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Small confusion regarding singular homology

Mathematics Asked by Const on January 2, 2021

This is the proof of Theorem 4.19. in Joseph J. Rotman „An Introduction to Algebraic Topology“. We want to show that $H_n(X)=0$ for all $n>0$ and $X$ convex bounded in some euclidean space. I refer to a similar question Homology group of convex sets – boundedness condition for used notation, in short, $gamma=partial cgamma+ cpartialgamma$ for a chain $gammain S_n(X)$. He then says that if $gamma$ is a cycle, we have $gamma =partial cgamma$, a boundary, so homology is trivial. But this I don‘t understand: is the cone over $0$ again $0$? In my eyes, I would see that the cone over $0$ is a boundary, which of course wouldn’t change the conclusion, but the argument confuses me a little.

Thanks in advance for any sort of clarification!

One Answer

I came to realise my mistake... the $0$-chain is not the zero simplex (which doesn't make any sense anyway, right?) but the neutral element in the group $S_n(X)$. I first deleted the question, but now I think one needs to stand by such wrong thoughts.

Answered by Const on January 2, 2021

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