In Munkres’ book, he proves Lemma 6.1 as follows:
Lemma $6.1$. Let $L$ be the complex whose underlying space is a rectangle. Let $BdL$ denote the complex whose space is the boundary of the rectangle. Orient each 2-simplex $\sigma_i$ of $L$ by a counterclockwise arrow. Orient the 1-simplices arbitrarily. Then:
(1) Every 1-cycle of $L$ is homogenous to a 1-cycle carried by $BdL$.
(2) If $d$ is a 2-chain of $L$ and if $\partial d$ is carried by $BdL$ then $d$ is a multiple of the chain $\Sigma \sigma_i$.
I am confused about his proof of (2) only, reproduced below:
$Proof$. The proof of (2) is easy. if $\sigma_i$ and $\sigma_j$ have an edge $e$ in common, then $\partial d$ must have value $0$ on $e$. It follows that $d$ must have the same value on $\sigma_i$ as it does on $\sigma_j$. Continuing this process, we see that $d$ has the same value on ever oriented $2-$simplex $\sigma_i$.
The part that has me confused is his assertion that the 2-chain $d$ has the same value on every simplex.
I believe I understand the assertion that the value of $\partial d$ on $e$ is zero; because, if e is a shared edge, it is necessarily carried by the boundary (and hence has value 0, by Munkres’ definition of 'carried by' a subcomplex).
I don’t quite get how he concludes that $d$ must be the same on every oriented 2-simplex; I believe it is something simple, but I can’t quite figure it out. Any clarity on why this assertion is true would be greatly appreciated.
Let the chain be $d=\sum a_i\sigma_i$. If $e$ is the edge between $2$-simplices $\sigma_i$ and $\sigma_j$, the coefficient of $e$ in $\partial d$ is $a_i-a_j$ or $a_j-a_i$ (depending on the orientation of $e$) so if $e$ is on in the edges of the rectangle then $a_i=a_j$ whenever $\partial d$ is carried by the boundary of the rectangle.
So any two adjacent triangles have the same coefficient $a_i$. As we can move from any triangle to any other triangle through a sequence of adjacent triangles, all the $a_i$ are the same.