# $\pi_0$ in the long exact sequence of a fibration and quaternionic projective space

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 $S^3$.
• This also means that the higher homotopy of $\mathbb{H}P^n$ will be somewhat more complicated. – Miha Habič Jul 7 '13 at 20:26