Orientation of points in cell complexes

Question

On page 268 - 269, Hatcher defines the notion of an orientation of a cell in his book on algebraic topology. He writes that for a cell complex $X$ with skeleton filtration $\varnothing =: X^{-1} \subseteq X^0 \subseteq X^1 \subseteq \dots \subseteq X$, we have that $$H_n(X^n,X^{n - 1})$$ is free abelian with basis in 1:1 correspondence with the $n$-cells of $X$. But there is a sign ambiguity for the cells, i.e. we could choose $e^n$ or $-e^n$ since both are generators for the $\mathbb{Z}$-summand. What I do not understand is, he moreover writes, that for $n = 0$, this choice is canonical. I mean, we have that $$H_0(X^0,X^{-1}) = H_0(X^0) \cong \bigoplus_{0 \text{ cells } e^0} \mathbb{Z}e^0$$ and by the dimension axiom we moreover know that $0$-cells $e^0$ are generators of the $\mathbb{Z}$-summands. But why should this choice be canonical? I mean, I could simply choose $-e^0$. Is this due to the fact that $e^0$ are points, i.e. $e^0 = x \in X$, and $-x$ does not make "sense" in the topological space (but in homology of course)?

Answer 1

You could choose $-e_0$, but the canonical choice is $e_0$. It's like choosing an orientation for $\mathbb{R}$, sure, you can choose -1, but the canonical choice is 1.

The thing is that for a vertex, you only have one way of mapping a singleton to your space with value your vertex. But for example, for an edge, you have two ways of mapping an interval to your space with value your edge: in one direction or the other. And if I just give you the space, you have no way of knowing which one I chose when I built the space if I don't tell you. The cell decomposition for the 0-skeleton is unique, but not for the higher skeletons.