I am doing a past paper for an introductory course in algebraic topology. The question is
Calculate the homology of the quaternionic projective space. What can you say about its homotopy groups?
I figured that we have a CW decomposition with one cell in every $4k$-th dimension. All boundary maps are 0, so that $$H^{4k}(\mathbb HP^n,\mathbb Z)=\mathbb Z\\H^i(\mathbb HP^n,\mathbb Z)=0,\quad i\neq 4k$$
In particular, if the fundamental group is Abelian, it must be trivial.
For the homotopy groups, there is a fibration $$\mathrm{Sp}(1)\to S^{4n+3}\to\mathbb HP^n$$ so that $\mathbb HP^n$ has higher homotopy groups isomorphic to those of $S^{4n+3}$.
But the long exact sequence of the fibration ends in $$\cdots\to\pi_1S^{4n+3}\to\pi_1\mathbb HP^n\to\pi_0\mathrm{Sp}(1)\to\pi_0S^{4n+3}\to\pi_0\mathbb HP^n\to 0$$
But $\mathrm{Sp}(1)$ has two path-components, so this seems to suggest that $\pi_1\mathbb HP^n=\mathbb Z_2$, which contradicts the fact that $H_1(\mathbb HP^n,\mathbb Z)=0$. Where have I gone wrong?
Sp(1) doesn't have two path components. It's just the unit quaternions, which are $S^3$.