Orientation of points in cell complexes

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)?

• I would write $-(x)$ rather than $x$...just to make the point one is dealing with a $0$-chain. You could of course choose some or all of the $-(x)$ rather than the $(x)$ to form a basis of the $0$-chain group, but why on earth would you bother? – Lord Shark the Unknown Jun 21 '18 at 18:56
• @LordSharktheUnknown Because Hatcher writes this choice is canonical. I would rather say it is a convention, not a canonical choice. – TheGeekGreek Jun 21 '18 at 18:57

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.