Let $f:X\rightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that 
$H\_{\ast}(f):H\_{\ast}(X,\mathbb{Z})\rightarrow H\_{\ast}(Y,\mathbb{Z})$ is an isomorphism for $\ast\leq n$


Is it true that the reduced homology of the fiber is $\tilde{H}\_{\ast}(F,\mathbb{Z})=0$ for $\ast\leq n$?



Homology of the fiber
As usual, there's no loss of generality in assuming that $f$ is the inclusion of a subspace $X\subset Y$, replacing $Y$ with the homotopy equivalent mapping cylinder of $f$ if necessary. By your assumptions and the five lemma, $H\_\*(Y,X)=0$ for $\*\leq n$, and the pair $(Y,X)$ is simply connected, therefore by the Hurewicz theorems $\pi\_\*(Y,X)=0$ for $\*\leq n$. If $F$ denotes the homotopy fiber of $f$, then $\pi\_\*(Y,X)=\pi\_{\*-1}(F)$ in all dimensions, hence the previous computation ensures that $F$ is $(n-1)$-connected, so $H\_\*(F)=0$ for $\*\leq n-1$. As @abx shows in the comment above, in general $H\_n(F)$ won't be trivial. The higher-dimensional Hopf fibrations provide further counterexamples, where even the fiber is simply connected.


