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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 $\ast\leq n$?

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    $\begingroup$ What about the Hopf fibration $f:\mathbb{S}^3\rightarrow \mathbb{S}^2$ with fiber $\mathbb{S}^1$? $H_1(f)$ is an isomorphism but $H_1(\mathbb{S}^1)=\mathbb{Z}$. $\endgroup$
    – abx
    Commented Mar 18, 2019 at 12:04
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    $\begingroup$ Besides the proof below, this (the vanishing of the reduced homology of fiber below dimension n) also admits an easy proof using the Serre spectral sequence. $\endgroup$
    – Nicholas Kuhn
    Commented Mar 18, 2019 at 22:02

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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.

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