Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 18 / 4

Let $p:E\to S^r$ be the Serre fibration. For $r\ge 1$, cover the base by $U=S^r\setminus \{s\}$ and a small open ball $V$ about $s$. Write $A=p^{-1}U$ and $B=p^{-1}V$. The contractible-base assumption gives $A\simeq F$ and identifies the pair $(B,A\cap B)$, up to fiberwise homotopy, with $(F\times V,F\times (V\setminus \{s\}))$.

Excision for the open cover $E=A\cup B$ gives $H_i(E,A)\cong H_i(B,A\cap B)$. The relative base pair hasby the deformation retraction of the punctured ball onto $S^{r-1}$ and its relative homology sequence. The relative Künneth theorem therefore givesThere is no Tor term because the only nonzero base group is free. The long exact sequence of a pair, replacing $A$ by its homotopy-equivalent fiber, now becomesThe last arrow is the relative boundary transported through these identifications; it can encode twisting, so it has not been set to zero. This proves the relative homology of a fibration over a sphere sequence. In the dimension-zero case, the total space is the disjoint union of the two fibers and the same formula is the split sequence for that union.

Use the based path space and the path-loop fibration $\Omega S^r\to P{S}^r\to S^r$, where paths begin at a fixed basepoint and the projection evaluates their final endpoint. The total space is contractible: the homotopy $\gamma (t)\mapsto \gamma (st)$, $s\downarrow 0$, contracts every path to the constant path. The fiber is path-connected because $S^r$ is simply connected. Hence $H_0(\Omega S^r)=\mathbb{Z}$ and all positive homology of $P{S}^r$ vanishes.

The exact sequence from part (i), for $i\ge 2$, gives $H_{i-r}(\Omega S^r)\cong H_{i-1}(\Omega S^r)$. Put $d=r-1$. Taking negative-degree homology to be zero, this says $H_j\cong H_{j-d}$ for every $j\ge 1$. Starting with $H_0=\mathbb{Z}$ determines every group:There are no torsion groups hidden in the recurrence, since each step is an actual integral isomorphism. This is the integral homology of the loop space of a sphere.

Choose the unit map $\eta :S^{r-1}\to \Omega S^r$ adjoint to the identity under $\Sigma S^{r-1}\cong S^r$. Loop adjunction gives ${\pi }_j(\Omega S^r)\cong {\pi }_{j+1}(S^r)$, so the loop space is $(r-2)$-connected and ${\pi }_{r-1}(\Omega S^r)\cong \mathbb{Z}$. The adjoint of $\eta$ is the degree-one identity, so $\eta$ sends the generator of ${\pi }_{r-1}(S^{r-1})$ to that generator. The Hurewicz theorem for both $(r-2)$-connected spaces, and its naturality, then show that ${\eta }_{\ast }$ is an isomorphism on $H_{r-1}$. It also is on $H_0$, since both spaces are connected. All other positive degrees below $2r-2$ vanish in both spaces. ThusThis proves the required existence with a specific map, without using the suspension theorem in advance.