The answer might be well-known, but I couldn't find it antwhere. Sorry for my lack of knowledge. Let

$\require{AMScd} \begin{CD} Y' @>i'>> X\\ @Vf'VV @VVfV \\ Y @>i>>Z \end{CD} $ be a fiber diagram of finite-type schemes over $\mathbb{C}$, where $i'$ and $i$ are closed immersions. Let $\mathcal{F}$ be a coherent sheaf on $Y$. Is it true that $i'_*f'^*(\mathcal{F})= f^*i_*(\mathcal{F})$?

My argument is as follows: To prove the statement, we can assume $Z$ is affine, and then choosing an affine open $U\subset X$, the fiber diagram above becomes a fiber diagram of affine schemes (since $i,i'$ are closed immersions). Then the question becomes a question of modules over rings, in which case it is straightforward to check.

Could someone kindly let me know if the statement is true or not?

  • $\begingroup$ In my answer below, I have rotated you fibre diagram to the more standard one. Also, the question is not local on $Z$. It is local on $X$ which is where the sheaves land after base change. $\endgroup$ Jul 26 at 12:35

Your statement is true in more generality than you write. (It is possible that it is true for more general situations than below, but I did not try for maximum generality.)

Let $i:Y\to Z$ be a closed immersion and let $f:X\to Z$ an arbitrary morphism, with all $X,Y$ and $Z$ locally Noetherian.

The consider the base change diagram

$$\require{AMScd} \begin{CD} Y' @>f'>> Y\\ @Vi'VV @VViV \\ X @>f>>Z \end{CD} $$

(Note that $i':Y'\to X$ is a closed immersion.)

We have two functors from $\operatorname{Coh}(Y)$ to $\operatorname{Coh}(X)$ given as compositions $$\operatorname{Coh}(Y)\xrightarrow{f'^*}\operatorname{Coh}(Y)\xrightarrow{i'_*} \operatorname{Coh}(X)$$ and $$\operatorname{Coh}(Y)\xrightarrow{i_*}\operatorname{Coh}(Z)\xrightarrow{f^*} \operatorname{Coh}(X).$$ The reason is that pushing forward by a finite morphism preserves coherence and so does pulling back along arbitrary morphisms.

There is a natural transformation $f^*i_*\to i'_*f'^*$.

This arises by looking at the unit $id \to f'_*f'^*$ and precomposing with $i_*$ (and noting $i\circ f'=f\circ i'$) to get $i_* \to f_* i'_*f'^*$ which by adjunction gives $f^* i_* \to i'_*f'^*$. To get a feel for what this is, assume that all schemes are affine and you'll get an explicit description of this morphism. In the non-affine case, it is the globalisation of this morphism.

Now the question is if it this canonically defined natural transformation is a natural isomorphism.

The question is local on $X$ so we may assume $X=\operatorname{Spec}A$ and so $Y'=\operatorname{Spec}(A/I)$ (since it is a closed subscheme of an affine scheme) where if $\mathcal{I}_Y$ is the ideal of $Y$ in $Z$, then $I=f'^*\mathcal{I}_Y A$.

Thus let $\mathcal{F}\in \operatorname{Coh}(Y)$, then $i_*\mathcal{F}\in \operatorname{Coh}(Z)$ is annihilated by $\mathcal{I}_Y$ and so $f^*i_*\mathcal{F}$ is annihilated by $f^*\mathcal{I}_Y A$.

Now $f'^*\mathcal{F}$ is in $\operatorname{Coh}(\operatorname{Spec}(A/I))$ so too is annihilated by $f^*\mathcal{I}_YA$.

In this case the natural morphism $f^*i_*\mathcal{F} \to i'_*f'^*\mathcal{F}$ corresponds to killing the ideal $f^*\mathcal{I}_YA =I\subset A$ (use the technique described above to understand this morphism explicitly).

But in this case this morphism is always an equivalence because of the following lemma (a commutative algebra fact phrased algebro-geometrically).

Lemma: Let $j:\operatorname{Spec}(A/I)\to \operatorname{Spec} A$ be closed immersion of affine schemes. Then $j_*:\operatorname{Coh}(A/I)\to \operatorname{Coh}(A)$ identifies $\operatorname{Coh}(A/I)$ with the subcategory $\operatorname{Coh}(A)_I\subset \operatorname{Coh}(A)$ consisting of sheaves with support in $V(I)$. The inverse is given by the restriction of $j^*:\operatorname{Coh}(A)_I\to \operatorname{Coh}(A/I)$.

  • $\begingroup$ thanks for your detailed answer. One doubt in your proof: by the line "in this case the natural morphism ... corresponds to killing the ideal...", do you mean we are considering an A-module as (A/I) - module? Is that what the map is doing? $\endgroup$ Jul 26 at 18:48
  • 1
    $\begingroup$ You are sending an $A$-module $M$ to $M\otimes A/I=M/IM$. In case $IM=0$, this is just regarding $M$ as an $A/I$ module, in which case the map is an isomorphism of $A$-modules. $\endgroup$ Jul 26 at 18:51

Question: "Could someone kindly let me know if the statement is true or not?"

Answer: Consider the diagram

$\require{AMScd} \begin{CD} Spec(A') @>i'>> Spec(A)\\ @Vf'VV @VVfV \\ Spec(k') @>i>> Spec(k) \end{CD} $

with $A':=k'\otimes_k A$ and let $E'$ be any $k'$-module. It follows

$$i_*f^*(E'):=E'\otimes_k A$$


$$(i')_*(f')^*(E')=E'\otimes_{k'}(k'\otimes_k A) \cong E'\otimes_k A \cong f^*i_*(E').$$

Hence the claim is true when the schemes are affine with no condition on $i,i'$ being closed immersions.

In general (Hartshorne,Prop. III.9.3) proves that $i$ is a separated morphism of finite type of Noetherian schemes, $E$ a quasi coherent module and $f$ a flat morphism of Noetherian schemes there are for any $j\geq 0$ isomorphisms

$$f^*\operatorname{R}^ji_*(E) \cong \operatorname{R}^j(i')_*((f')^*E).$$

The proof uses Cech cohomology. A closed immersion is not flat, but it may be the methods of proof can be "adapted" to your situation.

Note: If you want to check such general statements you should try the affine case first. There is the following general theorem:

Theorem: Usually a "completely general statement" has an elementary proof.

  • $\begingroup$ This statement is always true affinely, and for arbitrary maps. This most probably wasn't the OP's question. $\endgroup$ Jul 26 at 11:59
  • $\begingroup$ @hm2020 : Yes, this much I figured out, just didn't give the details as the question would become too long. I was mainly worried about whether it is true for arbitrary maps of schemes, as S.S. mentioned. More precisely, if restriction to open sets was being done properly. $\endgroup$ Jul 26 at 12:19
  • $\begingroup$ @Hajime_Saito - as you see from my argument: The claim is true for arbitrary maps of affine schemes with no condition on $i,i'$ being closed immersions. $\endgroup$
    – hm2020
    Jul 26 at 12:27
  • $\begingroup$ Yes. But the question was for maps of finite type $\mathbf{C}$ schemes, not affine schemes. In general, the base change maps are not isomorphisms for non-affine schemes without conditions on the maps eg. flatness etc $\endgroup$ Jul 26 at 12:34
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    $\begingroup$ Indeed. However, it is proved for flat maps, not closed immersions. The closed immersion case is not entirely trivial. Moreover my argument above doesn't work for higher pushforwards. $\endgroup$ Jul 26 at 12:38

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