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I'm trying to understand how the interpretations of $k$-dimensional holes comes from the quotient group definition of Homology Groups.

Taking $\partial_k$ as the $k$-dimensional boundary operator on a simplicial complex $X$. The $k$-th homology group of $X$ is: \begin{equation} $$ \tag{1} H_k(X) = \frac{\ker(\partial_k)}{\text{Im}(\partial_{k+1})}. $$ This is commonly said to be interpreted as the $k$-dimensional holes present in the simplicial complex. In this paper, it is stated that $H_k(X)$ is: $$ \tag{2} H_k(X) = \ker(\partial_k)\cap\big(\text{Im}(\partial_{k+1})\big)^C, $$ basically saying that the $k$-th homology group consists of the cycles that are not boundaries of a $(k+1)$-dimensional simplex.

This second interpretation of a $k$-dimensional hole makes sense to me but I'm having difficulty understanding how to go from (1) to (2) as my understanding is also that the elements in (1) are basically sets of cycles while the elements in (2) are simply cycles. Any help disambiguating this is appreciated.

D. Brito
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    Does the superscript $C$ mean some sort of complement? If so, what sort? Except in relatively simple situations, there is no canonical complement, so the quotient definition does not admit a reliable translation of this sort. Please clarify? – paul garrett Oct 30 '24 at 16:45
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    Ignore the paper you are quoting (what they wrote is simply wrong). My suggestion is to read an algebraic topology textbook instead. – Moishe Kohan Oct 30 '24 at 16:55
  • See related posts here and here. In brief: holes are an intuitive device for homology whose utility breaks down quickly once you start getting into the details. If you wish: keep them as an intuitive crutch as you limp through your early understanding of homology; dump them when you start to walk. – Lee Mosher Oct 30 '24 at 17:16

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It looks to me that (2) is not a good definition for homology. It often won't even be a group.

Example: take the simplicial complex $X$ with four vertices $X^0=\{0,1,2,3\}$, five edges $X^1 = \{[0, 1], [0, 2], [1, 2], [0, 3], [1, 3]\}$, and one triangle $X^2 = \{[0,1,2]\}$. Then $\alpha := [1, 3] - [0, 3] + [0,1]$ is a cycle but not a boundary. Also $\beta := [1, 3] - [0, 3] + [0,2] - [1, 2]$ is also a cycle but not a boundary. But these are really supposed to represent the "same hole" (they're homologous), so they ought to be identified in some sort of quotient. Note especially that $\alpha - \beta$ is a boundary!

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You might attempt to salvage this definition in certain circumstances by talking about bases for your chain groups: if $B$ is a basis for $\ker \partial_n$ and $B' \subset B$ is a basis for $\operatorname{im} \partial_{n+1}$, then a basis for $\ker \partial_n / \operatorname{im} \partial_{n+1}$ is given by $\{[b] : b \in B \setminus B'\}$. This actually can be made to always work when working with coefficients in a field, but for integer coefficients, there is no hope, since integral homology groups are not always free and thus wont always have a basis.

This actually will work in the above example, because $\ker \partial_1$ has a $\mathbb{Z}$-basis $\{\alpha, \partial_2[0,1,2]\}$, and $\operatorname{im} \partial_2$ has the sub-basis $\{\partial_2[0,1,2]\}$, so $H_1$ is free abelian with one generator: the equivalence class generated by $\alpha$. But we did need to be talking about the set-difference of sub-bases (and assuming those exist), NOT the set-difference of the groups they generate.

Dennis
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