TITLE: Deformation of a diagram preserve the homotopy limit
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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 $S^1$.