Let $f:X\to S$ be a morphism between algebraic varieties which are smooth over a field of characteristic zero. We define the (derived) direct image functor $f\_+:\mathsf{D}^b(\mathcal{D}\_X)\to \mathsf{D}^b(\mathcal{D}\_S)$ as in basically every textbook. (That's what [HTT] denotes by $\int\_f$ on page 40, for example.) This functor also restricts to $f\_+:\mathsf{D}^b\_\text{qc}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{qc}(\mathcal{D}\_S)$ and $f\_+:\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_S)$.


It is true that if $f$ is a closed embedding and $M\in \mathsf{D}^b(\mathcal{D}\_X)$ is concentrated in degree 0, then $\mathscr{H}^i(f\_+ M)=0$ for all $i\neq 0$. [HTT, Proposition 1.5.24.]


Given that this is very close to Artin vanishing on the context of perverse sheaves, I would expect the same result as above to be true more generally for affine + quasi-finite morphisms (at least when $M$ is holonomic).


**More precisely, is it true that the functor $f\_+:\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_S)$ is right t-exact (with respect to the usual t-structures) when $f$ is affine and t-exact when $f$ is also quasi-finite?**


Reference: [HTT] is R. Hotta, K. Takeuchi, T. Tanisaki - D-modules, Perverse Sheaves, and Representation Theory.



Artin vanishing for D-modules (i.e., when is $f_+$ t-exact?)
The functor $f\_+$ by definition is the composite of a right t-exact functor (tensor product with the transfer bimodule) and a left t-exact functor (the derived pushforward of sheaves). In the case $f$ is affine, the latter functor is t-exact, so $f\_+$ is indeed right-exact.


On the other hand, D-module pushforward under a quasi-finite morphism is not necessarily t-exact. For example, consider the (non-affine) open embedding $f:\mathbb A^2 - \{(0,0)\} \hookrightarrow \mathbb A^2$. I claim the pushforward $f\_+(\mathcal O)$ is not concentrated in degree 0.


EDIT: I didn't catch the ``also'' in the question. It is true that when $f$ is quasi-finite, then $f\_+$ is left t-exact, though I don't have an entirely straightforward argument for this.


By Zariski's main theorem, it is enough to check the cases:
(i) $f$ is an open embedding, and (ii) $f$ is finite. In case (i), $f\_+$ is easily seen to be left t-exact (as the transfer bimodule is a localization, or alternatively, as the left adjoint to $f\_+$ is t-exact). In case (ii), the result then follows from the main result of [Raskin](https://arxiv.org/abs/1611.04940), noting that when $f$ is finite $f\_+=f\_!$ (so in particular, the pro-category is unnecessary). For holonomic modules, the finite case can be dealt with by using that $f\_+$ is right exact (as $f$ is affine) and $f\_+ = \mathbb D f\_+ \mathbb D$. (I think the general proof follows similar lines, but requires a bit more care.)


