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I am hoping for a problem I am working on that the function defined above is an isomorphism of groups, but I cannot manage to prove it. Is $[a \cdot \sigma] = [a \cdot id: \Delta^n \to \Delta^n]$ for any $\sigma: \Delta^n \to \Delta^n$?

blacksmith
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    How much do you know about homology so far ? Do you know that it is homotopy invariant ? and do you know the homology long exact sequence of a pair ? – Maxime Ramzi Oct 12 '19 at 19:13
  • @Max yes to both questions. I know what excision is and that is pretty much it. In case you mean to use that $\partial \Delta^n$ is homotopic to $S^{n - 1}$ (homology group isomorphic to $A$) , $\Delta^n$ is homotopic to a point (trivial homology group) and then use the long exact sequence to show that the connecting homomoprhism is an isomorphism between $H_n(\Delta^n, \partial \Delta^n;A)$ and $H_{n - 1}(\partial \Delta^n;A)$, is there a way to show that that is this specific isomorphism? – blacksmith Oct 12 '19 at 19:23
  • Do you know perhaps that simplicial homology agrees with singular homology ? – Maxime Ramzi Oct 12 '19 at 20:01
  • @Max afraid not – blacksmith Oct 12 '19 at 20:20
  • Right, well I have a full solution except for one detail : my method of proof uses an explicit generator for $H_{n-1}(\partial \Delta^n;\mathbb Z) $ and I don't see how to get that it actually generates it without using simplicial homology. In other words, I need to deal explicitly with the $\mathbb Z$-case and I need something for that. I will try to think if one can get around it – Maxime Ramzi Oct 12 '19 at 20:25
  • Hi, blacksmith. This question was asked earlier today in a slightly different form: https://math.stackexchange.com/questions/3390923/isomorphism-h-ndn-sn-1a-and-a/3391211#3391211. – Matt Carr Oct 12 '19 at 23:20
  • Thank you, that answers my question. I'm guessing the second point is the way around simplicial homology Max was getting at. – blacksmith Oct 13 '19 at 09:21

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