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Let $X\rightarrow Y$ be a morphism (may not be smooth) of varieties such that the fibres are vector spaces. Are the $l$-adic cohomologies of $X$ and $Y$ equal?

If not, under what condition (other than smoothness of the map) is sufficient to ensure that the cohomologies are equal?

What did I mean to ask?: Let $f:X\rightarrow Y$ be a continuous map of manifolds such that the fibres are vector spaces and there is a "zero section" $s:Y\rightarrow X$. Then we can actually write a homotopy equivalence between $X$ and $Y$. Therefore it follows that the cohomologies of $X$ and $Y$ are same.

I was not sure how to pose this question in the $l$-adic cohomology setup. If you assume the map is smooth then etale locally it is actually trivial, which is an easy case. Therefore i asked is there a slightly weaker condition which ensure that the cohomologies are same?

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    $\begingroup$ "to be the same" is vague and can have several meanings. There's probably a more adequate way to formulate what you want to ask. $\endgroup$ – YCor Feb 22 at 19:14
  • $\begingroup$ @YCor I have no idea what do you mean? $\endgroup$ – user100841 Feb 22 at 19:24
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    $\begingroup$ My concern also applies to "be equal": even isomorphic varieties do not have "equal" cohomologies. Why not use the words "isomorphism/isomorphic"? Then again, "$H^*(X)$ and $H^*(Y)$ are isomorphic" is not equivalent to "$f$ induces an isomorphism $H^*(Y)\to H^*(X)$"; which (if either) of these do you mean? $\endgroup$ – YCor Feb 22 at 19:30
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    $\begingroup$ That is definitely not true. Consider the case that $Y$ is a nodal plane cubic and $X$ is the open complement in the normalization $Y^\text{nor}$ of one of the two preimages of the node. These two schemes have different etale fundamental groups, different etale $H^1$, etc. Yet every fiber is a single point (which is a vector space of dimension $0$). $\endgroup$ – Jason Starr Feb 22 at 20:23
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    $\begingroup$ The question does have a positive answer when the target $Y$ is normal by Koll'ar's generalization of Hironaka's Flatness Lemma. I added this to my answer below. $\endgroup$ – Jason Starr Feb 27 at 14:27
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The answer below is slightly reorganized to incorporate the edits. I thank @PiotrAchinger and @S.D. for their comments correcting and clarifying this answer.

Definition 1. An affine space morphism is a separated, smooth morphism whose geometric fibers are affine spaces.

Definition 2. A morphism of schemes is a étale cohomological equivalence if the pullback maps on étale cohomology are isomorphisms. The morphism is a universal étale cohomological equivalence if every base change of the morphism by a separated, finitely presented morphism to the target is an étale cohomological equivalence. A commutative diagram of morphisms of schemes, $$ \begin{array}{ccc} X' & \xrightarrow{f'} & Y' \\ \downarrow & & \downarrow \\ X & \xrightarrow{f} & Y \end{array}, $$ is an étale cohomological equivalence of $f$ and $f'$, resp. a universal étale cohomological equivalence of $f$ and $f'$, if the vertical arrows are étale cohomological equivalences, resp. universal étale cohomological equivalences.

Let $k$ be an algebraically closed field. Let $n$ be an integer that is prime to $\text{char}(k)$. Let $Y$ be a $k$-scheme. Let $f:X\to Y$ be an affine space morphism. Let $\mathcal{E}$ be a locally constant $\mathbb{Z}/n\mathbb{Z}$-module on the (big) étale site of $Y$. For every integer $q\geq 0$, denote the pullback map associated to $f$ by $$H^q(f^*,\mathcal{E}):H^q(Y,\mathcal{E})\to H^q(X,f^*\mathcal{E}).$$

Question 3. Under what additional hypotheses is every affine space morphism an étale cohomological equivalence? Does this still hold if we consider non-smooth morphisms $f$ whose geometric fibers are affine spaces?

The reference that I use below for étale cohomology is the following textbook by J. S. Milne.

J. S. Milne
Lectures on etale cohomology
https://www.jmilne.org/math/CourseNotes/LEC210.pdf

Pullback maps are isomorphisms for affine space morphisms to separated, finite type $k$-schemes. Although the application is to affine space morphisms, it is convenient to formulate the next results in terms of étale cohomological equivalence.

Lemma 4. A morphism $f$ is an étale cohomological equivalence if it is étale cohomologically equivalent to a morphism $f'$ that is an étale cohomological equivalence. Similarly, a morphism $f$ is a universal étale cohomological equivalence if it is universally étale cohomologically equivalent to a morphism $f'$ that is a universal étale cohomological equivalence.

Proof. This follows from the fact that pullback maps on étale cohomology are functorial. QED

Proposition 5. For a class of smooth morphisms between separated, finitely presented $k$-schemes that is stable for separated, finitely presented base change, to check that all such morphisms are universal étale cohomological equivalences, it suffices to check that the morphisms are étale equivalences whenever the target is $k$-smooth.

Proof. Since the closed immersion of the induced reduced scheme is an affine morphism that is universal homeomorphism, and since such morphisms are étale cohomological equivalences, by the previous lemma it suffices to consider the case of a reduced target $Y$ of a morphism $f:X \to Y$.

As mentioned in the comments below, using de Jong's Alterations of Singularities Theorem, for every separated, reduced, finitely presented $k$-scheme $Y$, there exists an proper hypercovering $Y_\bullet\to Y$ such that every term of $Y_\bullet$ is $k$-smooth. This is Theorem 3.1 of the following.

MR1748625 (2001f:14027)
Geisser, Thomas
Applications of de Jong's theorem on alterations.
Resolution of singularities (Obergurgl, 1997), 299–314
Progr. Math., 181, Birkhäuser, Basel, 2000.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.446.2452&rep=rep1&type=pdf

By Deligne, a proper hypercovering is of étale cohomological descent for constructible $\mathbb{Z}/n\mathbb{Z}$-modules with $n$ prime to the characteristic of $k$.

Since $f$ is smooth, by the Smooth Base Change Theorem, also the pullback $X_\bullet := X\times_Y Y_\bullet$ is a proper hypercovering of $X$, and thus of étale cohomological descent. Thus, we can compute the pullback map from the étale cohomology of $Y$ to the étale cohomology of $X$ in terms of the simplicial pullback maps $H^*(Y_\bullet,\mathcal{E}) \to H^*(X_\bullet, f^*\mathcal{E})$. For every $Y_i$ in the hypercovering $Y_\bullet$, the morphism $X_i\to Y_i$ is in the given class with smooth target. Thus, by hypothesis, the pullback map is an isomorphism. QED

Proposition 6. A smooth morphism to a separated, finitely presented $k$-scheme $Y$ is a universal étale cohomological equivalence if and only if for every point $y$ of $Y$ the fiber over $\text{Spec}\ \kappa(y)^{\text{sep}}$ is an étale cohomological equivalence.

Proof. Universal homeomorphisms are universal étale cohomological equivalences. Thus, the fiber over $\text{Spec}\ \kappa(y)^{\text{sep}}$ is an étale cohomological equivalence if and only if the fiber over $\text{Spec}\ \kappa(y)^{\text{alg}}$ is an étale cohomological equivalence. And then the base change by any further field extension also is an étale cohomological equivalence. Thus, both the hypotheses and conclusion are stable for separated, finitely presented morphisms. Therefore, to prove that such morphisms are universal étale cohomological equivalences, it suffices to prove that such morphisms are étale cohomological equivalences. Moreover, by Proposition 5, it suffices to check this when $Y$ is smooth.

Let $f:X\to Y$ be such a morphism. Consider the adjunction map, $$\alpha_{f,\mathcal{E}}:\mathcal{E} \to f_*(f^*\mathcal{E}).$$ Since $f$ is faithfully flat, the adjunction map is injective. Moreover, by the Finiteness Theorem, Chapter 19 of Milne's book, every $R^qf_*(f^*\mathcal{E})$ is a constructible sheaf. Thus, also $\text{Coker}(\alpha_{f,\mathcal{E}})$ is a constructible sheaf. By the Smooth Base Change Theorem, at the Henselized generic point $\text{Spec}\ \kappa$ of $Y$, $\kappa:= k(Y)^{\text{sep}}$, the stalk of $\text{Coker}(\alpha_{f,\mathcal{E}})$, resp. the stalk of $R^qf_*(f^*\mathcal{E})$ for $q>0$, equals the cokernel of the pullback map $H^0(f^*_{\kappa},\mathcal{E}_{\kappa})$, resp. equals $H^q(f^*_{\kappa},\mathcal{E}_{\kappa})$. By hypothesis, these all vanish. Thus, the generic point of $Y$ is not contained in the support of $\text{Coker}(\alpha_{f,\mathcal{E}})$ nor in the support of $R^qf_*(f^*\mathcal{E})$ for $q>0$.

In particular, if $Y$ is $0$-dimensional, this already establishes that the base change maps are isomorphisms. This is the base case of an induction proof for the dimension of $Y$. Thus, by way of induction, assume that $Y$ is smooth and integral of dimension $>0$, and assume that the result is proved for all separated, finitely presented $k$-schemes of strictly smaller dimension.

By the previous paragraphs, the stalk at the generic point of $Y$ vanishes for the constructible $\mathbb{Z}/n\mathbb{Z}$-module $\text{Coker}(\alpha_{f,\mathcal{E}})$ and $R^qf_*(f^*\mathcal{E})$ for $q>0$. Thus, the union of the supports of these finitely many constructible $\mathbb{Z}/n\mathbb{Z}$-modules is a locally closed subset of $Y$ that does not contain the generic point. If this union is the empty set, then we are done. Thus, by way of contradiction, assume that the union is nonempty.

Denote by $Z$ an irreducible component of the closure of this support. In particular, at least one of these constructible $\mathbb{Z}/n\mathbb{Z}$-modules has a nonzero stalk at the generic point $\eta_Z$ of $Z$. Thus, the situation is preserved by restricting to a dense Zariski open subscheme of $Y$ that contains $\eta_Z$.

Since $k$ is algebraically closed, and since $Z$ is integral, there exists a dense open subscheme of $Z$ that is $k$-smooth. Thus, there exists a dense Zariski open $Y^o$ of $Y$ such that $Z^o:= Z\cap Y^o$ is an integral smooth $k$-scheme. Also, up to shrinking $Y^o$ further, assume that $Z$ is the only irreducible component of the support of these sheaves that intersects $Y^o$. Therefore, on the open complement $U:=Y^o\setminus Z^o$, all of these sheaves are zero.

Denote by $c$ the codimension of $Z^o$ in $Y^o$. Denote $X\times_Y Y^o$ by $X^o$. Denote $X\times_Y Z^o$ by $W^o$. Finally, denote $X\times_Y U$ by $V$. Since $f$ is smooth, by the Going-Up and Going-Down Theorems of Cohen-Seidenberg, also the codimension of $W^o$ in $X^o$ equals $c$.

By construction, the pullback maps from $U$ to $V$ are isomorphisms, since all of the constructible $\mathbb{Z}/n\mathbb{Z}$-modules are zero on $U$ (and using the Leray spectral sequence). By the induction hypothesis, the pullback maps on cohomology from $Z^o$ to $W^o$ are also isomorphisms. Now consider the commutative diagram of relative cohomology long exact sequences, $$ \begin{array}{ccccccc} \to & H^r_{Z^o}(Y^o,\mathcal{E}) & \to & H^r(Y^o,\mathcal{E}) & \to & H^r(U,\mathcal{E}) & \to \\ & \downarrow & & \downarrow & & \downarrow & \\ \to & H^r_{W^o}(X^o,f^*\mathcal{E}) & \to & H^r(X^o,f^*\mathcal{E}) & \to & H^r(V,f^*\mathcal{E}) & \to \end{array}. $$ By hypothesis, the last vertical homomorphism is an isomorphism. By the Purity Theorem and Gysin Homomorphisms, Chapter 16 of Milne's book, there are natural isomorphisms, $$ H^{r-2c}(Z^o;\mathcal{E}(-c)) \xrightarrow{\cong} H^r_{Z^o}(Y^o,\mathcal{E}), \ \ H^{r-2c}(W^o;f^*\mathcal{E}(-c)) \xrightarrow{\cong} H^r_{W^o}(X^o,f^*\mathcal{E}). $$ Thus, by the induction hypothesis, also the first vertical homomorphism is an isomorphism. Therefore, by the Five Lemma, also the middle vertical homomorphism is an isomorphism.

This argument also holds after quasi-compact, étale base change of $Y^o$. Thus, the stalk at $\eta_Z$ of each of these constructible $\mathbb{Z}/n\mathbb{Z}$-modules vanishes, contrary to the hypothesis on $Z$. This proves vanishing of the cokernel of $\alpha_{f,\mathcal{E}}$ and every sheaf $R^qf_*(f^*\mathcal{E})$ for $q>0$.

By induction on the dimension of $Y$, the proposition holds for every smooth $k$-scheme $Y$. QED

Corollary 7. For every affine space morphism to a separated, finitely presented $k$-scheme $Y$, the natural injection $\alpha_{f,\mathcal{E}}:\mathcal{E} \to f_*f^*\mathcal{E}$ is an isomorphism, and $R^qf_*(f^*\mathcal{E})$ vanishes for every $q>0$. In particular, every pullback map $H^q(f^*,\mathcal{E})$ is an isomorphism.

Proof. By the proposition, it suffices to prove this when $Y$ is $\text{Spec}\ \kappa$ for a field $\kappa$. The higher étalae cohomology of affine space vanishes for locally constant $\mathbb{Z}/n\mathbb{Z}$-modules with $n$ prime to the characteristic. QED

Remark 8. As @S.D. notes, when the field $k$ equals $\mathbb{C}$, the Comparison Theorem for étale cohomology and singular cohomology of the underlying complex analytic space reduces this corollary to a result of differential topology: an affine space morphism is locally differentiably a fiber bundle with contractible fibers.

Remark 9. Since seminormalization of separated, finitely presented $k$-schemes is a universal homeomorphism, Lemma 4 also applies for seminormalization. Precisely, for a class of morphisms (not necessarily smooth) that is stable for separated, finitely presented base change, to prove that all morphisms in the class are universal étale cohomological equivalences, it is equivalent to prove that every such morphism whose target is reduced and seminormal is an étale cohomological equivalence.

Counterexample when the fiber dimension jumps. Consider the case when $Y$ is an integral, smooth $k$-scheme of dimension $d\geq 2$. Let $Z=\{z\}$ be a singleton set of a closed point in $\mathbb{A}^n_k$ considered as a reduced, closed subscheme of $Y$. Let $\overline{X}\to Y$ be the blowing up of $Y$ along $Z$. The exceptional divisor is isomorphic to $\mathbb{P}^{n-1}$. Let $X\subset \overline{X}$ be the open complement of a hyperplane in this $\mathbb{P}^{n-1}$. In this case, the restricted morphism, $$ f:X\to Y $$ is a separated, finitely presented morphism between smooth $k$-schemes such that every geometric fiber is an affine space.

Proposition 10. For the morphism $f$ above, not all of the pullback maps are isomorphisms. More precisely, for a $\mathbb{Z}/n\mathbb{Z}$-module $\mathcal{E}$ that is étale locally free of rank $1$, the pullback maps $H^2(f^*,\mathcal{E}(1))$ and $H^{2d-1}(f^*,\mathcal{E}(d))$ are injective with cokernel $\mathcal{E}_z$, and the pullback maps in all other degrees are isomorphisms.

Proof. This follows by analyzing the pullback commutative diagram of the Gysin sequence from the proof of Proposition 5. QED

Counterexample when the target is not normal. Let $\mathbb{P}^1_k$ denote $\text{Proj}\ k[s,t]$. Let $\mathbb{P}^2_k$ denote $\text{Proj}\ k[x,y,z]$. Denote by $\nu:\mathbb{P}^1_k\to \mathbb{P}^2_k$ the unique morphism with $\nu^*\mathcal{O}_{\mathbb{P}^2_k}(1)\cong \mathcal{O}_{\mathbb{P}^1}(3)$ defined by the following basepoint free linear system, $$\nu:\mathbb{P}^1_k \to \mathbb{P}^2_k, \ \ \nu^*x = s^2t, \ \nu^*y = s^3 + t^3, \ \nu^*z = st^2.$$ The image of this morphism is a nodal plane cubic, $$Y = \text{Zero}(z^3+xyz+x^3).$$ Define $X$ to be the distinguished open affine $D_+(s)\subset \mathbb{P}^1_k$. Denote by $f$ the restriction to $X$ of $\nu$, $$f:X\to Y.$$ This is an unramified, quasi-finite morphism that is bijective on points. Thus, every scheme-theoretic fiber is a $0$-dimensional affine space.

Proposition 11. The pullback map $H^1(f^*,\mu_n)$ is surjective with rank $1$-kernel $\mu_n(k)$, and the pullback map $H^2(f^*,\mu_n)$ is surjective with rank $1$-kernel $\mathbb{Z}/n\mathbb{Z}$.

Proof. This follows by comparing the pullback maps of long exact sequences arising from the Kummer sequence. Since $X$ is isomorphic to $\mathbb{A}^1$, all of its higher étale cohomology groups vanish. By the Kummer sequence, $H^1(Y,\mu_n)$ equals $\mu_n(k)$ and $H^2(Y,\mu_n)$ equals $\mathbb{Z}/n\mathbb{Z}$. QED

Question 12. Let $f:X\to Y$ be a separated, finitely presented morphism with affine space fibers of constant dimension between separated, finitely presented $k$-schemes that are normal. Is $f$ an étale cohomological equivalence? Is $f$ flat (if so, then $f$ is smooth, hence an affine space morphism)?

Kollár's Flatness Criteria for normal target. Question 12 has a positive answer by Kollár's Flatness Criteria.

MR1339664 (96j:14010)
Kollár, János
Flatness criteria.
J. Algebra 175 (1995), no. 2, 715–727.
https://ac.els-cdn.com/S0021869385712094/1-s2.0-S0021869385712094-main.pdf?_tid=c9850a97-a4a0-4482-84e6-2e5b863bce01&acdnat=1551275145_19dbfad83504d3662477ba2257220fb8

Proposition 13. Every separated, finitely presented morphism whose geometric fibers are affine spaces of constant dimension $n$, whose domain is a finitely presented, integral $k$-scheme, and whose target is a finitely presented, integral, normal $k$-scheme is an affine space morphism.

Proof. Such a morphism satisfies the hypotheses of Corollary 11 of Kollár's article. QED

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    $\begingroup$ Thanks for the answer! $\endgroup$ – Bombyx mori Feb 23 at 16:28
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    $\begingroup$ Why is the induced map on cohomology an isomorphism if $f$ is smooth? I am okay if $f$ is an etale-locally trivial bundle, but that doesn't always hold even in relative dimension one... $\endgroup$ – Piotr Achinger Feb 24 at 19:06
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    $\begingroup$ @JasonStarr Thank you. Regarding the first argument: is ${\rm Aut}(\mathbb{A}^1, 0) = \mathbb{G}_m$ over any base? There is an example (possibly due to Russell), of an inseparable form of $\mathbb{A}^1$: $x = y^p + tx^p$ over $\mathbf{F}_p(t)$. Over $\mathbf{A}^1_t$, it gives a smooth map with geom. fibers $\mathbf{A}^1$ but which is not etale-locally trivial. This is what made me worried in the first place. $\endgroup$ – Piotr Achinger Feb 25 at 4:43
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    $\begingroup$ P.S. I think the automorphism scheme of $\mathbf{G}_a$ is not even representable... (Feel free to continue this thread over email if you prefer, and thanks again for your answers.) $\endgroup$ – Piotr Achinger Feb 25 at 4:47
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    $\begingroup$ @S.D. the result in the reference doesn't apply "out of the box" to get what we want, as it only gives local acyclicity (vanishing of vanishing cycles), but since the map is not proper we do not know if the $R^q f_* \mathbf{Q}_\ell$ commute with base change... $\endgroup$ – Piotr Achinger Feb 25 at 4:50
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Jason Starr has already given a very nice answer, and this should just be seen as a small addition to what's already been said. I'll freely use the vocabulary from Jason's answer.

Proposition. Let $f:X \to Y$ be a smooth morphism of finite-dimensional Noetherian schemes. Suppose that for every geometric point $\overline{y} \to Y$, the fiber $f_\overline{y}$ is an etale cohomological equivalence. Then $f$ is a universal etale cohomological equivalence.

Proof. Fix $n$ invertible on $Y$. Given any $\mathcal{E} \in D^b_c(Y)=D^b_c(Y,\mathbf{Z}/n\mathbf{Z})$, write $K(\mathcal{E}) = K_Y(\mathcal{E}) = Cone(\mathcal{E} \to Rf_{\ast} f^\ast \mathcal{E})$. It's easy to check that $K(-)$ is a triangulated functor from $D^b_c(Y)$ to itself. If $g: Y' \to Y$ is any morphism, we define $K_{Y'}(-)$ analogously; note that there is a canonical base change map $g^\ast K_Y(\mathcal{E}) \to K_{Y'}(g^{\ast} \mathcal{E})$.

We now argue by induction on $\mathrm{dim}Y$ (when $\mathrm{dim}Y=0$, the result is easy). For any fixed $\mathcal{E}$, Deligne's generic base change theorem guarantees the existence of a dense open $U \subset Y$ such that the aforementioned base change map is an isomorphism for any map $Y' \to Y$ which factors over the inclusion $U \subset Y$. Applying this for $Y'$ running over all geometric points of $U$, and using our hypothesis on the geometric fibers of $f$, we deduce that $K(\mathcal{E})|U=0$. Write $j:U \to Y$ for the evident open immersion, and let $i:Z \to Y$ be the closed complement. Looking at the usual distinguished triangle $i_* Ri^! K(\mathcal{E}) \to K(\mathcal{E}) \to Rj_* j^*K(\mathcal{E})=0$, we deduce that $K(\mathcal{E}) \cong i_* Ri^!K(\mathcal{E})$. Moreover, $i_* Ri^!K(\mathcal{E}) \cong i_* K_Z (Ri^! \mathcal{E})$, by a combination of smooth and proper base change. By hard results of Deligne, Grothendieck, and Gabber, $Ri^! \mathcal{E}$ is still bounded and constructible, so (finally) we can use induction on the dimension of the target scheme to conclude that $K_Z(Ri^! \mathcal{E})=0$, as desired.

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  • $\begingroup$ Where are these "hard results of..." written down? Could you kindly provide a reference? I am just never able to find anything attributed to Gabber. $\endgroup$ – user138556 Apr 19 at 18:01

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