This should be such an elementary problem in algebraic topology that I'm almost too embarrassed to ask, but here goes.


Let $f: X\to Z$ be a surjective fibration, and let $g: Y\to Z$ be any map. Assume all spaces are path-connected, and base points $x,y,z$ chosen so that $f(x)=g(y)=z$. Form the pullback in the topological category,
$$
\begin{array}{ccc}
E & \to & X \newline
\downarrow & & \downarrow \newline
Y & \to & Z.
\end{array}
$$ 


Note that $E$ need not be path-connected.



> 
> Is it possible to express $\pi\_1(E,e)$ for a given choice of base point $e\in E$, in terms of $f\_\sharp: \pi\_1(X,x)\to \pi\_1(Z,z)$ and $g\_\sharp: \pi\_1(Y,y)\to \pi\_1(Z,z)$?
> 
> 
> 



Fundamental group of a topological pullback
There is a ``Mayer--Vietoris" sequence
$$\cdots \to \pi\_2(Z, z) \to \pi\_1(E, e) \to \pi\_1(X, x) \times \pi\_1(Y, y) \to \pi\_1(Z,z) \to \pi\_0(E) \to \cdots$$
that can be developed by fitting together the 4 long exact sequences of homotopy groups obtained from the 4 maps in your diagram.


With $X \simeq Y \simeq \*$ and $Z=S^2$, $E \simeq \Omega S^2$ has fundamental group $\mathbb{Z}$, whereas $X$, $Y$ and $Z$ are simply-connected, so you can't express $\pi\_1(E)$ solely in terms of the fundamental groups of the other spaces.


