Let $X \subset \mathbb{A}^n,~ Y \subset \mathbb{A}^m$ be affine varieties. Consider a regular map $f: X \to Y$. If $f$ is bijective, can we conclude that $f$ is an open mapping w.r.t the Zariski topology (so that $f$ is a homeomorphism)?
$\begingroup$
$\endgroup$
5
-
2$\begingroup$ What about the normalization map of the affine cuspidal cubic $y^2=x^3$? $\endgroup$– Francesco PolizziCommented Sep 29 at 11:52
-
2$\begingroup$ Welcome new contributor. No, that is not true. Let $Y$ be the product of a nodal affine curve with an affine curve $D$. Let $X$ be $C’$ times $D$, where $C’$ is the open complement in the normalization of $C$ of one of the two preimages of the node. The morphism is the evident one. An irreducible curve in $X$ that “should” intersect along the missing fiber $D$ will have non-closed image in $Y$. $\endgroup$– Jason StarrCommented Sep 29 at 11:56
-
2$\begingroup$ @FrancescoPolizzi. Actually that map is a (universal) homeomorphism. The open complement of one preimage of the node in a normalization of a nodal curve is the “standard” example of a bijective, unramified map that is not a universal homeomorphism (also it is not a disjoint union of locally closed immersions until we form an etale cover, another standard question). That is where the example above comes from. $\endgroup$– Jason StarrCommented Sep 29 at 12:16
-
2$\begingroup$ @FrancescoPolizzi It is actually very easy to see that any bijective map between (irreducible) curves is a homeomorphism, since any such curve is equipped with cofinite topology. So any example must be at least 2-dimensional. $\endgroup$– WojowuCommented Sep 29 at 18:19
-
1$\begingroup$ @Wojowu: yes, of course you are right. $\endgroup$– Francesco PolizziCommented Sep 29 at 19:54
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Welcome new contributor. I am just posting my comment as one answer. Let $k$ be any nonzero, unital, commutative ring (e.g., a field). Let $C'$ be $\text{Spec}\ k[u,v]/ \langle (u+v)^3 -uv\rangle$. The normalization map from $C:= \text{Spec}\ k[t]$ to $C'$ pulls back $u$ to $t^2(1-t)$ and pulls back $v$ to $t(1-t)^2$.
Let $C^o$ be the distinguished open affine $\text{Spec}\ k[t,t^{-1}]$ in $C$. Let $D$ be $\text{Spec}\ k[s]$. Let $Y$ be the product $C'\times_{\text{Spec}\ k} D = \text{Spec}\ k[u,v,s]/\langle (u+v)^3-uv\rangle$, let $X$ be the product $C^o\times_{\text{Spec}\ k} D= \text{Spec}\ k[t,t^{-1},s]$, and let $f$ be the product of the normalization morphism with the identity morphism of $D$ that pulls back $s$ to $s$, that pulls back $u$ to $t^2(1-t)$ and that pulls back $v$ to $t(1-t)^2$.
The morphism $f$ from $X$ to $Y$ is an unramified, universal bijection. Yet it is not a homeomorphism. The Zariski closed subset $Z$ of $X$ associated to $t-s$ maps to a nonclosed subset of $Y$ whose Zariski closure is the Zariski closed subset $W$ associated to $u-s^2(1-s),v-s(1-s)^2$. The Zariski closed subset $W$ contains the point $((u,v),s) = ((0,0),0)$, that is not in the image of $Z$.