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For a commutative square of spaces (of manifolds, or of simplicial sets): $$S=\left(\begin{array}{ccc} A & \to & B \newline \downarrow & & \downarrow \newline C & \to & D\end{array}\right)$$ I am trying to understand when the relative homotopy of $(B,A)$ is isomorphic up to a certain degree to the one of $(C,D)$. According to my poor understanding of algebraic topology, this condition is closely related with $S$ being a homotopy pullback square. Yet I am most interested in reformulating this condition cohomologically (or, maybe, homologically?). Certain examples and http://ncatlab.org/nlab/show/fiber+sequence#LongSequCoh seem to suggest that the cohomology of $D$ should be something like $H^*(B)\otimes_{H^*(A)}H^\ast(C)$ (in lower cohomological degrees). Yet I do not understand how to define the latter object. Is there a nice way to do this (in particular, I would not like to consider some weird spaces in the case when $A,B,C,D$ are manifolds)?

Moreover, I would like to have something like a Hurewicz theorem i.e. I would like this cohomological condition to be equivalent to the original homotopy one. Is there a way to do this if all the spaces are simply connected? Is there a nice way to deal with $\pi_1$ in this setting if $A,B,C,D$ are not simply connected (possibly, by passing to the universal covers)?

I would be deeply grateful for any hints, and for any references here (especially, for the ones that are not too abstract, and that 'do not deal with weird spaces'). The last remark: I would like to relate this matter with etale homotopy types (and so, with certain completions of homotopy types and homotopy groups).:)

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I don't quite see how your "commutative square" is a square. Could you explain your notation, please? –  Mark Grant Feb 28 '12 at 13:05
    
I am sorry; I would like to write $$S= \begin{CD} A@>{}>> B \\ @VV{}V @VV{}V \\ C@>{}>>D \end{CD}$$ but this doesn' seem to work. –  Mikhail Bondarko Feb 28 '12 at 13:14
    
You need to include an image --- I am afraid that the TeX interpreter on the forum cannot parse commutative diagrams. –  Federico Poloni Feb 28 '12 at 13:28
    
I can't read your comment. Some hints on how to include diagrams can be found here: tea.mathoverflow.net/discussion/871/… (I should say I haven't tried this yet myself). –  Mark Grant Feb 28 '12 at 13:29
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In K(Z-mod) (so not directly related to your question), Alberto Canonaco has found a commutative quadrangle with isomorphic cones that is not homotopy cartesian. Cf. Canonaco, K., "A sufficient criterion for homotopy cartesianess". –  Matthias Künzer Mar 11 '12 at 9:23

4 Answers 4

up vote 3 down vote accepted

I think that part of what you are looking for is from work of Eilenberg-Moore.

Suppose that all four spaces are simply-connected. On taking cohomology, you get a commutative diagram of graded-commutative rings $$ \begin{array}{ccc} H^*(A)&\leftarrow &H^*(B)\\ \uparrow & & \uparrow\\ H^*(C)&\leftarrow& H^*(D). \end{array} $$ The simplest possible criterion would be that $H^*(A)$ is a pushout (of graded-commutative rings) in this diagram, or equivalently that the map $H^*(B) \otimes_{H^*(D)} H^*(C) \to H^*(A)$ is an isomorphism.

Unfortunately, the answer is close but not quite that simple - you need a version that involves more homological algebra. Specifically, there is a commutative diagram of cochain algebras $$ \begin{array}{ccc} C^*(A)&\leftarrow &C^*(B)\\ \uparrow & & \uparrow\\ C^*(C)&\leftarrow& C^*(D). \end{array} $$ and Eilenberg-Moore used this to construct a map out of the derived tensor product $$ C^*(B) \mathop{\otimes}^{\mathbb L}_{C^*(D)} C^*(C) \to C^*(A). $$ In the simply-connected case, if $A$ is a homotopy pullback then this map is a quasi-isomorphism.

As one consequence, if $H^*(B)$ or $H^*(C)$ is a projective module over $H^*(D)$, then $A$ being a homotopy pullback implies that it satisfies $H^*(A) \cong H^*(B) \otimes_{H^*(D)} H^*(C)$.

So far as the rest of your questions:

  • You might be able to combine this with the universal coefficient theorem and with the homology Whitehead theorem to get a result. Specifically, if you check it with field coefficients for an arbitrary field then you get your desired result.

  • Unfortunately there's not a purely algebraic criterion, because being a homotopy pullback won't always be detected by the maps on cohomology; sometimes it will involve delicate analysis of secondary products.

  • I don't have a nice, complete answer for you about the non-simply-connected case other than carefully looking at what's going on with the fundamental group and universal covers. The Eilenberg-Moore result wlll have issues with "exotic" convergence.

  • Finally, I'm afraid that I can't really address your desire for references without "weird" spaces, since most of my friends fall into that category. Maybe someone else can help there.

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Thank you very much! I will try to understand all of this! –  Mikhail Bondarko Feb 28 '12 at 14:06

Letting $F$ denote the homotopy fibre of the map $f\colon A \to B$, and $G$ denote the homotopy fibre of $g\colon C\to D$, your diagram gives an induced map (up to homotopy) $F\to G$.

Then the map of pairs $(B,A)\to (D,C)$ is an $i$-equivalence if and only if the map $F\to G$ is an $(i-1)$-equivalence. (All this can be seen by converting the maps into fibrations and comparing the associated long exact homotopy sequences.)

Now, if you understand the map $F\to G$ well enough on homology, you can use the homology Whitehead Theorem, as in Tyler's answer. (There is a problem here though, in that $F$ and $G$ may not be simply-connected even if all four corners of your diagram are.)

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I think you are looking for a triadic Hurewicz theorem: at least that is one interpretation, and fortunately one in which an answer can be given. (This is relevant to Mark's answer.) So I give an account of part of the paper

R. Brown, ``Triadic Van Kampen theorems and Hurewicz theorems'', Algebraic Topology, Proc. Int. Conf. March 1988, Edited M.Mahowald and S.Priddy, Cont. Math. 96 (1989) 39-57.

particularly Theorem 7. (pdf from my publication list item 60.) For convenience I'll keep the notation of that paper, partly as I want to use $C$ for the cone construction!

It is easiest to assume we have a triad $T=(X;U,V)$, so that $U,V$ are subspaces of $X$ and $W=U \cap V$. We say the triad $T$ is $(t;r,s)$-connected if the pairs $(U,W), (V,W)$ are respectively $r,s$-connected and the triad $T$ is $t$-connected, in the usual sense of vanishing triad sets and groups.

The triadic Hurewicz Theorem then asserts:

Suppose the triad $T$ is $(p+q-2;p-1,q-1)$-connected. Let $Y=X\cup C(U \cup V)$ (cone construction). Then $Y$ is $(p+q-2)$-connected and $\pi_{p+q-1}(Y)$ is obtained from the triad group $\pi_{p+q-1}(T)$ by killing the action of $\pi_1(W)$ and the generalized Whitehead product elements.

Because $Y$ is given by a cone construction, you can immediately deduce homological information by the usual exact sequences.

This is a special case of a result which can be put in the framework of squares of spaces, or more generally $n$-cubes of spaces.

R. Brown and J.-L. Loday, ``Homotopical excision, and Hurewicz theorems, for $n$-cubes of spaces'', Proc. London Math. Soc. (3) 54 (1987) 176-192.

which is a consequence of the main theorem of our previous paper referenced there (Topology, 26 (1987) 331-334), namely a van Kampen Theorem for $n$-cubes of spaces.

Also relevant in this situation is the paper

Ellis, G.~J. and Steiner, R. "Higher-dimensional crossed modules and the homotopy groups of $(n+1)$-ads".J. Pure Appl. Algebra 46~(2-3) (1987) 117--136.

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Here are some comments in the form of an answer, which can be viewed as the Koszul dual to Tyler's approach via the Eilenberg-Moore spectral sequence.

Firstly, in the special case when $D$ is a point (or even contractible) the map $$ A \to B \times C $$ is a weak equivalence, so you can use the Künneth formula to compute the (co-)homology of $A$ interms of that of $B$ and $C$.

Secondly, if $D$ is path connected, we can choose a basepoint and pull everything back along the path fibration $\tilde D \to D$ to get a homotopy pullback $$ \tilde A \to \tilde B $$ $$ \downarrow \quad \quad \downarrow $$ $$ \tilde C \to \tilde D $$ and because $\tilde D$ is contractible, the map $\tilde A \to \tilde B \times \tilde C$ is a weak equivalence. To recover $A$ from this, note that $\Omega D$ acts diagonally on $\tilde B \times \tilde C$ (in some $A_\infty$-sense, but we can rigidify this if we want to using the Kan loop group). Taking orbits we obtain $$ A \to \tilde B \times_{\Omega D} \tilde C, $$ which is a weak equivalence (here the right side is really the Borel construction of $\Omega D$ acting on $\tilde B \times \tilde C$).

In particular, The above suggests that there's a quasi-isomorphism of DG coalgebras $$ C_\ast(\tilde A) \simeq C_\ast(\tilde B) \otimes_{C_\ast(\Omega D)} C_\ast(\tilde C) . $$

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