2
$\begingroup$

Suppose we have a fiber sequence of connected spaces $A\rightarrow B\rightarrow C$ and suppose we know the homology of A and B, is there a homological spectral sequence converging to the homology of $C$.

$\textbf{Edit}:$ I think a closely related question is the following. Suppose we have a map of fiber sequence $[A\rightarrow B\rightarrow C]\rightarrow [A^{'}\rightarrow B^{'}\rightarrow C^{'}]$ where all involved spaces are connected and the maps $A\rightarrow A^{'}$, $B\rightarrow B^{'}$ induce homology isomorphism, could we conclude that the map $C\rightarrow C^{'}$ induces a homology isomorphism.

$\endgroup$
  • 2
    $\begingroup$ Often the information on the homology of $C$ can be understood from the Serre spectral sequence. But this is nit what you want? $\endgroup$ – Thomas Rot Jul 12 '18 at 7:34
4
$\begingroup$

Here is one example to bear in mind, although it does not strictly answer your question. There is a Hopf fibration $$ \Omega S^3 \to \Omega S^2 \to S^1\to S^3\to S^2 = \mathbb{C}P^1 \to \mathbb{C}P^\infty = BS^1 $$ There is also a unit map $S^1\to\Omega S^2$, and we can use this together with the first three terms of the above sequence to get an equivalence $$ \Omega S^2\simeq S^1\times\Omega S^3 \simeq \Omega(\mathbb{C}P^\infty\times S^3). $$ Thus, the fibrations $$ \Omega S^2 \to * \to S^2 $$ $$ \Omega(\mathbb{C}P^\infty\times S^3) \to * \to \mathbb{C}P^\infty\times S^3 $$ have homotopy equivalent fibre and total space, but the bases have non-isomorphic homology. However, there is no map between these fibrations inducing an equivalence of fibres.

$\endgroup$
3
$\begingroup$

There are spectral sequences of this sort for certain fiber sequences. For example, if $A$ is a topological group and $B \rightarrow C$ is a principal $A$--bundle, one has the Rothenberg--Steenrod spectral sequence (and maybe some cases are due to Eilenberg and Moore). With a bit of luck, the $E^2$ term is an appropriate Tor group. There are also variants, e.g. with $A=\Omega X$. Try looking at original papers by Rothenberg, and also, maybe, original papers by Larry Smith.

$\endgroup$

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.