Let us consider a homotopy fibre sequence of connected spaces $A\rightarrow B\rightarrow C$
and let $K$ be a fixed field. Assume that the homology $H\_{\ast}(A, K)$ is trivial and that $C$ is a nilpotent space (but not simply connected). 


Does $H\_{\ast}(B, K)\rightarrow H\_{\ast}(C, K)$ have to be an isomorphism ? 



Homologically trivial fibre
In your situation, the Serre spectral sequence looks as follows


\[H\_p(C,H\_q(A,K))\Rightarrow H\_{p+q}(B,K).\]


The left hand side is the $E^2$ term. There, $H\_q(A,K)$ carries an action of $\pi\_1(C)$ induced by the fibration, and $H\_p(C,H\_q(A,K))$ is the homology with local coefficients. Now, if $A$ has the homology of a point then then $H\_q(A,K)=0$ for $q>0$ and $H\_0(A,K)=K$ with the trivial action of $\pi\_1(C)$ (any self-map of a connected space induces the identity on $H\_0$). Therefore $E^2\_{p,q}=0$ for $q>0$ and $E^2\_{n,0}=H\_n(C,K)$ is the ordinary homology. Moreover, the spectral sequence ends in the second step and the edge morphism $H\_n(B,K)\twoheadrightarrow E^\infty\_{n,0}\subset E^2\_{n,0}=H\_n(C,K)$, which is induced by the map $B\rightarrow C$, is an isomorphism.


