Homotopy Exact Sequence and Relative Homotopy Groups

The long exact sequence of a fibration is the primary tool for computing higher homotopy groups, connecting the homotopy groups of a total space, base, and fiber.

Relative Homotopy Groups

Definition

Let $(X, A, x_0)$ be a pointed pair with $x_0 \in A \subseteq X$ . The relative homotopy group $\pi_n(X, A, x_0)$ for $n \geq 1$ is the set of homotopy classes of maps $f : (I^n, \partial I^n, J^{n-1}) \to (X, A, x_0)$ where $J^{n-1} = \partial I^n \setminus \operatorname{int}(I^{n-1} \times \{0\})$ is the "top face complement." For $n \geq 2$ , this carries a natural group structure; for $n \geq 3$ it is abelian.

Definition

The boundary homomorphism $\partial : \pi_n(X, A, x_0) \to \pi_{n-1}(A, x_0)$ is defined by restricting a representative $f : (I^n, \partial I^n, J^{n-1}) \to (X, A, x_0)$ to the bottom face $I^{n-1} \times \{0\}$ , which maps to $A$ with boundary mapping to $x_0$ .

The Long Exact Sequence

Theorem9.3Long Exact Sequence of a Pair

For any pointed pair $(X, A, x_0)$ with $A$ nonempty, there is a long exact sequence $\cdots \to \pi_n(A) \xrightarrow{i_*} \pi_n(X) \xrightarrow{j_*} \pi_n(X, A) \xrightarrow{\partial} \pi_{n-1}(A) \to \cdots \to \pi_1(X, A) \xrightarrow{\partial} \pi_0(A) \to \pi_0(X)$ where $i_*$ is induced by inclusion and $j_*$ by the identity on representatives.

ExampleHomotopy groups of $S^n$

Consider the pair $(D^n, S^{n-1})$ . Since $D^n$ is contractible, $\pi_k(D^n) = 0$ for all $k$ . The long exact sequence gives isomorphisms $\pi_k(D^n, S^{n-1}) \cong \pi_{k-1}(S^{n-1})$ for all $k \geq 2$ . Since $\pi_n(D^n, S^{n-1}) \cong \pi_n(S^n) \cong \mathbb{Z}$ (by excision-like arguments), this recovers $\pi_{n-1}(S^{n-1}) \cong \mathbb{Z}$ .

Fibrations and the Long Exact Sequence

Theorem9.4Long Exact Sequence of a Fibration

Let $p : E \to B$ be a Serre fibration with fiber $F = p^{-1}(b_0)$ . Then there is a long exact sequence $\cdots \to \pi_n(F) \to \pi_n(E) \to \pi_n(B) \xrightarrow{\partial} \pi_{n-1}(F) \to \cdots \to \pi_0(F) \to \pi_0(E) \to \pi_0(B)$

RemarkComputing via fibrations

The Hopf fibration $S^1 \to S^3 \to S^2$ gives the exact sequence $\cdots \to \pi_n(S^1) \to \pi_n(S^3) \to \pi_n(S^2) \to \pi_{n-1}(S^1) \to \cdots$ . Since $\pi_n(S^1) = 0$ for $n \geq 2$ , this yields $\pi_n(S^3) \cong \pi_n(S^2)$ for $n \geq 3$ , explaining why $\pi_3(S^2) \cong \pi_3(S^3) \cong \mathbb{Z}$ .