Given a smooth map of schemes $f:X\to Y$ of relative dimension $d$, then there is a natural isomorphism $f^!\simeq f^\*[d](2d)$ (in any context where the six operations are defined; see Cesinski-Deglise).


If $f$ is a smooth map of *Artin stacks* I imagine the same is true (there is a notion of the six operations for Artin stacks: <https://arxiv.org/abs/1211.5948>).



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> **Question:** Let $f:X\to Y$ be a **quasismooth** map of Artin stacks (say also flat to rule out things like closed embeddings). Is it still true that $f^!\simeq f^\*[d](2d)$?
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$f^!=f^*[d]$ for quasismooth maps?
In general, I think this should fail for most non-smooth local complete intersections.
For a specific example, let $X = \mathbb V(xy) \subset \mathbb A^2$ and $Y = \*$. Then the stalk at the dualizing complex of $X$ at the origin is $\mathbb Q\_l[1] \oplus \mathbb Q\_l[2]^{\oplus 2}$, contradicting the local constancy of $f^! \mathbb Q\_l = \omega\_X$.


