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It is known that the nth cohomology group can be viewed as a left quillen adjunction from $Top$ to $Ab$. (It is a composition of a series of adjunction from $Top$ to $sSet$, $sAb$, $Ch_{\geq}(Ab)$, $Ab$. So the nth cohomology group maps coproducts of two CW complexes to the coproduct of their nth cohomology group.

Similar situations happen in cohomology rings (even better). For example, $H^*(\coprod X_{\alpha})=\prod H^*(X_{\alpha})$; Furthermore, if R is PID and $H_p(X)$ is a finitely generated free R-module for all p, then $H^*(X\times Y;R) = H^*(X;R)\otimes H^*(Y;R)$.

Since there are so many phenomenon which seems to indicate that for good enough situations, the coefficient ring preserve finite limits and colimits, I wondered if there exists any explanation for this.

To simplify the situation, we only consider in coefficient $\mathbb{Z}$ here.

The sad thing is that it is even not obvious that we still have left Quillen adjunction like the situation for nth homology group, since $Ch_\geq(Ab)$ has lost too much information for cohomology ring. Furthermore, it is hard to explain the condition "$H_p(X)$ is finitely generated for all p" (not always true even for CW complexes) if we can derive the conclusion simply by an adjunction.

Xingzhi Huang
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1 Answers1

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The cohomology ring functor

$$X \mapsto H^{\bullet}(X) : \text{Top}^{op} \to \text{CRing}^{\bullet}$$

does not preserve either finite limits or colimits. (Note that since it's contravariant, the meaning of "preserves" is "sends finite limits resp. colimits to finite colimits resp. limits.")

For a counterexample involving pushouts take

$$S^n \cong D^n \sqcup_{S^{n-1}} D^n$$

and for a counterexample involving pullbacks take

$$\mathbb{Z} \cong \mathbb{R} \times_{S^1} \mathbb{R}.$$

The issue is that limits and colimits in $\text{Top}$ don't respect homotopy equivalences (see what happens when we replace the contractible spaces $D^n$ and $\mathbb{R}$ in the above counterexamples by points), so there's no hope for them to have nice behavior with respect to a homotopy invariant functor. To fix this we replace them with homotopy limits and colimits, which you can think of as the "nonabelian derived functors" of limits and colimits, and which do respect homotopy equivalences. (The above pushout resp. pullback is in fact a homotopy pushout resp. pullback, but this is no longer true if we replace $D^n$ resp. $\mathbb{R}$ by a point.)

If we think of cohomology as producing not a graded ring but a ring spectrum, then it becomes a representable functor $[X, H \mathbb{Z}]$ in a homotopy sense, and then we expect it to send homotopy colimits to homotopy limits. After passing to the individual cohomology groups this becomes a kind of Grothendieck spectral sequence.

Dually, if we lift homotopy to take values in spectra rather than graded abelian groups, it ends up being a kind of tensor product $H \mathbb{Z} \otimes X$ which is in particular a left adjoint, so it sends homotopy colimits to homotopy colimits; after passing to the individual homology groups we get another spectral sequence.

As far as I know, neither homology nor cohomology have good behavior with respect to homotopy limits in general, although some things can be said in some cases, e.g. the Serre spectral sequence.

Edit: The notation I'm using is apparently causing confusion. To be precise, by $[X, H \mathbb{Z}]$ I mean the mapping spectrum of maps from $\Sigma^{\infty} X$ to $H \mathbb{Z}$, and by $H \mathbb{Z} \otimes X$ I mean the smash product of $\Sigma^{\infty} X$ and $H \mathbb{Z}$. The reason I write $X$ instead of $\Sigma^{\infty} X$ is that I am thinking of spectra as being tensored and cotensored over spaces; see the nLab for more on this.

The goal of this notation is to be as directly analogous to the $0$-truncated case as possible; in this case $X$ is a set, cohomology means the set of functions $X \to \mathbb{Z}$, and homology means the free $\mathbb{Z}$-module $\mathbb{Z}[X]$. These can be described in terms of a tensoring and cotensoring of abelian groups over sets.

Qiaochu Yuan
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    There seems to be some confusion in the last paragraph: Homology does not "end up being a kind of tensor product," it ends up being the homotopy groups $\pi_(H\mathbb{Z} \otimes X)$ of this tensor product, taken in $\mathrm{Sp}$. So, yes, $H\mathbb{Z} \otimes {{-}}$ is a left adjoint and commutes with homotopy colimits, but no, homology in general doesn't because $\pi_$ doesn't. – Ben Steffan Mar 20 '25 at 20:10
  • Actually, perhaps I'm not understanding your last paragraph correctly: What model (or $\infty$-)category are you taking as the codomain of $[{{-}}, H\mathbb{Z}]$? – Ben Steffan Mar 20 '25 at 20:24
  • @Ben: I had in mind lifting homology to take values in spectra rather than graded abelian groups, so I mean $H \mathbb{Z} \otimes X$ and not its homotopy. I'll edit to clarify this. I can't claim to be familiar with the rigorous details but $[-, H \mathbb{Z}]$ should take values in any of spectra, $H \mathbb{Z}$-module spectra, ring spectra, commutative ring spectra, etc. and I believe homotopy limits should be computed the same in each of these (I am working entirely based off of analogy with the $1$-categorical situation here). – Qiaochu Yuan Mar 20 '25 at 20:52
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    I'm being a bit careless with the term "representable functor" here; it doesn't suffice to think of $[-, H \mathbb{Z}]$ as taking values in spaces / homotopy types / $\infty$-groupoids / anima only, if that's the issue you see. – Qiaochu Yuan Mar 20 '25 at 20:55
  • Oh, you want $[{{-}}, H\mathbb{Z}]$ to be the (internal) mapping spectrum? Please don't write it like that. The bracket notation has a standard meaning, and that's mapping sets (or abelian groups, in this context) in the homotopy category. There's no truly standard notation for what you mean, but $F({{-}}, H\mathbb{Z})$ is relatively common (exponential notation is also sometimes seen, if that's your thing). – Ben Steffan Mar 20 '25 at 21:00
  • Thank you very much for your answer but I'm still a little confused about the notation. When you write $\otimes$ and $[,]$, do you mean the symmetric monoidal structure on $Sp$? Also, Is $X$ an abbrivation of $\Sigma^{\infty}X$? – Xingzhi Huang Mar 20 '25 at 22:14
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    @Xingzhi: yes, by $\otimes$ I mean the smash product. You can think of $X$ as an abbreviation of $\Sigma^{\infty} X$ if you want, but I am really thinking of spectra as being tensored and cotensored over spaces (see e.g. https://ncatlab.org/nlab/show/powering). Working with $\Sigma^{\infty} X$ makes it less clear why cohomology should be a ring spectrum. I want the notation to be as directly analogous to the $0$-truncated case as possible (where $X$ becomes a set and cohomology becomes the set of functions $X \to \mathbb{Z}$). – Qiaochu Yuan Mar 20 '25 at 22:28
  • @Ben: I don't have a strong opinion or a strong sense of what the convention for the meaning of $[-, -]$ is. You can see it used to refer to the mapping spectrum e.g. on the nLab here: https://ncatlab.org/nlab/show/function+spectrum – Qiaochu Yuan Mar 20 '25 at 22:31
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    @QiaochuYuan Well, I do, and given that this very notation is what confused me above, I do recommend you change it :) As for the nlab article, that's unfortunately worded: "[...] homotopy groups of the mapping spectrum $[\Sigma_\infty X, E]$" should be read as "(homotopy groups of the mapping spectrum) $[\Sigma_\infty X, E]$ ($\cong \pi_* F(\Sigma_\infty X, E)$)", given that they use the $F$-notation just one paragraph above. It also doesn't exactly speak to the quality of the source that they mistype $\Sigma^\infty$ as $\Sigma_\infty$. – Ben Steffan Mar 20 '25 at 22:40