Classically, the (non-local-coefficients) homology Whitehead theorem says that if $X \xrightarrow f Y$ is a map of simple spaces, and if the induced map $H_\ast(X;\mathbb Z) \to H_\ast(Y;\mathbb Z)$ is an isomorphism, then $f$ is a weak homotopy equivalence.

Conceptually for me, the essence of this theorem is that we have a stable invariant ($H_\ast(-;\mathbb Z)$), and we identify a (reasonably large) class of spaces (the simple spaces) such that our invariant detects equivalences when restricted to this class. I'm wondering how generally a statement of this form holds in a fairly general $\infty$-category $\mathcal C$ in place of $Spaces$.

For my purposes, I'm not particularly concerned with which stable invariant we use, so we might as well restrict attention to the universal case. Moreover, there are two general forms of "stabilization" which come to mind -- the category of spectrum objects $Sp(\mathcal C) = \varprojlim (\cdots \xrightarrow \Omega \mathcal C_\ast \xrightarrow \Omega \mathcal C_\ast)$, and the Spanier-Whitehead category $SW(\mathcal C) = \varinjlim (\mathcal C_\ast \xrightarrow \Sigma \mathcal C_\ast \xrightarrow \Sigma \cdots)$ (I use $\mathcal C_\ast$ to denote the $\infty$-category of pointed objects in $\mathcal C$). But we're eventually passing to some subcategory anyway, so we can reduce the $SW$ notion to the $Sp$ notion if we start out by replacing $\mathcal C$ with $Ind(\mathcal C)$ via the equation $Ind(SW(\mathcal C)) = Sp(Ind(\mathcal C))$.

Thus we are led to the following formulation:

Question: Let $\mathcal C$ be a presentable $\infty$-category. Can we identify a (reasonably large) full subcategory $\mathcal D \subseteq \mathcal C$ such that the composite functor $\mathcal D \to \mathcal C \xrightarrow {\Sigma^\infty_+} Sp(\mathcal C)$ is conservative? In particular, is this the case for $\mathcal D$ being one of the following?

  • The 1-fold suspension objects?

  • The 1-fold loop objects?

  • The 1-connected objects?

Here, a 1-fold suspension object is simply an object of the form $X = \Sigma Y$ for some $Y \in \mathcal C$; a 1-fold loop object is an object of the form $X = \Omega Y$ where $Y \in \mathcal C_\ast$ is a pointed object of $\mathcal C$. A 1-truncated morphism $W \to Z$ is a morphism such that for every $C \in \mathcal C$, the map $\mathcal C(C,W) \to \mathcal C(C,Z)$ has 1-truncated fibers, a morphism is 1-connected if it is left orthgonal to the 1-truncated morphisms, and an object $X$ is 1-connected if the map $X \to 1$ is 1-connected, where $1$ is the terminal object.

As a sanity check, I think each of my candidates for $\mathcal D$ are trivial when $\mathcal C$ has discrete hom-spaces, which is a good thing because in this case $Sp(\mathcal C)$ is also trivial.

  • 3
    $\begingroup$ Is it important to you that $\mathcal{C}$ be an arbitrary locally presentable $(\infty,1)$-category, or would it be sufficient to consider the case when it is an $(\infty,1)$-topos? In the latter case, the result would follow from a proof of the homology Whitehead theorem in homotopy type theory; I don't know if that exists in the literature yet, but it's at least not too far off. $\endgroup$ Feb 25, 2020 at 17:31
  • 8
    $\begingroup$ Taking $\mathcal{C}$ to be $n$-groupoids (aka $n$-truncated spaces), it seems to me that all of your criteria fail (because the stabilization is trivial). $\endgroup$ Feb 25, 2020 at 17:32
  • 1
    $\begingroup$ @LennartMeier That's a very good point. Offhand, I can't think of a natural restriction to put on $\mathcal C$ to rule this out, other than simply asking $\mathcal C$ to be an $\infty$-topos... $\endgroup$
    – Tim Campion
    Feb 25, 2020 at 17:36
  • 1
    $\begingroup$ Another question is whether hypercompleteness will matter. I looked back over my notes and I believe I know how to prove a "homology Whitehead theorem" in HoTT saying that if a map of 1-connected types induces an isomorphism on integral homology, or on cohomology with coefficients in all abelian groups, then the map is $\infty$-connected (i.e. induces an isomorphism on all homotopy groups). If this works, then it would imply a positive answer in the case of $\infty$-toposes where $\mathcal{D}$ is the hypercomplete 1-connected objects. But I don't know how to do better right now. $\endgroup$ Feb 27, 2020 at 0:49
  • 2
    $\begingroup$ The cohomology version of @MikeShulman's argument was written up by my student Luis Scoccola on pages 20 and 21 of arxiv.org/abs/1903.03245 . He cites Mike's blog post, where these ideas originated. $\endgroup$ Feb 27, 2020 at 1:08

1 Answer 1


Putting together my and Dan's comments deserves to be called an answer. Namely:

If $\mathcal{C}$ is an $(\infty,1)$-topos, then the statement is true when $\mathcal{D}$ is the class of hypercomplete, pointed, nilpotent objects. Hypercompleteness is the usual $(\infty,1)$-categorical notion, pointed is obvious, while "nilpotent" here means that the internal group object $\pi_1(X)$ is nilpotent and acts nilpotently on each internal abelian group object $\pi_n(X)$, in an appropriate internal sense.

For a proof, see section 3 of Nilpotent Types and Fracture Squares in Homotopy Type Theory by Luis Scoccola, which proves in homotopy type theory that any cohomology isomorphism between pointed nilpotent types is $\infty$-connected — hence an equivalence if the types are hypercomplete. Then we get the result for $(\infty,1)$-toposes by interpreting homotopy type theory internally therein, as shown here for universes and here for higher inductive types. (Those papers don't yet quite complete the interpretation by showing that the universe is closed under HITs, but I doubt that this proof depends crucially on that.)

Note that the assumption used here is weaker than yours, namely that an isomorphism is induced only on internal cohomology group objects with coefficients in all internal abelian group objects. It seems likely that with your stronger assumption hypercompleteness could be removed from the result, but probably a different method would be required.


Your Answer

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