I have been a bit sloppy in the title, but let me be specific. I stepped again into the subtle difference between homotopy limit and limit in the homotopy category, in the following version.

Suppose you have two diagrams of spaces $X_{\bullet}, Y_{\bullet}$ and a homotopy equivalence $f_i : X_i \to Y_i $ such that for any morphism $i\to j$, the morphisms $Y(i\to j)\circ f_i$ and $f_j\circ X(i\to j): X_i\to Y_j$ are homotopic. Is it true that $\text{holim} X_{\bullet} \simeq \text{holim} Y_{\bullet} $ ?

My favorite diagram is $\Delta$, i.e. cosimplicial objects, but I suspect the result should not depend on this. Let me remark that if the $f_i$ would commute strictly the result would be true.

Bonus question. If the answer is "No, they can be different", are there a higher compatibility constraints one can impose so that it becomes true? For example, by specifying what the homotopy $f_{ij}$ between $Y(i \to j) f_i $ and $f_j X(i \to j)$ should respect, and then bla bla..

Remark. I suspect that a "not so higher" version should be true for the following reason. If I have a cosimplicial object, I take the chain complex pointwise, and I apply the Dold-Kan correspondence, I get a bicomplex $C_*(X_\bullet)$: the horizontal differential is the differential of chains, while the vertical differential is the alternated sum of the cosimplicial (co)faces. I have tried to apply the definition of multicomplex homotopy equivalence between $C_*(X_{\bullet})$ and $C_*(Y_{\bullet})$, but there a few more maps to find and I don't want to mess with so many equations. In particular I should find an homotopy inverse to $f$. I hope that the "not so high" is low enough that one homotopy suffices, without higher constraints.


2 Answers 2


This is false.

Consider the two $C_2$-spaces $S^{2\sigma}$ and $S^2$, where $\sigma$ is the sign representation and $S^V$ denotes the one-point compactification. Then the two underlying spaces are the same and the action looks trivial in the homotopy category. However, the homotopy fixed points differ. (After 2-completion, these look like the same as the actual fixed points- by the Sullivan conjecture- and the fixed points in $S^{2\sigma}$ are $S^0$ which is different than $S^2$.)

The `higher' version would amount to unwinding the definition of a morphism in the $\infty$-category $\mathsf{Fun}(K, \mathcal{C})$ (if you have such a morphism, and it is pointwise an equivalence, then the diagrams are equivalent and hence give equivalent homotopy limits). In the case $\mathcal{C}=\mathsf{Spaces}$, this goes back a ways to work of Dwyer and Kan (see, e.g., their paper "Realizing diagrams in the homotopy category by means of diagrams of simplicial sets.")

In the case of cosimplicial objects in stable $\infty$-categories, we can do better. By the $\infty$-categorical Dold-Kan correspondence, the $\infty$-category of cosimplicial objects is equivalent to the $\infty$-category of cofiltered objects. If $X$ and $Y$ are cofiltered objects, then it really is true that the two diagrams are equivalent you can build equivalences $f_i$ from $X(i)$ to $Y(i)$ making the diagrams commute up to homotopy. This is because the poset $\mathbb{N}$ can be built as a coequalizer of two maps from $\coprod [1]$ to $\coprod [0]$, so it's easy to build maps out of $\mathbb{N}$.

  • $\begingroup$ Thank you for the really complete answer! Would you mind providing a reference for the last part? That is, if I have two (strict) cosimplicial objects in a stable infinity category, and a pointwise equivalence between them that makes the diagram commute up to homotopy, then the homotopy limits are the same. Infact, I am interested in the spectral sequence on chains $C_*(X_{\bullet})$, and your statement would imply the spectral sequences are the same (or at least they converge to the same thing). $\endgroup$ Apr 22, 2021 at 8:41
  • $\begingroup$ I think I forgot to tag: @DylanWilson. $\endgroup$ Apr 23, 2021 at 14:12

Let me try to complement Dylan Wilson's great answer with an example which is easier to analyze at the point-set level.

Take $I = B\mathbb N^2$, so that a functor $I\to\mathcal S$ is a space $X$ together with two commuting endomorphisms $\alpha_{0},\alpha_1:X\to X$. An explicit model for the homotopy limit is the space of tuples $(x,\gamma_i,\Gamma)$, where $x\in X$ is a point, $\gamma_i$ is a path from $x$ to $\alpha_i(x)$, and $\Gamma:[0,1]^2\to X$ is a square with sides $\gamma_0,\alpha_0\circ\gamma_1,-\alpha_1\circ\gamma_0,-\gamma_1$.

For the counterexample, set $X = S^1\times\mathbb R, \alpha_0(z,t) = (z,t+1),\alpha_1(z,t) = (e^{2\pi i t}z,t)$; one can check that for any choice of $(x,\gamma_i)$ the winding number of the concatenation of the four paths $\gamma_i$ and $\alpha_i\circ\gamma_{1-i}$ is $1$, so that the square $\Gamma$ can't exist, i.e. the homotopy limit is empty. Then set $Y = S^1$ with both endomorphisms the identity; since the limit of this diagram is nonempty, so is the homotopy limit (in fact, one easily computes the homotopy limit as $S^1\times \mathbb Z^2$); in particular the homotopy limits of $X$ and $Y$ are not homotopyequivalent. The homotopy equivalence $f$ is the projection to the first factor, and the required homotopy between $(z,t)\mapsto z$ and $(z,t)\mapsto e^{2\pi i t}z$ is conctructed by rescaling the second variable.

The reason that this counterexample works is that $X$ is a point-set model for the $\infty$-functor $B\mathbb N^2\to \mathcal S$ which sends the point to $S^1$ and the morphisms to the identity, but fills the resulting commutative square with a nontrivial self-homotopy of the identity of $S^1$.

  • $\begingroup$ Thank you Bertram! This is a very nice example and I will undoubtedly recycle it for geometric examples of homotopy limits :)) It is also clear from your last sentence how the higher things get into play. However, I think I'll accept Dylan answer because it has references to the cosimplicial part, which I am more interested in. Best! $\endgroup$ Apr 22, 2021 at 8:45

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