I'm reading Hatcher's treatment of cellular homology, and on pages 140 and 141 he shows how the cellular boundary maps $H_n(X_n,X_{n-1}) \to H_{n-1}(X_{n-1},X_{n-2})$ can be computed in terms of degrees. There are a lot of details to this discussion that would be too tedious to rewrite here so I'll just give a link to the book: http://pi.math.cornell.edu/~hatcher/AT/ATch2.pdf. I feel like this whole discussion suffers from the fact that Hatcher ignores how different choices affect the overall picture. For example, when we consider the map $S^{n-1}_\alpha \to X^{n-1} \to S_\beta^{n-1}$, since the source and target spheres are different, the degree of the map is only well-defined up to a sign. We have to make choices about how we identify the homology of each sphere with $\mathbb{Z}$, and even worse, we have to make choices for generators of each summand of $H_n(X^n,X^{n-1})$ for each value of $n$. All this sign ambiguity becomes a complete mess in the formula.

Rather than go through the arduous process of trying to figure out how to make all these choices compatibly, I want to ask whether or not this affects homology calculations. It seems that when building the matrix corresponding to each cellular boundary map, the entries could differ by a sign in any slot depending on how we choose our generators. But does switching signs arbitrarily in the matrices of maps between free abelian groups affect the kernels and images enough to change the homology we end up with? Or am I free to ignore signs the whole way through?

  • $\begingroup$ I know that's not your question, but let me just point out that the map $S^{n-1}_\alpha\to S^{n-1}_\beta$ are well-defined, not up to a sign, if you fix the cell-structure of $X$. Indeed $S^{n-1}_\alpha\to X^{n-1}$ is given by the attaching map, and then you also have $D^{n-1}_\beta\to X^{n-1}$ given by the characteristic map, and the latter induces a homeomorphism $S^{n-1}_\beta \to X^{n-1}/\text{what you know}$ : but if you fix (once and for all, before beginning algebraic topology) the homeomorphism $S^{n-1}\cong D^{n-1}/\partial D^{n-1}$, this is again strictly defined (1/2) $\endgroup$ Sep 26 '20 at 22:00
  • $\begingroup$ I was actually thinking about this myself right when you commented. So I guess that removes one of the complications, but we still have the problem of choosing generators for each summand of the free abelian groups. $\endgroup$
    – Nick A.
    Sep 26 '20 at 22:02
  • $\begingroup$ so you get a zigzag $S^{n-1}_\alpha \to X^{n-1}\to X^{n-1}/\text{what you know} \leftarrow S^{n-1}_\beta$ where the only wrong-directional arrow is a homeomorphism. This provides a map $S^{n-1}_\alpha\to S^{n-1}_\beta$ which is well-defined, and not up to a sign. Similarly, the choices of generators can be made "canonically", if you fix (once and for all of time) generators of $H_n(D^n,\partial D^n)$ (2/2) $\endgroup$ Sep 26 '20 at 22:02
  • $\begingroup$ As for your specific question of whether it matters in practice, one would need to write down a precise statement, but now that the ambiguity at the level of maps is removed, this is easier : an automorphism of each of the terms of a complex, compatible with the differentials, induces an automorphism of the homology ! $\endgroup$ Sep 26 '20 at 22:04

The definition of a CW complex requires the existence of a set of characteristic maps $\chi_{n,\alpha} : D^n \to X^n$, where $n$ is the dimension and $\alpha$ is an index for the set of open $n$-cells.

The ambiguity that concerns you (regarding ambiguous signs in the matrix of the boundary map) is resolved once you fix the collection of characteristic maps whose existence is required by the definition, because the attaching map $S^{n-1}_\alpha \mapsto X^{n-1}$ is then simply the restriction of $\chi_{n,\alpha}$ to $\partial D^n = S^{n-1}$.

But you're missing an important point, namely the theorem guaranteeing the existence of an isomorphism between the singular homology groups of $X$ (defined solely in terms of the topology) and the CW homology groups (defined in terms of the skeleta, the cells, and the choice of characteristic maps). So no matter how you make all of the the latter choices --- not only can you change the characteristic maps, you can even change the number of cells in each dimension! --- what you get from the calculation is guaranteed to be isomorphic to the singular homology groups. It's the homology groups that matter, not the chain groups or the boundary maps or the choice of a basis for the chain groups.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.