Is inverse image along finite group quotient $t$-exact for the perverse $t$-structure?

Question

Let $q: \mathbb{C}\to \mathbb{C}$ be the quotient by $\mathbb{Z}/n\mathbb{Z},$ i.e. the map taking $z\mapsto z^n$. In the accepter answer to Operations on perverse sheaves on disk the inverse image of a perverse sheaf along this map is described, and it turns out that the functor $^pq^*: \mathrm{Perv}(\mathbb{C})\to\mathrm{Perv}(\mathbb{C})$ is exact. Indeed, if we write a perverse sheaf as a pair of vector spaces $\Psi,\Phi$, this inverse image functor simply does nothing to the spaces itself and the morphisms of such pairs, and it only changes the canonical maps between $\Phi$ and $\Psi$.

It is also true that $D^b(\mathrm{Perv}(\mathbb{C}))\simeq D^b_{\mathrm{cons}}(\mathrm{Sh}(\mathbb{C}))$, where on the right we have the derived category of sheaves on $\mathbb{C}$ with constructible cohomology, and on both sides "constructible" means constructible with respect to the two-stratum stratification of $\mathbb{C}$. So in fact, the functor $q^*$ should be the left derived functor of $^pq^*$, which we have established to be exact already. All in all, the functor $q^*: D^b_{\mathrm{cons}}(\mathrm{Sh}(\mathbb{C}))\to D^b_{\mathrm{cons}}(\mathrm{Sh}(\mathbb{C}))$ is $t$-exact for the perverse $t$-structure.

A possibly more natural way to see it is this: for a perverse sheaf $F$, the only reason why $q^*F$ could fail to be a perverse sheaf is if $H^{-1}(q^*F)$ somehow had non-zero sections supported at 0, and this clearly doesn't happen.

Is the situation the same more generally? It seems like something that is easy to prove once you figure out the correct statement. Say, we have a geometric quotient morphism $q: X\to X/G$, where $X$ is a smooth complex algebraic variety, and the group $G$ is either finite or acts on $X$ with only finite stabilizers, is the inverse image functor $q^*$ $t$-exact? Or maybe, even more generally, it is enough for $q: X\to Y$ to be an open morphism with smooth fibers of constant dimension? Another possibly relevant property in the one-dimensional example above is that the stratification is invariant under the group action and the inclusion morphisms of the strata are affine.

Answer 1

The inverse image functor is not actually exact in this case.

Let $Q$ be the quadratic cone, i.e. the quotient of $X=\mathbb{C}^2$ by the antipodal action of $G=\mathbb{Z}/2\mathbb{Z}$. Let $U$ be the complement of the origin in $Q$, $j: U\to Q$, and let $L$ be the local system (placed in degree -2) on $U$ corresponding to the $1$-dimensional representation $V$ of $G$ with the action by the antipodal map.

I claim that $j_!L$ is a perverse sheaf on $X$. Indeed, we know that intermediate extension functor $j_{!*}$ takes perverse sheaves to perverse sheaves, and by Deligne's formula $j_{!*}L\cong \tau^{\leqslant -1} j_* L$. It turns out that $j_!L\cong j_{!*}L$, because $\tau^{\leqslant -1} j_* L$ has zero stalk at the origin. $Q$ is a conic set, so all small neighborhoods of zero are homotopy-equivalent to $U$ itself and we may compute the stalk of $j_*$ by computing the cohomology $H^*(U,L)$. We are interested in $H^k(U,L)$ for $k=-2$ and $k=-1$, and at least for these two values of $k$, this cohomology group coincides with the group cohomology $H^{k+2}(G,V)$, and both $H^{0}(G,V)$ and $H^1(G,V)$ vanish.

However, for the quotient map $q: X\to Q$ the inverse image $q^*L$ is the extension by zero of the constant local system on $\mathbb{C}^2\setminus\{0\}$, and this sheaf is not perverse, since the complement of $0$ is not affine.