Is there a cojoin, the dual construction of the join of topological spaces?

Question

The join $X\star Y$ of topological spaces $X$ and $Y$ can alternatively be written as a homotopy pushout of the canonical diagram $X\leftarrow X\times Y\rightarrow Y$: \begin{equation} X\star Y =X\stackrel{h}{+}_{X\times Y}Y. \end{equation} What about the dual construction, the homotopy pullback of the canonical diagram $X\rightarrow X+Y\leftarrow Y$, written as $X\stackrel{h}{\times}_{X+Y}Y$? I haven't found anything about it yet. I suspect it to be empty (and therefore uninteresing) because of disjoint images in the mentioned diagram, but is that true and if yes, how can it be argued elegantly? I think, there should be a short argument involving cofibrations.

Answer 1

The construction comes up in the pointed category. Thus for based spaces $X,Y$ let $X\mathop{\hat\ast} Y$ be the homotopy pullback of the triad $$X\xrightarrow{in_X}X\vee Y\xleftarrow{in_Y}Y.$$ I guess we can call this the cojoin of $X,Y$.

This space is generally non-trivial and is related to the so-called flat product $X\mathop{\flat} Y$, which is defined be the homotopy fibre of the inclusion $$X\vee Y\hookrightarrow X\times Y.$$ In fact, for path-connected, well-pointed spaces $X,Y$ there is a homotopy equivalence $$X\mathop{\hat\ast} Y\simeq \Omega(X\mathop{\flat} Y)\simeq \Omega\Sigma(\Omega X\wedge\Omega Y).$$

The cojoin construction comes up when $X=Y$ and $X$ is a co-H-space with comultiplication $c:X\rightarrow X\vee X$. In this case $c$ induces a morphism of triads from $$\ast \rightarrow X\leftarrow \ast$$ to $$X\xrightarrow{in_1}X\vee X\xleftarrow{in_2}X,$$ and taking homotopy pullbacks yields the coHopf construction $H'(c)$ as a map $$H'(c):\Omega X\rightarrow X\mathop{\hat\ast} Y.$$ The coprojective plane of $(X,c)$ is defined to be the homotopy fibre of $H'(c)$. In case $c$ is suitably coassociative, this construction can be repeated so as to yield higher coprojective planes.

This is nicely dual to the theory of projective planes for H-spaces and higher projective planes for $A_n$-spaces. However, unlike the dual theory, things quickly breaks down, and I don't know of any real use for the coprojective planes.

Answer 2

Yes, the pullback is empty, in fact any $Z \to X + Y$ that factors as both $Z \to X \to X+ Y$ and $Z \to Y \to X +Y$ must have $Z$ empty: for any $z \in Z$, $z \to X \to X + Y$ is homotopic to $z \to Y \to X + Y$. But $[0,1]$ is connected so there is no such $z$.