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I recently noticed that I could mostly prove a special case of the following statement. I think it's true in general, though perhaps only for nice spaces.

Let $X$ be a topological space, and suppose $X=X_0\cup X_1$, where $X_0$, $X_1$, and $Y:=X_0\cap X_1$ are all nonempty and path-connected. Then $$ C_*(\Omega X)\cong C_*(\Omega X_0)\otimes_{C_*(\Omega Y)}C_*(\Omega X_1) $$ where $\otimes$ is the homotopy pushout of dg-algebras and $\cong$ is $A_\infty$ quasi-isomorphism.

For disconnected spaces, it should still be true, where one replaces $C_*(\Omega X)$ with the dg-category of chains on the path space.

This seems like a natural mostly-extension of the classical Seifert--van Kampen theorem, which makes me think it exists somewhere in the literature, but I couldn't find it. Does anyone know where to find this statement?

Thanks!

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  • $\begingroup$ I think that the right statement should be the following $C_{\ast}(\Omega X)\simeq C_{\ast}(\Omega X_{0})\odot^{\mathbb{L}}_{C_{\ast}(\Omega Y)}C_{\ast}(\Omega X_{1})$ where $\odot^{\mathbb{L}}$ is the derived coproduct in the $\infty$-category of $A_{\infty}$-dg algebras. $\endgroup$
    – Ilias A.
    Sep 21, 2015 at 18:12
  • $\begingroup$ Illustration: take the homotopy push out $BG\leftarrow\ast\rightarrow BH$ which is $B(G\star H)$. The homotopy push out of $G\leftarrow\ast\rightarrow H$ in the category of $A_{\infty}$-spaces is $G\star H$ and $C_{\ast}(G\star H)=k[G\star H]\simeq C_{\ast}(G)\odot C_{\ast}(H)=k[G]\odot k[H]$ which is different from $k[G]\otimes k[H]$. I think the prove of the statement should be as follows, first you prove that you have a Quillen adjunction between monoids in simplicial sets and monoid in the category of simplicial abelian groups, $\endgroup$
    – Ilias A.
    Sep 21, 2015 at 20:10
  • $\begingroup$ ... then the left functor commutes with homotopy push out and you can identify (up to homotopy) with $C_{\ast}(-)$. $\endgroup$
    – Ilias A.
    Sep 21, 2015 at 20:10
  • $\begingroup$ Simplicial abelian groups should be repaved by simplicial k-vector spaces. it depends on what coefficients you chose. $\endgroup$
    – Ilias A.
    Sep 21, 2015 at 20:17
  • $\begingroup$ I also don't believe you're going to get the tensor product on the right. That would introduce too many commutativity relations on the left (up to homotopy). The cobar construction (which models chains on the loops) is a left adjoint, so it preserves pushouts, but the coproduct of associative algebras is not the tensor product. $\endgroup$
    – Fernando Muro
    Sep 21, 2015 at 20:46

2 Answers 2

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Sketch of proof. I will use the following ingredients

0) the category of spaces will be the category of simplicial sets. All the computation are in the derived sense.

1) use the Quillen adjunction $$F: sSet^{\otimes}\longleftrightarrow sMod_{k}^{\otimes}: U$$ between the category of monoids in simplicial sets and monoids in simplicial $k$-modules .

2) the category of simplicial $k$-algebras is Quillen equivalent to the category of differential graded $k$-algebras (i.e. monoids in the category of chain complexes of $k$-modules in positive degree)

$$N: sMod_{k}^{\otimes}\longleftrightarrow Ch^{\otimes}_{k}: S$$

Now suppose that $T$ is a homotopy push out $X\leftarrow Y\rightarrow Z$ of connected pointed simplicial sets (Kan complexes), then $\Omega T$ is the homotopy pushout of $\Omega X\leftarrow \Omega Y\rightarrow \Omega Z$ in the category in $sSet^{\otimes}$ (notice that the model that I'm using for loop space gives me a honest simplicial monoid). The functor $F$ commutes with homotopy push out, it means that that $F(\Omega T)$ is a homotopy push out of $F(\Omega X)\leftarrow F(\Omega Y)\rightarrow F(\Omega Z)$ in $sMod_{k}^{\otimes}$, therefor the $NF(\Omega X)\leftarrow NF(\Omega Y)\rightarrow NF(\Omega Z)$ is a homotopy push out in the category $Ch^{\otimes}_{k}$ and the functor $NF$ can be identified with $C_{\ast}(-,k)$. It means that $$C_{\ast}(\Omega T)\simeq C_{\ast}(\Omega X)\sqcup_{C_{\ast}(\Omega Y)}^{h}C_{\ast}(\Omega Z)$$ as differential graded algebras, where $\sqcup$ is the coproduct in the category of $DGA$. This statement is true for any pointed connected spaces $X, Y$ and $Z$.

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  • $\begingroup$ Thanks for the proof. It seems like the key fact is that $\Omega$, as a functor to $sSet^\otimes$, preserves homotopy pushouts. To me, that feels like a better version of the statement I had. Is there an easy way of seeing why it's true, or can you give a reference? Sorry if this is a basic question, I'm very much not a homotopy theorist. $\endgroup$
    – Zack
    Sep 24, 2015 at 14:59
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    $\begingroup$ @Zack, There is a pair of functors $B: sSet^{\otimes}\longleftrightarrow sSet_{\ast}: \Omega$ which restricts to an equivalence of categories (at homotopy level) when you take the restriction to the category of simplicial monoids group like $sSet^{\otimes}_{group}$ and the category of pointed connected simplicial sets $sSet^{0}_{\ast}$. It means that the derived functor $\Omega: sSet^{0}_{\ast}\rightarrow sSet^{\otimes}_{group}$ commutes with homotopy colimits and homotopy limits. $\endgroup$
    – Ilias A.
    Sep 24, 2015 at 15:47
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Zack, the coproduct in the category of associative algebras is quite different from the tensor product. For instance, the coproduct of two free associative algebras on one generator, $k\langle X\rangle$ and $k\langle Y\rangle$, is free on two generators, $k\langle X\rangle\amalg k\langle Y\rangle=k\langle X,Y\rangle$, while the tensor product is not $k\langle X\rangle\otimes k\langle Y\rangle=k[X,Y]$ since it is the commutative algebra of polynomials in two variables.

If you replace tensor product with homotopy push-out in your equation, and if all spaces are simply connected, then the result is true. In this case, the chain coalgebra $C_*(X)$ is the push-out $C_*(X_0)\oplus_{C_*(Y)}C_*(X_1)$ (this push-out is the same in the category of chain complexes and in the category of coalgebras). The cobar construction $\Omega$ is a left adjoint, so it preserves pushouts, $\Omega C_*(X)=\Omega C_*(X_0)\amalg_{\Omega C_*(Y)}\Omega C_*(X_1)$. Now use everywhere the natural quasi-isomorphism $\Omega C_*(X)\simeq C_*(\Omega X)$ for simply connected spaces.

For non-simply connected spaces the result sketched in the previous paragraph is probably false. I think a counterexample can be found in the following way, although I haven't been brave enough to tackle the final step of the computation. Take $X_0=X_1=D^2$, $Y=X_0\cap X_1=S^1$ and $X=X_0\cup_YX_1=S^2$, in this case $C_*(\Omega S^2)\simeq k\langle x\rangle $ where $x$ has degree $1$ and trivial differential. The homotopy push-out $C_*(\Omega X_0)\amalg_{ C_*(\Omega Y)} C_*(\Omega X_1)$ is quasi-isomorphic to the suspension of $C_*(\Omega Y)$ in the category of augmented chain algebras, since $D^2$ is contractible. The augmented chain algebra $C_*(\Omega Y)$ is quasi-isomorphic to $k\langle y,z\rangle/(yz-1)$ concentrated in degree $0$. In order to compute its suspension, we have first to obtain a graded-free DG-resolution. This can be done by means of curved Koszul duality theory. I think that this algebraic suspension is going to be quite different to $k\langle x\rangle $.

In any case, in the non-simply connected case the natural replacement of your statement could maybe use the Baues-Tonks twisted cobar construction.

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  • $\begingroup$ Fernando, this is a silly question but: in the second paragraph, you're saying that the pushout in dg coalgebras and in chain complexes is the same, that is, the forgetful functor preserves pushouts? $\endgroup$
    – Pedro
    Dec 1, 2018 at 17:00
  • $\begingroup$ @PedroTamaroff Yes, that's exactly what I meant. $\endgroup$
    – Fernando Muro
    Dec 2, 2018 at 21:13

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