The Residue Theorem via Fundamental Groups and Homology Group: Integration as a Group Homomorphism

Consider a simply connected domain $U\subseteq \mathbb{C}$$U\subseteq\mathbb C$ and $X=U-\{a_1,\dots ,a_n\}$$X = U - \{a_1,\ldots,a_n\}$ with a function $f$$f$ holomorphic in $X$$X$.

Firstly, notice that integration is a group homomorphism:

$$\begin{array}{}\text{(1)}& f_{\ast }(\gamma ):={\int }_{\gamma }f(z)dz:{\pi }_1(X)\to (\mathbb{C},+).\end{array}$$

This is well-defined since integration is invariant under path homotopy.

Remark. $f(z)dz$$f(z)dz$ is a closed form by Cauchy-Riemann Condition, hence for ${\gamma }_1\sim {\gamma }_2,$$\gamma_1\sim\gamma_2,$

$$\begin{array}{}\text{(2)}& f_{\ast }({\gamma }_1{\gamma }_2^{-1})={\int }_{\partial D}f(z)dz={\int }_{D}d(f(z)dz)=0\end{array}$$

This is a group homomorphism because:

$$\begin{array}{}\text{(3)}& f_{\ast }(\gamma \circ {\gamma }^{\prime })={\int }_{\gamma }f(z)dz+{\int }_{{\gamma }^{\prime }}f(z)dz=f_{\ast }(\gamma )+f_{\ast }({\gamma }^{\prime }).\end{array}$$

Note that $X$$X$ is homotopy equivalent to $\underset{i=1}{\overset{n}{\bigvee }}{S}^1$$\bigvee_{i=1}^n S^1$​。

Here is a picture drawn by Magritte, in the case $n=3$$n=3$.

and the fundamental group of $\underset{i=1}{\overset{n}{\bigvee }}{S}^1$$\bigvee_{i=1}^n S^1$ is $F[n]$$F[n]$, where $F$$F$​ is the free group functor .

Notice that the codomain of $f_{\ast }$$f_*$ is $\mathbb{C}$$\mathbb C$, whcih is a abelian group, hence it uniquely pass through its abelianization via $h$$h$ and unique determined a $f_{\ast }!:F[n]/[F,F]\to \mathbb{Z}$$f_*!:F[n]/[F,F]\to\mathbb Z$ such that

$$\begin{array}{}\text{(4)}& f_{\ast }=f_{\ast }!\circ h\end{array}$$

Notice that the abelianization functor compose with fundamental group functor is $H_1(-)$$H_1(-)$​

So indeed, what we get is $f_{\ast }!:H_1(X)\to (\mathbb{C},+)$$f_*!:H_1(X)\to(\mathbb C,+)$.

Hence The Abelianization morphism:

$$\begin{array}{}\text{(5)}& h:{\pi }_1\left(\underset{i=1}{\overset{n}{\bigvee }}{S}^1\right)\to H_1\left(\underset{i=1}{\overset{n}{\bigvee }}{S}^1\right)\cong {\mathbb{Z}}^n\end{array}$$

can be written as:

$$\begin{array}{}\text{(6)}& h(\gamma )={\text{rot}}_{a_1}(\gamma )e_1+\cdots +{\text{rot}}_{a_n}(\gamma )e_n\end{array}$$

where ${\text{rot}}_{a_i}(\gamma )$$\text{rot}_{a_i}(\gamma)$ denotes the winding number of path $\gamma$$\gamma$ around $a_i$$a_i$ and $e_i=h({\gamma }_i)$$e_i=h(\gamma_i)$, where ${\gamma }_i$$\gamma_i$ is the loop around $a_i$$a_i$, whcih is the generator of ${\pi }_1(X)$$\pi_1(X)$​.

Remark. The $\begin{array}{c}{\int }_{\gamma }f(z)dz\end{array}$$\int_{\gamma} f(z)dz\\$, or maybe we should write as $⟨h(\gamma ),f(z)dz⟩$$\langle h(\gamma),f(z)dz\rangle$ is a bilinear form $H_1(X)\times H^1(X)\to \mathbb{C}$$H_1(X)\times H^1(X)\to\mathbb C$ .

Now let's analyze this group homomorphism:

$$\begin{array}{}\text{(7)}& f_{\ast }!:H_1(X)\cong {\mathbb{Z}}^n\to \mathbb{C}.\end{array}$$

We know that for $f\in {\text{Hom}}_{\text{Ab}}({\mathbb{Z}}^n,A)$$f\in\text{Hom}_{\text{Ab}}(\mathbb{Z}^n, A)$, it is equivalent selecting $n$$n$ elements (possibly with repetition) from $A$$A$.

For elements $s_1,s_2,\dots ,s_n\in A$$s_1, s_2, \ldots, s_n \in A$, we can define a homomorphism through:

$$\begin{array}{}\text{(8)}& \sum _{i=1}^n{\alpha }_i{e}_i⟼\sum _{i=1}^n{\alpha }_i{s}_i.\end{array}$$

Conversely, defining a homomorphism by assigning $f(e_i)$$f(e_i)$ gives us $n$$n$ elements (possibly with repetition) from $A$$A$, which is essentially a linear function...

Therefore, any $f:{\mathbb{Z}}^n\to A$$f: \mathbb{Z}^n \to A$ can be written as ${\displaystyle \sum _{i=1}^n{\alpha }_{i}f(e_i)}$$\displaystyle \sum_{i=1}^n \alpha_i f(e_i)$.

Looking back at this part:

$$\begin{array}{}\text{(9)}& f_{\ast }(\gamma ):={\int }_{\gamma }f(z)dz:{\pi }_1(X)\to (\mathbb{C},+),\end{array}$$

we know it can be decomposed into expressions of the form:

$$\begin{array}{}\text{(10)}& {\int }_{\gamma }f(z)dz=f_{\ast }(\gamma )=f_{\ast }!\circ h(\gamma )=f_{\ast }!(\sum _{i=1}^n{\mathrm{rot}}_{a_i}(\gamma )h({\gamma }_i))=\sum _{i=1}^n{\mathrm{rot}}_{a_i}(\gamma )f_{\ast }({\gamma }_i),\end{array}$$

Define the residue as:

$$\begin{array}{}\text{(11)}& \text{Res}(f,a_i):=\frac{1}{2\pi i}{f}_{\ast }(e_i)=\frac{1}{2\pi i}{\int }_{{\gamma }_i}f(z)dz.\end{array}$$

The reason that I do not define $\text{Res}(f,a_i)$$\text{Res}(f, a_i)$ as $f_{\ast }(e_i)$$f_*(e_i)$ is to agree with the classical residue definition.

The classical residue definition states that there exists a unique $R$$R$ where $f(z)-\frac{R}{z-a}$$f(z) - \frac{R}{z-a}$ has an analytic antiderivative in $0<|z-a|<\delta$$0 < |z-a| < \delta$, hence:

$$\begin{array}{}\text{(12)}& {\int }_{{\gamma }_i}f(z)-\frac{R}{z-a_i}dz=0⟹R=\frac{1}{2\pi i}{\int }_{{\gamma }_i}f(z)dz.\end{array}$$

While there are many ways to calculate $\text{Res}(f,a_i)$$\text{Res}(f, a_i)$, that's not important here. What's significant is that:

$$\begin{array}{}\text{(13)}& {\int }_{\gamma }f(z)dz=2\pi i\sum _{i=1}^n{\text{rot}}_{a_i}(\gamma )\text{Res}(f,a_i).\end{array}$$