# interpreting a long exact sequence of homotopy groups

It is a well-known result in homotopy theory that a fibration $F \rightarrow E \rightarrow B$ induces a long exact sequence in the homotopy groups; namely,

$$\pi_n(F) \rightarrow \pi_n(E) \rightarrow \pi_n(B) \rightarrow \pi_{n-1}(F)\rightarrow \cdots \pi_1(B) \rightarrow \pi_0(F) \rightarrow \pi_0(E) \rightarrow \pi_0(B).$$

My concern is, what does exactly mean being exact at the level of the $0$-th Homotopy groups? In general these are not groups. In particular, is there a geometric interpretation of the connecting map $\pi_1(B) \rightarrow \pi_0(F)$?

## 1 Answer

$F, E, B$ are all supposed to be pointed spaces here, and so their $\pi_0$ are pointed sets. The definition of exactness for a sequence of pointed sets is the same as for a sequence of groups: it means the kernel of one map (defined as the preimage of the distinguished point) is the image of another.

The map $\pi_1(B) \to \pi_0(F)$ is given by applying the monodromy action of $\pi_1(B)$ on $\pi_0(F)$ to the distinguished point of $\pi_0(F)$.

• This isn't really related to the original question, but if $F$ is a cogroup object in $\mathsf{hTop}_{\bullet}$, so that $\pi_0(F)$ is naturally a group, is the map $\pi_1(B) \to \pi_0(F)$ a group homomorphism? – Michael Albanese Oct 17 '17 at 21:40
• @Michael: for $\pi_0(F)$ to naturally be a group you want $F$ to be a group object. If this group structure is unrelated to the structure of the bundle $F \to E \to B$ then there's no reason this ought to be true. The basic way to get a relation between these things is to assume that $E \to B$ is itself the inclusion of a (homotopy) fiber, or equivalently that the bundle $F \to E \to B$ deloops to a fiber sequence $F \to E \to B \to X$. In this situation $F$ can be identified with the loop space $\Omega X$ and the natural map $\pi_1(B) \to \pi_0(F)$ is a group homomorphism because it is the... – Qiaochu Yuan Oct 17 '17 at 21:49
• ...map on $\pi_1$ induced by the bundle map $B \to X$, keeping in mind that $\pi_0(F) \cong \pi_0(\Omega X) \cong \pi_1(X)$. – Qiaochu Yuan Oct 17 '17 at 21:50
• Thanks. I was about to change cogroup to group as your comment appeared. – Michael Albanese Oct 17 '17 at 21:50
• In terms of fiber sequences you can think of the map $\pi_1(B) \to \pi_0(F)$ itself as coming from extending the fiber sequence $F \to E \to B$ to $\Omega B \to F \to E \to B$, and then applying $\pi_0$. In fact the entire long exact sequence in homotopy is just $\pi_0$ applied to the "long fiber sequence" $\dots \to \Omega F \to \Omega E \to \Omega B \to F \to E \to B$ given by repeatedly taking homotopy fibers; once you know this then the proof of exactness at every point in the long exact sequence is exactly the same. – Qiaochu Yuan Oct 17 '17 at 21:51