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Given a pushout $P=B\cup_AC$ which we represent as a commutative diagram

$$ \begin{matrix} A & \stackrel{f}{\rightarrow} & B\\ \downarrow{g} & &\downarrow{k} \\ C &\stackrel{h}{\rightarrow} & P \end{matrix} $$ the euler characteristic is given by $\chi(P)=\chi(C)+\chi(B)-\chi(A)$.

Do we have a similar situation when a space is constructed from a pullback, i mean what can be said of $\chi(X)$ when a space $X$ is given by a pullback : $$ \begin{matrix} X&\stackrel{f}{\rightarrow}&Y\\ \downarrow{g}&&\downarrow{k}\\ Z&\stackrel{h}{\rightarrow}&T \end{matrix}$$

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At the best you mean homotopy pullback, since pullback isn't always homotopy invariant. For example, the pullback of $* \to X \gets *$ is $*$, but the pullback of $PX \to X \gets PX$, where $P$ denotes path space is $\Omega X$, the loop space on $X$. –  Thomas Belulovich Dec 2 '11 at 20:14
if the spaces/maps are very nice then $\chi(X)=\chi(Y)\chi(Z)/\chi(T)$ (I don't specify what "very nice" means, so this is not an answer) –  user8268 Dec 2 '11 at 20:18

2 Answers 2

Since Euler char is multiplicative in fibrations, if $Y\to T$ is a fibration, $\chi(X)=\frac{\chi(Y)\cdot\chi(Z)}{\chi(T)}$ (and hence the same formula always holds for homotopy pullbacks) — if, say, all Euler characteristics are defined and non-zero.

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This will heavily depend on the maps. For instance, you could have $k(Y)=\{p\}$ and $h(Z)=\{q\}$ two distinct points $p\ne q$, in which case you have $X=\emptyset$. You can always choose such maps unless $T=\{\ast\}$. In the latter case, computing the euler characteristic $\chi(Y\times Z)=\chi(Y)\cdot\chi(Z)$ is trivial. Thus, I do not think that there is a very satisfying formula in general, but you have that

$$X = \left\{\, (y,z) \in Y\times Z \::\: h(z)=k(y) \,\right\}.$$

In specific cases, you might be able to compute $\chi(X)$ from that by means of excision.

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