4
$\begingroup$

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 $\mathbb{Z}_n$.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.