MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


If $X$ is a based space then the James construction on $X$ is the space $J(X)$ given by $$ X \quad \cup \quad X^{\times 2} \quad \cup \quad X^{\times 3} \quad \cup \quad \cdots $$ in which we identify a $k$-tuple of points $(x_1,\dots ,x_k)$ whose $j$-th coordinate is the basepoint with the $(k-1)$-tuple given by dropping the $j$-th coordinate. Then $J(X)$ is the (reduced) free monoid on the points of $X$ and for $X$ a connected based CW complex, one has a natural weak equivalence $$ J(X) \quad \simeq \quad \Omega \Sigma X . $$ Let $J_k(X) \subset J(X)$ be the subspace represented by $k$-tuples of points of $X$ or less. Then $J(X) = \cup_k J_k(X)$ and $J_k(X)$ is obtained from $J_{k-1}(X)$ by means of the pushout of $$ J_{k-1}(X) \quad \leftarrow \quad X(k) \quad \overset{\subset} \to \quad X^{\times k} , $$ where $X(k) \subset X^{\times k}$ be the set of tuples such that at least one of the coordinates is the basepoint. Since is $X$ a CW complex, the inclusion $X(k) \subset X^{\times k}$ is a cofibration.

If it hadn't been a cofibration, we could have instead taken a homotopy pushout of the diagram with $X(k)$ replaced by the space $X(k)'$ which is given as a homotopy colimit of the punctured $k$-cube of spaces $X^T$, where $T \subsetneq \lbrace 1,\cdots k \rbrace$, where the maps $X^T \to X^S$ for $T \subset S$ are given by inserting the basepoint in those coordinates corresponding to those indices $i \in T-S$.

This will result in a derived version of $J(X)$. It seems to me that the derived version has the same homotopy type when $X$ is CW since in this case $X(k)' \simeq X(k)$ and the homotopy pushout has the same homotopy type as the pushout of the above displayed diagram.

More Background

I have an urge to Hilton-Eckmann dualize the above. For a based space $X$, we can define a tower of spaces $$ \cdots \to L_3(X) \to L_2(X) \to L_1(X) = X , $$ in which $L_k(X)$ is obtained from $L_{k-1}(X)$ as a homotopy pullback of a diagram of the form $$ X^{\vee k} \to X\langle k \rangle \leftarrow L_{k-1}(X) $$ where $X\langle k \rangle$ is the homotopy inverse limit of the punctured cube given by $T \mapsto X^{\vee T}$ (for $T \subsetneq \lbrace 1,\cdots k \rbrace$) where the maps $X^{\vee T} \to X^{\vee S}$ are given by projections. Here $X^{\vee T}$ means those functions $T \to X$ wish are supported on a single point in $T$ (the point is not fixed and all other points map to the basepoint of $X$), i.e., it is identified with the $|T|$-fold wedge of copies of $X$.

The map $L_{k-1}(X) \to X\langle k \rangle$ in the diagram can be given a simple description of this map here, but I will omit it for reasons of space. For example, when $k=2$, it amounts to a map $X \to X \times X$. This is just the diagonal.

In particular, $L_2(X) = \text{holim}(X \to X \times X \leftarrow X \vee X)$. It is well known that $$ L_2(X) \quad \simeq \quad \Sigma \Omega X $$ when $X$ is connected.

The Question

Define $L(X)$ to be the homotopy inverse limit $\text{holim}_k L_k(X)$. Then $L(X)$ is Hilton-Eckmann dual to the derived version of the James construction. It is a kind of "derived free comonoid on the points of $X$."

Question: What can be said about the homotopy type of $L(X)$?

That is, does $L(X)$ have a simpler description in terms of the functors we know and love?

(Disclaimer: I once asked an abbreviated version of this question on Don Davis' list about a decade ago, but received no identifiable answers. The post can be found here: )

share|cite|improve this question
This is a disturbing question. If you assume that the cobar construction on $\Sigma X$ recovers $X$, then it should follow pretty formally that the cofree comonoid on $X$ is $\Sigma \Omega X$. But that shows up as an intermediate construction! – Ben Wieland Aug 12 '11 at 2:50
It was a long time ago and I don't know that I have records, but Frank Quinn followed essentially this path in trying to dualize my recognition principle for n-fold loop spaces. I found a mistake right away and his preprint never saw the light of day. Probably at least 30 years ago, but the moral is that this is not something that you can expect to work: it's easy to dualize formally, but the dual claim will likely fail if there is any non-formal, calculational, content to the proof. – Peter May Apr 18 '12 at 18:28
@Peter: However, Vogt, Schwaenzl and I did Hilton-Eckmann dualize the Stasheff's theorem on recognizing one-fold loop spaces. We showed that a space with a co-action by the Stasheff cells is a suspension whenever the space is 2-connected. You are right though: it involved a calculational aspect to it (essentially the Blakers-Massey excision theorems for n-cubical diagrams of spaces). – John Klein Apr 18 '12 at 22:33
@Peter: the story you told about the Mighty Quinn is related, as I recall, to the bulletin announcement:… – John Klein Apr 19 '12 at 0:46
@Ben: I am confused, why would we assume that? the cobar construction is not typically used as a model for desuspension is it? – Sean Tilson Apr 19 '12 at 5:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.