It seems to me that although homological localization (i.e. formally inverting $E$-homology equivalences for some $E$) is a reasonable thing to do to a spectrum, it's a pretty brutal thing to do to a space, particularly if that space is not simply connected. To put a finer point on it: shouldn't there be a better notion of localization for non-simply-connected spaces? Shouldn't it at least take into account local coefficients, for example?

As soon as one says "local coefficients" and has general cohomology theories in mind, one should start thinking in terms of parameterized homotopy theory. Recall that for $X$ a space, the $\infty$-category of $X$-parameterized spectra is just the functor category $Fun(X,Sp)$ where $Sp$ is spectra. For $E \in Sp$, let $Sp_E$ be the category of $E$-local spectra, and define the category of $E$-local $X$-parameterized spectra to be $Fun(X,Sp_E)$.

Definition: A map $f: Y \to X$ of spaces is a strong $E$-local equivalence if the induced functor $f_\ast: Fun(Y,Sp_E) \to Fun(X,Sp_E)$, between categories of parameterized $E$-local spectra, is an equivalence.

(Note that $f_\ast$ has a left adjoint $f^\ast$ which has a further left adjoint $f_!$; it's equivalent to ask for any one of these to be an equivalence).

I'm curious to what extent this notion agrees with just Bousfield localizing at $E$-homology equivalence. So, some test cases:


  1. Let $E = H \mathbb Q$. Is a rational equivalence between simply-connected spaces a strong $H\mathbb Q$-equivalence?

  2. Let $E = H\mathbb Z_{(p)}$. Is a $p$-local equivalence between (sufficiently connected?) spaces a strong $H\mathbb Z_{(p)}$-equivalence?

  3. Let $E = M(p)$. Is a $p$-complete equivalence between sufficiently connected spaces a strong $M(p)$-equivalence?

  4. Let $E = K(n)$. Is a $K(n)_\ast$-equivalence between $(n+1)$-connected spaces a strong $K(n)$-equivalence?

Guiding analogy: My perspective is that a space is like a scheme which is etale over a point. A parameterized spectrum is like a quasicoherent sheaf over this scheme. So we're asking for an equivalence of localized categories of quasicoherent sheaves. From this perspective, an $E_\ast$-homology equivalence $f: Y \to X$ looks like pretty a crude thing -- it just means that $t_! f_! \mathbb S_Y \to t_! \mathbb S_X$ is an $E$-local equivalence, where $t: X \to pt$ is the unique map. In the algebraic geometry analogy, this is like saying that the map induces an $E$-local isomorphism of the cohomology-with-compact-support of the structure sheaves, which seems like a pretty weak condition.

So perhaps strong $E$-equivalences are truly a stronger notion than $E$-equivalences...


1 Answer 1


Let $E \neq 0$ be a spectrum. Here is a classification of the strong $E$-local equivalences, a proof that the localization with respect to them exists, and an analysis of some special cases.

Classification of strong $E$-local equivalences:

Claim 1: A map of spaces $f: X \to Y$ is a strong $E$-local equivalence if and only if it is a $\pi_0$-equivalence and each map of connected components is a strong $E$-local equivalence.

The proof is straightfoward.

Claim 2: If $X,Y$ are connected spaces, then $f: X \to Y$ is a strong $E$-local equivalence if and only if $E_\ast(\Omega X) \to E_\ast(\Omega Y)$ is an isomorphism.

Proof: Note that $f^\ast: Sp_E^Y \to Sp_E^X$ is conservative and has a left adjoint $f_!$, so is an equivalence if and only if $M \to f^\ast f_! M$ is an equivalence for all $M \in Sp_E^Y$. Note that $Sp_E^X$ is equivalent to the category of modules for $L_E \Sigma^\infty_+ \Omega X \in Alg^{E_1}(Sp_E)$, and likewise $Sp_E^Y$ is the category of modules for $L_E \Sigma^\infty_+ \Omega Y$. So $f_!(M) = L_E \Sigma^\infty_+ \Omega Y \wedge_{L_E \Sigma^\infty_+ \Omega X} M$. So if $L_E \Sigma^\infty_+ \Omega X \to L_E \Sigma^\infty_+ \Omega Y$ is an equivalence, then $f^\ast$ is an equivalence, and taking $M = L_E \Sigma^\infty_+ \Omega X$, we see that the converse also holds. And this can be tested by taking $E$-homology, yielding the claim.

The local objects with respect to strong $E$-local equivalences form an accessible localization of all spaces:

Claim 3: Strong $E$-local equivalences are closed under 2-out-of-3 and under filtered colimits and coproducts in the arrow category.

The proof is straightforward.

Claim 4: Strong $E$-local equivalences are closed under cobase change (i.e. pushout along an arbitrary map).

For reference, consider a pushout square and its image under $\Omega$:

$$\require{AMScd} (\ast) \begin{CD} X @>>> Y\\ @VVV @VVV\\ W @>>> Z \end{CD} \qquad (\ast\ast) \begin{CD} \Omega X @>>> \Omega Y\\ @VVV @VVV\\ \Omega W @>>> \Omega Z \end{CD}$$

Observation 5: The monad for $E_1$-spaces commutes with $\pi_0$, and preserves $E_\ast$-equivalences.

As a result,

Lemma 6: If $(\ast)$ is a pushout, all spaces are connected, and $X \to Y$ is a $\pi_1$-isomorphism, then $(\ast\ast)$ is also a pushout of $E_1$-spaces.

Proof: By the equivalence of pointed connected spaces and grouplike $E_1$-spaces, it suffices to check that the pushout of the relevant $E_1$ spaces is grouplike. This can be checked on $\pi_0$, and so follows from the observation.

Lemma 7: $E_\ast$-equivalences of $E_1$-spaces are stable under pushout.

Proof: Because the $E_1$ monad commutes with $E_\ast$-equivalences, it descends to the Bousfield localization at $E_\ast$-equivalences, and the result follows.

Proof of Claim 4: Suppose that $X \to Y$ is a strong $E$-local equivalence in $(\ast)$; we want to show that $W \to Z$ is as well. It suffices to consider the case where $W$ is connected. We may take the pushout in two stages: first we take the pushout along $\amalg_i X_i \to \vee_i X_i$ where the $X_i,Y_i$ are the connected components of $X,Y$. Second, we take the pushout along $\vee_i X_i \to W$, i.e. we consider the case where $X$ is connected.

The first case reduces to showing that if $X \to Y$ is a strong $E$-local equivalence of connected spaces, then so is $X \vee U \to Y \vee U$ where $U$ is connected, which reduces to the second case. Thus we may assume that $X$ is connected. But then by the two lemmas, because $\Omega X \to \Omega Y$ is an $E_\ast$-equivalence, so is $\Omega W \to \Omega Z$ as desired.

Claim 8: Every strong $E$-local equivalence is a highly filtered colimit of a small set of strong $E$-local equivalences.

Proof: This is clear in the category of pointed connected spaces, since $E_\ast$ and $\Omega$ are accessible functors. Then an arbitrary strong $E$-local equivalence is a coproduct strong $E$-local equivalences of pointed connected spaces, and because filtered colimits are computes the same in both categories the result follows.

Thus the strong $E$-local equivalences are an accessible subcategory of the arrow category of spaces closed under 2-out-of-3, colimits, and cobase change. It follows that the objects local with respect to them form an accessible localization of the category of spaces.

Some special cases:

Case A: Suppose that $f: X \to Y$ is a map of simply-connected spaces and an $H_\ast(-;R)$-equivalence, where $R$ is a PID. Then $\Omega f: \Omega X \to \Omega Y$ is also an $H_\ast(-;R)$-equivalence, i.e. $f$ is a strong $HR$-equivalence.

Proof: We have a map of Serre spectral sequences:

$$ H_p(X;R) \otimes_R H_q(\Omega X; R) \oplus Tor_1^R(H_{p-1}(X;R),H_q(\Omega X; R)) \Rightarrow R \\ \qquad \qquad \qquad \qquad \qquad\qquad \downarrow \qquad \qquad \qquad \qquad \qquad\qquad \qquad \quad = \\ H_p(Y;R) \otimes_R H_q(\Omega Y; R) \oplus Tor_1^R(H_{p-1}(Y;R),H_q(\Omega Y; R)) \Rightarrow R$$

Let $q$ be minimal such that $(\Omega f)_q: H_q(\Omega X;R) \to H_q(\Omega Y; R)$ is not an isomorphism. Then $q \geq 1$ because $\Omega X, \Omega Y$ are connected. If $(\Omega f)_q(\xi) = 0$ with $\xi\in H_q(\Omega X; R)$, then there are no differentials that can kill $1 \otimes \xi$, a contradiction. If $\zeta \in H_q(\Omega Y;R)$ is not in the image of $(\Omega f)_q$, then there are no differentials which can kill $1 \otimes \zeta$, another contradiction. So $(\Omega f)_\ast$ is an isomorphism as desired.

Case B: Let $\Phi$ be the Bousfield-Kuhn functor (for a fixed prime $p$ and $n\geq 1$). Suppose that $f: X \to Y$ is a map of connected spaces and $\Phi f$ is an equivalence. Then $f$ is a strong $T(n)$-equivalence (a similar statement holds for $L_{K(n)} \Phi$ and strong $K(n)$-equivalences).

Proof: $\Phi$ commutes with fiber sequences, so $\Phi(\Omega f)$ is an equivalence, so $T(n)_\ast \Omega f$ is an equivalence as desired.

  • $\begingroup$ This is very cool! Have you thought about the converse to "Case B"? $\endgroup$ Jul 27, 2021 at 14:20
  • $\begingroup$ @SaalHardali I'm sure I thought a little bit about it around the time I posted this, but I forget what I thought :). Note that the same argument as in Case B shows that if $\Phi f$ is an equivalence, then $T(n)_\ast \Omega^d f$ is an equivalence for every $d \in \mathbb N$, which is a stronger condition than "strong $T(n)$-equivalence" -- I presume that it's strictly stronger, but I don't really know enough chromatic homotopy theory to say for sure! $\endgroup$
    – Tim Campion
    Jul 27, 2021 at 14:38
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    $\begingroup$ I think bousfield showed that this stronger version of the converse you stated is true for n=1 (i mean combining what bousfield proved for KU with the telescopic conjecture at height 1) $\endgroup$ Jul 27, 2021 at 14:42
  • $\begingroup$ I think this question was motivated by your question. $\endgroup$
    – Tim Campion
    Jul 27, 2021 at 14:42
  • $\begingroup$ I anticipate that if there's a correct converse to Case B, there should be a connectivity hypothesis involved stronger than simple-connectivity -- something like $n$-connectivity. In the case $n=1$, this is the same I suppose... $\endgroup$
    – Tim Campion
    Jul 27, 2021 at 14:47

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