From Rotman's Algebraic Topology:
Let $X$ be a space. A $1$-chain $\gamma = \sum m_i \sigma_i$ is a cycle if and only if $\gamma$ is homologous to a linear combination of polygons.
Proving $\Rightarrow$:
Let $\gamma = \sum m_i \sigma_i$ be a cycle. If some $m_i \lt 0$, then a theorem says that $m_i \sigma_i$ is homologous to $(-m_i)\sigma_i^{-1}$. We may assume that each $m_i \ge 0$. The proof proceeds by induction on $\sum m_i \ge 0$; the induction does begin when $\sum m_i = 0$, for now $\gamma = 0$. For the inductive step, we may assume that each $m_i \gt 0$. (The proof then constructs a polygon $\pi = \sum_{j=p}^n \sigma_{i_{j}}$ made from a selection of $\sigma_i$'s from $\gamma$.) Thus $\gamma – \pi$ is a $1$-cycle to which the inductive hypothesis applies. Therefore $\gamma – \pi$ and hence $\gamma$ is homologous to a linear combination of polygons.
How exactly is induction used here? From my understanding of an inductive proof. You start with a base case, make an inductive assumption on some $n$-th case, and then show that it implies the $n+1$-th case also follows. But I don't understand how this proof follows this format.
Can someone explain what I'm missing here?
Best Answer
I will just spell out Ben Steffan's comment more explicitly. I'm assuming that "polygon $\implies$ cycle" has been proven separately, because it should be less complicated.
Denote $\gamma = \sum_{i = 1}^k m_i \sigma_i$ where $\sigma_i$ are singular simplices. WLOG we assume all coefficients are positive, and proceed to prove "$\implies$" by induction on their sum $\sum m_i$.
Base case: $\sum m_i = 0$.
Since all coefficients are positive, the sum must be empty and $\gamma = 0$, so it is homologous to a sum of zero polygons.
Inductive step: For $n\geq 0$ suppose that the theorem is true for all $\gamma'$ such that $\sum m_i' \leq n$; then establish the theorem for cycles $\gamma$ such that $\sum m_i = n+1$.
Then you say the proof constructs a polygon $\pi = \sum_{j=1}^l m_{i_j}\sigma_{i_j}$ (and hence a cycle) for some non-empty subset of indices $I = \{i_1,\dots, i_l\} \subset \{1,\dots, k\}$. If $k = l$ then $\gamma = \pi$ is already a polygon so we are done, so suppose $k < l$. Since $\gamma$ was assumed to be a cycle the difference $\gamma - \pi$ is then also a cycle, but $\gamma - \pi = \sum_{i\notin I} m_i \sigma_i$ and since $l \neq k$ the sum of these coefficients must be $\leq n$, so by "complete induction" (also known as "strong induction") $\gamma - \pi$ is homologous to a linear combination of polygons, and therefore so is $\gamma = (\gamma - \pi) + \pi$.