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    $\begingroup$ Well you don't need $\pi_n$, you can directly look at $H_n$ :) In fact I think the only way I know how to prove that transpositions are sent to $-1$ without being circular uses $H_n$ rather than $\pi_n$ (and for $H_n$, you can use explicit generating cycles in the singular chain complex of $S^n$, which get mapped to their opposite) $\endgroup$ Commented Mar 9, 2022 at 17:03
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    $\begingroup$ If you do simplicial homology, then the sign is built into the definition of the action on chains. If you do singular homology, I'm not sure, but I am skeptical that you can get far enough to prove that $H^n_{sing}(S^n)$ is nontrivial without ever introducing the sign of a permutation. $\endgroup$ Commented Mar 9, 2022 at 17:32
  • $\begingroup$ @DavidESpeyer Yes, I guess you are correct about this. I really love and hate this question. $\endgroup$ Commented Mar 9, 2022 at 18:02
  • $\begingroup$ @DavidESpeyer : what do you mean ? It seems to me like neither the definition of singular homology nor its computation in the case of $S^n$ involves the sign, can you specify where it would come up ? (I'm probably just missing something) $\endgroup$ Commented Mar 9, 2022 at 18:14
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    $\begingroup$ I don't think it requires an orientation, you just have ordered simplices and you plug in a $(-1)^i$ when you remove the $i$th one. What you have is way stronger than an orientation, it's an ordering (I agree that otherwise, you might as well just talk about oriented simplices) $\endgroup$ Commented Mar 9, 2022 at 18:42