6
$\begingroup$

Some authors introduce singular homology as a family of homology functors indexed by $n\in\mathbb Z$: $$H_n\colon \textbf{hTop} \to \textbf{Ab}$$

However, other authors define one functor $$H\colon \textbf{hTop} \to \textbf{GradedAb}$$ Clearly both points in some sense "agree" on the resulting objects – a sequence $H_n(X)_{n\in \mathbb Z}$ is "essentially the same" as $\bigoplus_n H_n(X)$.

However, there is a slight difference in morphisms. A continuous map (or rather its equivalence class) $f\colon X\to Y$ induces a family of maps $H_n(f)\colon H_n(X)\to H_n(Y)$. These can be "glued" into a morphism

$$\prod_n H_n(X) \to \bigoplus_m H_m(Y) = H_m(Y)$$

however usually $\prod_n H_n(X)$ is bigger than $H(X) = \bigoplus H_n(X)$ and such "glued" morphism does not need to be a graded morphism.

For spaces homotopy equivalent to finite CW-complexes almost all homology groups vanish and both point of views seem to be equivalent.

Which is the preferred notion of singular homology? Do they agree for all spaces, not only CW-complexes?

I prefer the first definition – not only it obviously works for all topological spaces, but also it homology can be defined in the setting of homological algebra using the language of derived functors. (And I'm not aware of any way of glueing resulting homologies in an arbitrary abelian category – for these only finite sums are guaranteed to exist).

$\endgroup$
  • 2
    $\begingroup$ I don’t understand what the problem is. To give a graded homomorphism $\bigoplus A_n\to\bigoplus B_n$ is exactly the same as giving a family of homomorphisms $A_n\to B_n$. $\endgroup$ – Jeremy Rickard Mar 28 at 18:55
  • $\begingroup$ I always thought that $\mathsf{GradedAb}$ meant the category $\coprod \limits_\mathbb{Z} \mathsf{Ab}$ in which case the notions should agree. I did not study graded algebra though, so I don’t know about „proper“ graded abelian groups though... $\endgroup$ – PrudiiArca Mar 28 at 18:59
3
$\begingroup$

A family of maps $H_n(f):H_n(X) \rightarrow H_n(Y)$ induce a map $\bigoplus_i H_i(X) \rightarrow \bigoplus_j H_j(Y)$. I don't know why you're introducing $\Pi_i H_i(X)$.

There is also no canonical map $\Pi_i H_i(X) \rightarrow \bigoplus_j H_j(Y)$, the canonical map would be $\bigoplus_i H_i(X) \rightarrow \Pi_j H_j(Y)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Counter question: why not? For example, the Chern character of the tautological line bundle on $\mathbb{C}P^\infty$ is non-zero in infinitely many degrees. Granted, the Chern character lives in cohomology, not homology, but still. If you look around the literature, you'll find that said cohomology ring is sometimes described as a polynomial ring and as a power series ring at other times. It's usually what people need it to be. I wouldn't mind having a convincing argument for why it's one or the other or at least why it doesn't really matter. $\endgroup$ – Daniel Kruse Mar 28 at 20:42
  • 3
    $\begingroup$ @DanielKruse I think that discussion is way beyond the scope of this post. He claims that two categories are not equivalent ($\textbf{Ab}^{\mathbb N}$ and $\textbf{GradedAb}$). The fact is that the categories are equivalent. $\endgroup$ – Noel Lundström Mar 28 at 21:44
  • $\begingroup$ @NoelLundström Thanks a lot, both your answer and comment are excellent. I was thinking about the total Chern class (similarly to Daniel Kruse's comment) that lives not in $\bigoplus H^k(X)$ but in $\prod H^k(X)$ (that happens to be the same for a finite CW-complex) and I confused the directions of arrows. $\endgroup$ – Paweł Czyż Mar 29 at 20:05
  • $\begingroup$ @PawełCzyż Thank you! I'm always happy to help :) $\endgroup$ – Noel Lundström Mar 29 at 21:12

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.