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All varieties are over the complex numbers. Given a smooth variety $X$, write $T^* X$ for its cotangent bundle. For a morphism of smooth varieties $f: X \to Y$ write $f_{\pi}: T^*Y \times_Y X \to T^*Y$ for the projection map, and let $f_d: T^*Y \times_Y X \to T^* X$ be the map dual to the derivative.

Now let $f\colon X\to Y$ be a smooth morphism. Let $M$ be a coherent $D_Y$-module. I am trying to understand the proof of:

$Ch(f^*M) \subseteq f_df_{\pi}^{-1}Ch(M)$,

(where $Ch$ means characteristic variety) as presented in J.P. Schneider's notes (see pages 34-35).

I need a bit more notation before I get to my exact question: we know $f^*M = O_X\otimes_{f^{-1}O_Y}f^{-1}M$ as an $O_X$-module. Endow $f^*M$ with the good filtration $F_i(f^*M) = O_X\otimes_{f^{-1}O_Y}f^{-1}F_iM$, where the $F_i$ on the right hand side is some good filtration of $M$. Then we have a surjection

$O_X\otimes_{f^{-1}O_Y}f^{-1}gr(M) \twoheadrightarrow gr(f^*M)$

of $gr(D_X)$-modules, where $gr$ denotes associated graded. Write $\tilde{gr}(-)$ for the sheaf on the cotangent bundle associated to $gr(-)$. Then the claim (last two lines of the proof on p35 of the aforementioned notes) is that:

since $f_d$ is finite we obtain a surjection

$f_{d*}f_{\pi}^*\tilde{gr}(M) \twoheadrightarrow \tilde{gr}(f^*M)$

I don't understand this claim. Any clarifying thoughts would be most appreciated.

Unless I made a mistake:

$f_{d*}f_{\pi}^*\tilde{gr}(M) = f_{d*}(O_{T^*Y\times_Y X}\otimes_{f_{\pi}^{-1}\pi_Y^{-1}gr D_Y}f_{\pi}^{-1}\pi_Y^{-1}gr(M))$,

where $\pi_Y\colon T^*Y\to Y$. What is required is that this equal

$O_{T^*X}\otimes_{\pi_X^{-1}gr D_X}(O_X\otimes_{f^{-1}O_Y}f^{-1}gr(M))$

I don't see how to proceed. Presumably I am making this way too complicated and there is a very simple explanation?

Also, it is mentioned at the end of the proof that the estimate of the characteristic variety is in fact an equality. Does anyone know of a reference for this?

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