Labeling edges of a cube with + and - so each face has an odd number of +s.

Question

I am looking for a specific proof, using tools from cellular homology, of the following theorem.

Let $I^n$ be the standard $n$-dimensional hypercube with its standard cellular structure. There exists a labeling of the edges in $I^n$ from the set $\{+,-\}$ such that there are an odd number of $+$-labelled edges incident to each face.

I recall seeing a proof a while ago now which used the fact that $H_\bullet(I^n,\mathbb{Z}_2)$ is trivial, where $H_\bullet(-,\mathbb{Z}_2)$ is mod $2$ cellular homology. Presumably by considering the image of the boundary map $\partial\colon C_2\rightarrow C_1$ on the $2$-cells, although it's not obvious to me how to restate the theorem in terms of this boundary map.

I wonder if anyone could reproduce this particular proof? (I'm aware of more basic methods for proving this theorem, such as induction on $n$.)

Answer 1

Recall that $I^n$ has trivial cohomology in all dimensions as it is a contractible space. In particular, $H^i(I^n,\mathbb{Z}_2)=0$ for all $i\geq 0$. We'll rename the $+$ and $-$ in the question to $1$ and $0$ respectively and consider these as labels from the coefficient group $\mathbb{Z}_2$ attached to the various face-types of the cell complex defined by $I^n$.

Consider the cochain $\alpha\in C^2$ which assigns to each face in $I^n$ the coefficient $1$. Because the second cohomology group of $I^n$ over $\mathbb{Z}_2$ coefficients is trivial, the coboundary map $\delta^1\colon C^1\to C^2$ has its image equal to the kernal of the coboundary map $\delta^2\colon C^2\to C^3$. We note that $\delta^2(\alpha)=0$ because there are an even number of $2$-cells (for $n\geq 3$) and so there must exist a $\gamma\in C^1$ such that $\delta^1(\gamma)=\alpha$. We see that $\gamma$ is an assignment of $0$s and $1$s to the edges of $I^n$, and because $\delta^1(\gamma)=\alpha$, there must be an odd number of $1$s on the boundary of each face. It follows that $\gamma$ is the required assignment.