Conditions under which a bijective morphism of quasi-projective varieties is an isomorphism

Question

I'm currently reading a paper by Nakajima (Quiver Varieties and Tensor Products), and I'm having a hard time understanding a very specific step in his proof of Lemma 3.2. Essentially, we have two (quasi-projective) varieties, say $X$ and $Y$, that we would like to show are isomorphic. The proof uses the following argument:

(1) Construct a bijective morphism $f:X \to Y$.

(2) Show that $\mathrm{d}f: T_x(X) \to T_{f(x)}(Y)$ is an isomorphism for all $x \in X$, where $T_x(X)$ is the tangent space of $X$ at $x$.

From (1) and (2), he concludes that $f$ is an isomorphism. My question is, how can one conclude that f is an isomorphism from (1) and (2)?

In doing a little research, I found this topic. To summarize Theorem 14.9 and Corollary 14.10 in Joe Harris' Algebraic Geometry: A First Course:

If $f:X \to Y$ is a morphism of varieties and either $f$ is finite or $X$ and $Y$ are projective, then $f$ is an isomorphism if and only if $f$ is bijective and $\mathrm{d} f: T_x(X) \to T_{f(x)}(Y)$ is injective for all $x \in X$.

In the context of Lemma 3.2 in Nakajima's paper, neither variety is projective, which leaves two possibilities. Either the morphism constructed in the proof is in fact finite (but this, to me at least, is not at all obvious), or he is using some other fact that allows him to conclude that $f$ is an isomorphism given (1) and (2) (potentially something about the symplectic form on the quiver varities). Does anyone have any insights as to what facts Nakajima might be using in his proof?

Answer 1

As CamSar notes, if you assume that $Y$ is smooth, the result is true. In fact, if you even just assume that $Y$ is normal and that $df$ is an isomorphism at the generic points, then Zariski's Main Theorem implies that $f$ is an isomorphism if $X$ is reduced and $f$ is bijective. As Nakajima explains on the top of p. 12, his space $\mathfrak{M}(\mathbf{v},\mathbf{w})$ is smooth. Moreover, the group action is by the linearly reductive group $\mathbb{G}_m$. Thus the fixed point set $Y$ is smooth by "Iversen's theorem". So that explains the result in Nakajima's paper.

Of course there are counterexamples when $Y$ is singular. If you allow $X$ and $Y$ to be nonreduced, it is trivial to construct examples, as explained in the answers here. However, your wording suggests that you are interested in the case when $X$ and $Y$ are integral varieties. None of the previous answers seems to address that case. So, just for the record, following is an example where $X$ and $Y$ are integral varieties, $f$ is bijective and isomorphic on Zariski tangent spaces, yet $f$ is not an isomorphism. Note, the following morphism $f$ is not finite -- if it were, then the theorems mentioned in your comment would imply that $f$ is an isomorphism.

Beginning with $\mathbb{A}^2$ with coordinates $(x,y)$, let $Y$ be the plane curve $$ Y = \{(x,y)\in \mathbb{A}^2 : y^4+y^2x-x^3 = 0 \}.$$ This is an irreducible curve with a unique singular point at $(x,y)=(0,0)$. Next, beginning with $\mathbb{A}^3$ with coordinates $(u,v,w)$, let $X$ be the curve $$X = \{(u,v,w)\in \mathbb{A}^3 :(v+1)w-1 = v^3+v^2-u^2=0\}. $$
Projection from $X$ to the $(u,v)$-plane is a locally closed immersion with image the complement of $(0,-1)$ in the plane curve with equation $v^3+v^2-u^2$. In particular, $X$ is an irreducible curve with a unique singular point at $(0,0,1)$. Now consider the morphism, $$ F :\mathbb{A}^3 \to \mathbb{A}^2, \ \ f(u,v,w) = (u,v^2+v).$$

I claim that $F(X)$ equals $Y$, and the restriction morphism, $$ f :X \to Y,$$ is a bijection that induces isomorphisms on all Zariski tangent spaces. The simplest way to prove this is to first normalize $X$ and $Y$. In the function field of $X$, observe that the following monic equation holds, $$ t^2 - (v+1) = 0, \ \ t = u/v.$$ Thus, $k[t]$ is a subring of the integral closure of the fraction field. But, of course, $v = t^2 - 1$ and $u = t(t^2-1)$ are already in $k[t]$. Finally, $w = 1/t^2$ is in the integral closure. Thus, the integral closure of $k[X]$ equals the integral closure of $k[t][1/t^2]$, which is already the integrally closed ring $k[t][1/t]$. So the normalization of $X$ is just $$\nu: \mathbb{G}_m \to X, \ \ \nu(t) = (t(t^2-1),t^2-1,1/t^2).$$
In particular, the composition with $F$ is, $$ F\circ \nu: \mathbb{G}_m \to \mathbb{A}^2, \ \ F(\mu(t)) = (t(t^2-1),t^2(t^2-1)).$$ By the same argument as above, the normalization of $Y$ is, $$ \mu: \mathbb{A}^1 \to Y, \ \mu(t) = (t(t^2-1),t^2(t^2-1)).$$ Thus, $F\circ \nu$ is just the restriction of the normalization $\mu$ to the open $\mathbb{G}_m\subset \mathbb{A}^1$.

Since $F\circ\nu$ factors through the normalization of $Y$, in particular $F(X)$ is contained in $Y$. Denote by $f$ the restriction. Moreover, since $\nu$ is a bijection except over $(0,0,1)$, also $f$ is injective at that point. Similarly, since the domain of $F\circ \nu$ contains every point except $t=0$, the image of $f$ contains every point, except possibly $\mu(0)=(0,0)$. But by direct computation, the inverse image under $f$ of $(0,0)$ is precisely $(0,0,1)$. Therefore $f$ is bijective.

Since $f$ matches up normalizations, $f$ is an isomorphism at every point where the normalization morphisms $\nu$ and $\mu$ are isomorphisms, i.e., everywhere except $(0,0)$ in $Y$ and $(0,0,1)$ in $X$. In particular, except possibly at those points, $df$ is an isomorphism. Finally, by direct computation, $df$ is an isomorphism at the point $(0,0,1)$: both Zariski tangent spaces are two dimensional, and $df$ is an isomorphism between them (just linear projection, really).

Finally, $f$ is not an isomorphism. If $f$ were an isomorphism, it would induce an isomorphism of normalizations. But the induced morphism of normalizations is the non-surjective inclusion, $\mathbb{G}_m\subset \mathbb{A}^1$.

Answer 2

First of all, note that $\pi:X\rightarrow Y$ bijective is not enough to conclude that $\pi$ is an isomorphism. For instance consider $C\subset\mathbb{A}^2$ a cubic with a cusp and let $\pi:\mathbb{A}^1\rightarrow C$ given by $t\mapsto (t^2,t^3)$. Then $\pi$ is bijective, it is indeed a topological homomorphism. However, $C$ is singular and $\mathbb{A}^1$ is smooth. Therefore $\pi$ can not be an isomorphism. As you see the differential $d\pi_{0}$ of $\pi$ in zero is zero.

Let $\pi:X\rightarrow Y$ be a morphism or relative dimension $r$ of smooth varieties over an algebraically closed field. Assume that the relative cotangent sheaf $\Omega_{X/Y}$ is locally free of rank $r$ on $X$. We have an exact sequence $$\pi^{*}\Omega_{Y}\rightarrow\Omega_{X}\rightarrow\Omega_{X/Y}\mapsto 0.$$ Let $k(x)$ be the residue field at a closed point $x$. Tensorizing we get
$$\pi^{*}\Omega_{Y}\otimes k(x)\rightarrow\Omega_{X}\otimes k(x)\rightarrow\Omega_{X/Y}\otimes k(x)\mapsto 0.$$ Since $X$ and $Y$ are smooth and $\Omega_{X/Y}$ is locally free of rank $r$ these three vector spaces are of dimension $dim(Y),dim(X),r$ respectively. So the first map is injective and we have $$0\mapsto\pi^{*}\Omega_{Y}\otimes k(x)\rightarrow\Omega_{X}\otimes k(x)\rightarrow\Omega_{X/Y}\otimes k(x)\mapsto 0.$$ For any closed point $x\in X$ we have $k(x)\cong k$. Therefore we can identify the injective map $\pi^{*}\Omega_{Y}\otimes k\rightarrow\Omega_{X}\otimes k$ with the map between the cotangent spaces $\mathfrak{m}_y/\mathfrak{m}^2_y\rightarrow\mathfrak{m}_x/\mathfrak{m}^2_x$ where $y = \pi(x)$. Dualizing we have that the differential $T_\pi(x):T_xX\rightarrow T_yY$ is surjective. Therefore $\pi$ is a smooth morphism.

In particular, if any fiber of $\pi$ is just one point, then $\Omega_{X/Y}$ is locally free of rank $r = 0$ and $\pi$ is a smooth morphism of relative dimension zero. Finally, since $\pi$ is surjective it is an isomorphism.