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).:)



(Co)homological characterization of homotopy pullbacks
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.


