Let $\Delta^*$ be the unit disc minus 0 in the complex plane with parameter $\lambda$ and consider the fibration $X \to \Delta^*$ given by the (projectivization of) the equation $y^2 = x(x-1)(x-\lambda)$.
A generator of the fundamental group of $\Delta^*$ (say a counter clockwise loop around $|\lambda| = 1/2$ induces an action on the (singular) cohomology of the fiber over a point, say at $\lambda = 1/2$.
This cohomology is a free group of rank $2$ and I know for abstract reasons that the action has to be unipotent. How can I compute this explicitly (ie, pick a nice basis for the cohomology and show that the action is the standard unipotent)?