Let $E, X$ be path-connected and suppose I have a fibration $p\colon E\to X$ which admits a section $s$. 
For a generalised homology theory $\mathcal{E}\_\ast$, there is a splitting $\mathcal{E}\_\ast (E)\cong \mathcal{E}\_\ast(X)\oplus R\_\ast$ induced by $p$ and $s$.


Choosing a basepoint $x\colon \ast \to X$, we have a homotopy fibre sequence
$$
\begin{array}{ccc} F & \to & E \\ \downarrow & & \downarrow \\ \ast & \to & X \end{array}
$$
which implies a map $\psi\colon \mathcal{E}\_\ast(F) \to \mathcal{E}\_\ast(X)\oplus R\_\ast$.
Commutativity of the diagram implies that projecting $\psi$ to the first factor results in the zero map.


**Question:** is the map $\mathcal{E}\_\ast (F)\to R\_\ast$ an isomorphism?



Generalised homology of a split fibration
This is not true. Consider the following (split) homotopy fiber sequence
$$S^1\to S^1\times S^1\to S^1$$
Then, by a standard argument, we have 
$$\Sigma(S^1\times S^1)=S^2\vee S^3\vee S^2$$
so for every spectrum $E$
$$E\_\*(S^1\times S^1)=E\_\*(S^1)\oplus E\_\*(S^2)\oplus E\_\*(S^1)$$
In particular the sequence
$$E\_\*(S^1)\to E\_\*(S^1\times S^1)\to E\_\*(S^1)$$
is not exact (the first map is the inclusion of the first summand and the second one is the projection onto the third summand).


