$\DeclareMathOperator{\Nat}{Nat}$In a current project, I am trying to "commute" $!$ and $*$ functors that are both upper or both lower. (Sheaf-theoretic context: constructible étale sheaves.) The fact that they commute when we have one of each ultimately comes down to proper base change: that is, if we have maps $f \colon X \to Z$, $g \colon Y \to Z$, and their fiber product with projections $p, q \colon P \to X, Y$, then we have a natural isomorphism $$g^* f_! \cong q_! p^*.$$ From each direction of this isomorphism we can derive arrows between un-mixed compositions: $$\begin{align} g^* f_! \to q_! p^* \implies f_! \to g_* q_! p^* \implies f_! p_* \to g_* q_! \qquad &(1)\\ q_! p^* \to g^* f_! \implies p^* \to q^! g^* f_! \implies p^* f^! \to q^! g^* \qquad &(2) \end{align}$$ (the second implications are valid! Work it out: if $L$ and $R$ are left- and right-adjoints, then for functors $F$ and $G$ we have $\Nat(F, GL) \cong \Nat(FR, G)$.)

It is clear that (1) is an isomorphism when $f$, and hence $q$, are proper, and that (2) is an isomorphism when they are open immersions. It is also easy to prove directly that (1) is an isomorphism when $f$ is an open immersion, since we can check that both sides have the same restriction to the open subscheme and their restrictions to its closed complement are zero. So (1) is an isomorphism by Nagata's compactification theorem, which is the basis for defining the lower-$!$ functors anyway. Here's my question:

Is (2) an isomorphism when $f$ is a proper map?

My one trick, which was a direct computation, is no good here. I normally avoid like the plague dealing with the upper-$!$ functor directly, reducing it to something better by adjunction or duality, but the question is self-dual and I already exhausted my options with adjunction in deriving it.


1 Answer 1


(1) is not always an isomorphism when $f$ is an open immersion. (Take $X=Y$ equal to an open subscheme of $Z$, with the obvious maps.) Here is why : when you try to show that the restriction of $g_*q_!$ to the closed complement is $0$, you will want to use a base change isomorphism which is not true in general (it is true if $g$ is proper, but then (1) is trivially an isomorphism).

Anyway, so (1) is not always an isomorphism.

Neither is (2), and it doesn't matter whether $f$ is proper or not. Take $Z=\mathbb{A}_2$ (over some field, say), $X=$ one of the axes, $Z=$ the origin, with the obvious maps (so $P=Z$, and $f$ is a closed immersion, hence proper), and look what happens for the constant sheaf $\Lambda$ on $Z$ ($\Lambda$ is, for example, a finite ring of torsion prime to the characteristic of the base field). Then $f^!\Lambda=\Lambda(-1)[-2]$ by purity, so the rleft-hand side of (2) is $\Lambda(-1)[-2]$, while the right-hand side is $\Lambda$.

  • $\begingroup$ You're right; I glossed over the base-change business because it does work when checking restriction to the open part. As for (2): thanks, somehow I missed the possibility of the fiber product changing the codimension. (I think you mean $Y = \mathbb{A}^2$ and $P = Y$, though.) $\endgroup$
    – Ryan Reich
    Jan 16, 2012 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.