I'm working on Topology by Armstrong, and I'm struggling to complete problem 2 in section 8.2.
The problem statement is as follows:
Show the elementary 1-cycles, mentioned in Section 8.1, generate $Z_1(K)$ for any complex $K$.
In section 8.1, an elementary 1 cycle is defined as follows:
An oriented, simple closed polygonal curve in K, when thought of as the sum of its oriented edges...may be referred to as an elementary 1-cycle.
My approach was essentially to do the following:
1] show that cycles are precisely those 1-chains where each vertex appears the same number of times as the initial and as the final element of an edge
2] show that if a chain has that property, it's generated by an elementary 1 cycle
My proof of 1]:
Suppose $s = \sum_{i} \lambda_i(v_i, u_i) \in Z_1(K)$, that is, $\sum_{i} \lambda_i \partial (v_i, u_i) = 0$. Then $\sum_{i} \lambda_i(u_i - v_i) = 0$ so $\sum_i \lambda_i u_i = \sum_i \lambda_i v_i$. Reduce the sums on both sides so that each distinct vertex appears only once. Then we have $\sum_k \alpha_k u_k = \sum_k \beta_k v_k$. In particular, note that because we consider formal combinations of distinct vertices, by definition this equality holds if and only if for every vertex it appears (weighted by the $\lambda_i$) an equal number of times as the final and as the initial vertex of an edge.
My proof of 2] (this is where I got stuck):
We proceed by induction on the number of terms in the reduced sum form of $s$, say $t$. Note that we don't need to consider $t = 1$ or $t = 2$; such linear combinations will not be cycles. [some work omitted - here I just explained why combinations of 1 and 2 edges aren't cycles]. For the base case, we consider $t = 3$; [I'm also relatively confident in my proof for t = 3, so I'm omitting it for brevity].
For the induction step, my idea was essentially to let $c$ be a maximal subsum of $s$ that is still generated by elementary 1 cycles [maximal in the number of terms in the sum]. Then I thought I could show $s-c$ has less terms than $s$, and apply the induction hypothesis. Here, I got stuck, because I couldn't figure out how to show that $s-c$ has less terms than $s$.
I'm aware this is VERY likely not the simplest way to tackle this problem but I figured I'd give my full thought process anyway. If anyone has any simpler ideas or preferably a way to salvage my own proof, they'd be much appreciated. Thank you!
Ok, after talking with some people I came to a solution that is very similar in spirit to what I'm trying to do, but works a little better.
We can WLOG assume the coefficients of $s$ are positive; for any negative coefficient, we can swap the order of the 1 simplex and omit it from the sum. Now instead of doing induction on the number of terms, we proceed by induction on the total number of edges (more precisely, the sum of coefficients of s).
The base cases are essentially the same, so I'll omit them here.
For the induction hypothesis, the idea is similar, but we don't care about taking a maximal term generated by one cycles. Rather, we build an elementary one cycle explicitly, subtract it from $s$, and apply the induction hypothesis. To build an elementary one cycle explicitly, start with some edge $(v_0, v_1)$. Because $v_1$ appears an equal number of times as an initial vertex, there is some $v_2$ such that $(v_1, v_2)$ is an edge in $s$. Because $s$ is written in reduced form, we can assume $v_2 \neq v_0$. Continue this process, building a sequence of vertices $v_0, ..., v_k$ such that $v_i \neq v_j$ if $i \neq j$ and $(v_i, v_{i+1})$ gives a distinct edge of $s$ for all $i$.
Because $s$ has finitely many edges, eventually the only edge who's initial vertex is $v_k$ will be $(v_k, v_i)$ where $v_i$ is already on the list and $i < k$. Then $c = (v_i, v_{i+1}) + (v_{i+1}, v_{i+2}) + ... + (v_{k}, v_i)$ is an elementary 1 cycle contained in $s$.
Now $c$ has the desired property [every vertex in $c$ appears an equal number of times as the final and as the initial vertex] and so does $s$, and therefore so does $s-c$. But $s-c$, when reduced, contains strictly fewer edges. Thus, by the induction hypothesis, $s-c$ is generated by elementary 1 cycles. Because $c$ is itself an elementary 1 cycle, it follows $s = s-c+c$ is generated by elementary 1 cycles.