# homology of a base space of a a fiber sequence

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.

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

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.

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.