Show that a fibration sequence \(\cdots \to \Omega B \to F \to E \to B\) induces a long exact sequence \(\cdots \to \langle X, \Omega B \rangle \to \langle X, F \rangle \to \langle X, E \rangle \to \langle X, B \rangle\), with groups and group homomorphisms except for the last three terms, abelian groups except for the last six terms.
Solution
For a fibration \[ F\to E\xrightarrow p B, \] the sequence of pointed sets \[ \langle X,F\rangle\longrightarrow\langle X,E\rangle \longrightarrow\langle X,B\rangle \] is exact: if \(pf\) is nullhomotopic, lift a chosen nullhomotopy of \(pf\) to homotope \(f\) into the fiber. Applying this to every looped fibration in the fibration sequence \[ \cdots\to\Omega B\to F\to E\to B \] gives the long exact sequence \[ \cdots\to\langle X,\Omega B\rangle \to\langle X,F\rangle \to\langle X,E\rangle \to\langle X,B\rangle. \]
For every space \(K\), the set \[ \langle X,\Omega K\rangle \] is a group using pointwise concatenation of loops, and maps induced by loop maps preserve this operation. Thus every term except the last three is a group and every corresponding map is a homomorphism. After two loopings, the loop multiplication is homotopy-commutative by the Eckmann–Hilton argument, so \[ \langle X,\Omega^rK\rangle \] is abelian for \(r\ge2\). Reading this along the fibration sequence gives exactly Hatcher’s count: all terms except the last six are abelian groups. Hence the sequence has the asserted group and abelian-group structure.