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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)?

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  • $\begingroup$ 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? $\endgroup$ – Lord Shark the Unknown Jun 21 '18 at 18:56
  • $\begingroup$ @LordSharktheUnknown Because Hatcher writes this choice is canonical. I would rather say it is a convention, not a canonical choice. $\endgroup$ – TheGeekGreek Jun 21 '18 at 18:57
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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.

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  • $\begingroup$ Exactly what I thought...but then it is more of a convention rather than a canonical choice, isn't it? Or do you have some insights, why it is actually canonical? $\endgroup$ – TheGeekGreek Jun 21 '18 at 18:59
  • $\begingroup$ I expanded my answer a bit. $\endgroup$ – Najib Idrissi Jun 21 '18 at 19:01
  • $\begingroup$ Very nice! Thank you. $\endgroup$ – TheGeekGreek Jun 21 '18 at 19:02

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