Example to show that the inverse image under a finite morphism is not t-exact with respect to the perverse t-structure According to Chapter 4 of Beilinson, Bernstein, and Deligne's "Faisceaux Pervers" (Asterisque 100, 1980) the inverse image $Rf^*$ with respect to a finite morphism $f$ is right t-exact with respect to the perverse t-structure. Is there an example that shows that it is not necessarily t-exact? Is there a class of finite morphisms other than the etale morphisms where it is t-exact?
 A: By Noether normalisation any affine variety admits a finite map $f$ to affine space. Hence it is enough to find an example of an affine variety for which the constant sheaf is not perverse. (In this case $f^* \mathbb{Q}_{\mathbb{A}^n}[n]$ will provide a counter-example.)
Now it is easy to find examples where the constant sheaf is not perverse. For example any affine cone $C_X$ over a smooth projective variety $X$ provides a counter-example as long as the cohomology of $C_X \setminus \{ 0 \}$ doesn't look like that of a sphere. So I guess the cone over an elliptic curve would do it.
Edit (...not quite:) see Daniel's comment below.
A: My idea was to consider $f \colon X \to Y$ of dimension $d$ such that $a_Y^\ast \mathbf Q[d] \cong IC_Y$ but $a_X^\ast \mathbf Q[d] \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q[d] = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact. 
(Incorrect example removed)
For a concrete example let $X$ be two copies of $\mathbb A^2$ glued at a point, let $f\colon X \to Y \cong \mathbb A^2$ be the quotient by the involution switching the two copies. If it were true that $IC_X \cong a_X^\ast \mathbf Q[2]$ then the same should be also true for $\mathbf P^2$ with two points identified, as perversity is a local property. But the cohomology of $\mathbf P^2$ with two points identified does not have Poincaré duality.
