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The question is in the title. Suppose that $X$ and $Y$ are two pointed connected CW-complexes. I was wondering if there exists a spectral sequence computing the homology of the function space $$H_{\ast}(map_{\ast}(X,Y);k) $$ where $k$ is a fixed field. Could we say something interesting in the case when $H_{\ast}(X;k)$ is trivial.

For simplicity we can assume that $X$ and $Y$ are simply connected.

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2 Answers 2

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An alternative spectral sequence for $H_∗(map_∗(X,Y);k)$ can be constructed using the approach in this paper of mine (apols for self-promotion). Actually, this spectral sequence is discussed already in an earlier paper of Bendersky and Gitler. This spectral sequence converges in the same cases as the Anderson spectral sequence that is alluded to in John Klein's answer. An explicit comparison of the two spectral sequences was given by Podkorytov. See also the paper of Ahearn and Kuhn that gives a nice introduction to the spectral sequence, and studies its structure in detail.

I think that the approach in this paper can answer the question on what happens when $H_*(X;k)=0$. Indeed, the spectral sequence is based on the following model for the chains on $map_*(X, Y)$. Let $C_*(Y)$ denote the (reduced) complex of singular chains on $Y$ and $C^*(X)$ denote the reduced singular cochains on $X$. Let $\mathcal E$ be the category of non-empty finite sets and surjective functions, and let $\mathcal E^t$ be the "twisted arrow category" of $\mathcal E^t$. An object of $\mathcal E^t$ is a surjection of sets $i\twoheadrightarrow j$ and a morphism from $i\twoheadrightarrow j$ to $s\twoheadrightarrow t$ consists of sujections $i\twoheadrightarrow s$ and $t\twoheadrightarrow j$ that make the evident square commute (note the twist in the directions of the arrows).

There is an evident contravariant functor from $\mathcal E^t$ to chain complexes that sends $i\twoheadrightarrow j$ to $C^*(X^{\wedge j}) \otimes C_*(Y^{\wedge i}) \otimes k$. The point is that the homotopy limit of this functor is quasi-isomorphic to $C_*(map_*(X,Y);k)$ when $X$ is a finite q-dimensional complex and $Y$ is $q-1$-connected. Now, suppose $H_*(X;k)=0$. Then $C^*(X^{\wedge j})\otimes k$ is an acyclic complex for all $j$, and therefore the homotopy limit is acyclic as well. So we may conclude the following

Claim: Suppose $X$ is a finite $q$-dimensional complex and $Y$ is a $q-1$-connected space. Suppose furthermore that $H_*(X;k)$ is trivial. Then $H_*(map_*(X,Y);k)$ is trivial.

Remark 1: it is not necessary to assume that $X$ is simply connected.

Remark 2: The conclusion should hold whenever either one of the spectral sequences converges. The condition that $X$ is q-dimensional and $Y$ is $q-1$-connected is the standard condition that guarantees convergence. But when $k={\mathbb Z}/p$ there also are results about "exotic convergence" of these types of spectral sequences that may be relevant.

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  • $\begingroup$ Hi Greg, if I'm reading your paper right the correct bound for convergence is that $X$ is $q$-dimensional and $Y$ is $q$-connected, or am I missing something? $\endgroup$ Commented Nov 20, 2021 at 21:06
  • $\begingroup$ Dan, you are right. In my paper I assumed that the connectivity of $Y$ is at least the dimension of $X$. In this case strong conversion follows from Goodwillie's theory of analytic functors. In my answer here I was following the other answer, which said that the Anderson-Bousfield spectral sequence converges when $X$ is $s$-dimensional and $Y$ is $s-1$-connected. I am rather confident that the two spectral sequences have the same convergence properties. This follows for example by Podkorytov's work. But I did not know it when I wrote that paper. (continued) $\endgroup$ Commented Nov 21, 2021 at 14:29
  • $\begingroup$ Furthermore, I checked the papers of Anderson and Bousfield. Anderson actually says that the connectivity of Y has to be at least the dimension of X, but Bousfield attributes the slightly better statement to Anderson! This can't be right if $Y$ is not nilpotent. For example, if $Y=S^1\vee S^1$, and $X=S^1$, then the spectral sequence will converge to the nilpotent completion of $F_2$. But: Bousfield proves the stronger convergence result when $Y$ is simple. As I said, I am pretty sure that the Goodwillie tower will converge in these cases as well, even though I did not prove it in my paper. $\endgroup$ Commented Nov 21, 2021 at 14:37
  • $\begingroup$ Thanks Greg. Is it possible that Bousfield just misquoted Anderson? In the result Bousfield states as being more general in §4.2 he has the assumption that $\widetilde H^m(X,\pi_{m+n}(Y))=0$ for $n \leq 0$ which would fit better with $\dim(X) \leq \mathrm{connectivity}(Y)$. $\endgroup$ Commented Nov 21, 2021 at 20:06
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Let $X$ be a simplicial set and $Y$ a space (or simplicial set) . Then $F_\ast(X,Y)$ is a cosimplicial space and we can consider its homology spectral sequence. Bousfield gave conditions for when this will converge to the homology of $F_\ast(X,Y)$ with field coefficients. See here:

On the Homology Spectral Sequence of a Cosimplicial Space A. K. Bousfield American Journal of Mathematics Vol. 109, No. 2 (Apr., 1987), pp. 361-394

[This is not my area of expertise, but if I remember correctly, convergence is guaranteed if, say, $X$ is $s$-dimensional and finite, and $Y$ is $(s-1)$-connected.]

Added:

See page 371 of Bousfield's paper, where he attributes the spectral sequence and its convergence to Anderson. Bousfield also claims the $E_2$-term may be identified when $k=\Bbb Z/p$ in terms of the graded vector space $H_\ast(X;k)$ and the algebra $H^\ast(Y;k)$ provided that the latter is "Hopf-like:" i.e., a tensor product of exterior and polynomial algebras.

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  • $\begingroup$ What can we say concretely in the case when $H_{\ast}(X;k)$ is trivial ? $\endgroup$
    – tictac
    Commented Jul 25, 2018 at 21:51
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    $\begingroup$ @tictac I think you'll have to go back and dig up Bousfield's paper list as [5] in the paper I cited. Look at what I "added" in my answer. Judging from what Bousfield wrote, when $k =\Bbb Z/p$, $H_\ast(X;k)$ is trivial and $H^\ast(Y;k)$ is Hopf-like, it looks to me that in this case $H_\ast(F_\ast(X,Y);k)$ will be trivial when $X$ is $q$-dimensional and $Y$ is $(q-1)$-connected. $\endgroup$
    – John Klein
    Commented Jul 26, 2018 at 1:34

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