From Hatcher's Spectral Sequences:

Compute the homology of the homotopy fiber of a map $S^k → S^k$ of degree $n$, for $k,n > 1$.

Here's where I am: For $k > 1$, the sphere $S^k$ is connected, so we have a Serre fibration. Since $H_i(S_k)$ is $\mathbb{Z}$ for $i=0,k$ and $0$ elsewhere, we use the Serre fibration and the Universal Coefficient Theorem to find that the $E^2$ page contains the homology of the fiber in the $p=0$ and $p=k$ columns, and is zero outside of these columns.

Furthermore, we have a deformation retraction of $E_f$, the pathspace, onto $S^k$. Since the skeletons of $S^k$ are all trivial until the kth skeleton, we use the Serre fibration to find that the only nonzero terms on the $E^{\infty}$ page are $E_{0, 0}^{\infty} = E_{k, 0}^{\infty}=\mathbb{Z}$. Then the only possible nonzero $d$ map is $d_k$, which tells us that each $H_{b(k-1)}(Fiber)=H_{(b+1)(k-1)}(Fiber)$ for positive $b$, and the remaining homologies of the fiber are $0$. Furthermore, the kernel of the map $d_k: \mathbb{Z}\rightarrow H_{k-1}(Fiber)$ is $\mathbb{Z}$, so $H_{k-1}(Fiber)$ must be a finite cyclic group.

So: how do I find $H_{k-1}(Fiber)$? Should I be attempting to unravel the transgression? The obvious guess for the homology of this $H_{k-1}(Fiber)$ is $\mathbb{Z}_n$.


You're almost there. Try using the long exact sequence for homotopy to find $\pi_{k-1}(Fiber)$, and then use the Hurewicz theorem. I'll see you after class on Wednesday, Kevin.

  • $\begingroup$ How lucky I am, for my professor to appear with the solution. Thanks for the help! $\endgroup$ – Kevin Yin May 27 '15 at 6:05

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