don't quite see how to finish this line of reasoning rigorously, assuming it actually makes sense. Any help would be appreciated!


To find a counterexample, we can use the van Kampen theorem. Let $D$ be a small closed disk in $U$ containing $z_0$. Then the van Kampen theorem says that $πœ‹_1(U)$ is the quotient of $πœ‹_1(U\setminus D)$ by the normal closure of $πœ‹_1(βˆ‚D)$. We know that $𝛾 ∈ πœ‹_1(U \setminus D)$ is in this normal closure, but it may not be in $πœ‹_1(βˆ‚D)$. The simplest example of this situation is when $U = β„‚ \setminus \{0\}$ and $𝛾$ is a loop going one times around $0$, then one times around $z_0$, then $-1$ times around $0$. In $πœ‹_1(U)$, we have $[𝛾]=0$, but it is not any power of a loop going one time around $z_0$.

So we should not consider the situation homotopically but homologically. Here, we should use instead the Mayer-Vietoris sequence. Since the groups are abelian, we get that $[𝛾]$ is in $H_1(βˆ‚D)$.

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