TITLE: Euler characteristic in pullbacks
QUESTION [5 upvotes]: 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}$$

REPLY [3 votes]: 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.