Null-homologous cycles and homologous cycles in an open set

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open and let $\mathrm{\Gamma }$ be a complex chain which is a cycle and whose trace lies in $\mathrm{\Omega }$ (Complex chains, their traces, and cycles).

$\mathrm{\Gamma }$ is null-homologous in $\mathrm{\Omega }$ when

$n(\mathrm{\Gamma },p)=0\text{for every }p\in \mathbb{C}\setminus \mathrm{\Omega },$

the index being that of Integration over a complex chain and the index of a chain; the values are defined because ${\mathrm{\Gamma }}^{\ast }\subseteq \mathrm{\Omega }$, so every $p\mathrm{\notin }\mathrm{\Omega }$ lies off the trace, and they are integers by The index of a cycle about a point off its trace is an integer.

Two cycles ${\mathrm{\Gamma }}_1,{\mathrm{\Gamma }}_2$ with traces in $\mathrm{\Omega }$ are homologous in $\mathrm{\Omega }$ when ${\mathrm{\Gamma }}_1-{\mathrm{\Gamma }}_2$ is null-homologous in $\mathrm{\Omega }$. By Chain integration and the index are additive in the chain, and reverse with it the chain ${\mathrm{\Gamma }}_1-{\mathrm{\Gamma }}_2$ is again a cycle with trace inside ${\mathrm{\Gamma }}_1^{\ast }\cup {\mathrm{\Gamma }}_2^{\ast }\subseteq \mathrm{\Omega }$, and its index at a point $p$ off that union is $n({\mathrm{\Gamma }}_1,p)-n({\mathrm{\Gamma }}_2,p)$; so the condition says exactly that

$n({\mathrm{\Gamma }}_1,p)=n({\mathrm{\Gamma }}_2,p)\text{for every }p\in \mathbb{C}\setminus \mathrm{\Omega }\mathrm{.}$

Both notions depend on $\mathrm{\Omega }$, not on the cycle alone. The same cycle can be null-homologous in one open set and not in another: enlarging $\mathrm{\Omega }$ removes points from $\mathbb{C}\setminus \mathrm{\Omega }$ and so weakens the requirement. Every statement below that uses these words names the open set it uses them in, and $\mathrm{\Omega }$ is not omitted anywhere.

Null-homologous does not mean equal to the empty chain. It is a condition on the numbers $n(\mathrm{\Gamma },p)$ for $p$ outside $\mathrm{\Omega }$, and a cycle with a large trace can satisfy it. In particular, being homologous is a relation between two cycles and never an assertion that the two lists coincide; chains here are lists and equality of chains is equality of lists.

Taking $\mathrm{\Omega }=\mathbb{C}$ makes the condition vacuous, since $\mathbb{C}\setminus \mathbb{C}$ is empty, so every cycle is null-homologous in the plane. The content of the notion appears when $\mathrm{\Omega }$ omits points, and it is those omitted points that the index has to ignore. When a nonempty connected $\mathrm{\Omega }$ is wanted it is called a complex domain (A complex domain is a nonempty connected open subset of $\mathbb{C}$).