Suppose I have two maps of topological spaces, $f:X\rightarrow B$ and $g:Y\rightarrow B$, such that $f$ induces a homology isomorphism and $g$ is a fibration and $B$ is connected. Is it true that the natural map $X\times\_{B}Y\rightarrow Y$ induces an isomorphism in homology?



Pullback and homology
This is not necessarily true. For example, there is a space $X$ constructed by attaching a 3-dimensional cell to $S^1 \vee S^2$, which serves as a standard counterexample to several questions. The map $X \to S^1$ produces isomorphisms $\pi\_1(X) \to \pi\_1(S^1)$ and $H\_\* X \to H\_\* S^1$, but $\pi\_2 X \to \pi\_2 S^1$ is not an isomorphism. As a result, the map of universal covers $\tilde X \to \Bbb R$ does not give an isomorphism $H\_2(\tilde X) \to H\_2(\Bbb R)$.


By letting $B = S^1$ and $g: \Bbb R \to S^1$ be the universal covering, we get a counterexample.


However, it turns out that this is almost completely an issue with fundamental groups. If $B$ is simply-connected, the answer to your question is yes. A heavy hammer that can be used to prove this is the [Eilenberg-Moore spectral sequence](https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Moore_spectral_sequence).


