Homology of the fiber

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 $$\ast\leq n$$?

• 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 $H_1(\mathbb{S}^1)=\mathbb{Z}$. – abx Mar 18 '19 at 12:04
• 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. – Nicholas Kuhn Mar 18 '19 at 22:02

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.