Let $X$ be a smooth algebraic variety over $\mathbb{C}$ and let $M^\bullet\in\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)$ be a complex of D-modules with holonomic cohomologies. We define the support of $M^\bullet$, as usual, as
$$\bigcup\_{i\in \mathbb{Z}}\operatorname{Supp}(\mathscr{H}^i(M^\bullet)),$$
where $\operatorname{Supp}(\mathscr{H}^i(M^\bullet))$ has its usual interpretation. (The set of points in which the stalk is non-zero.)


Given a morphism of algebraic varieties $f:X\to Y$, we can form functors $f\_+:\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_Y)$ and $f^!:\mathsf{D}^b\_\text{h}(\mathcal{D}\_Y)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_X)$. The former is the usual direct image of left D-modules and the latter is the functor which, up to a shift, coincides with the derived inverse image of O-modules.



> 
> I wonder if it's true that $\operatorname{Supp}(f\_+ M^\bullet)=f(\operatorname{Supp}(M^\bullet))$ and $\operatorname{Supp}(f^! P^\bullet)= f^{-1}(\operatorname{Supp}(P^\bullet))$.
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> 
> 


By a remark in section VI.4 in Borel's et al book on algebraic D-modules, I know that $\operatorname{Supp}(f^! P^\bullet)\subset f^{-1}(\operatorname{Supp}(P^\bullet))$ always holds. Is there a counter-example for the equality? And what about the direct image? Do we always have $\operatorname{Supp}(f\_+ M^\bullet)=f(\operatorname{Supp}(M^\bullet))$? If not, is one side always contained in the other?



About the support of a holonomic D-module
Regarding $f\_+$: neither inclusion holds in general, as the following two examples show.


* Let $j$ be the inclusion of $U:=\mathbb A^1 - \{0\}$ into $\mathbb
 A^1$. Then $j\_+(\mathcal O\_U)$ has support all of $\mathbb A^1$,
which contains $U=j(Supp(\mathcal O\_U))$ as a proper subset.
* On the other hand, consider the projection $p: U \to pt$, and let $M$
be the $D$-module $\mathcal O\_U z^{\lambda}$ corresponding to a rank 1 local  

system with non-trivial monodromy. Then $p\_+(M) =0$, so its support is empty (which is a proper subset of $p(Supp(M))=pt$).




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Here is a counter example for the equality in the $f^!$ case.


* Let $i: \{0\} \hookrightarrow \mathbb A^1$, and let $N=j\_+(\mathcal
 O\_U)$, using the notation above. Then $i^!(N)=0$ so has empty support
(whereas $i^{-1}(Supp(N))=\{0\}$).


