# Is a bijective etale morphism from a quasi-projective $\Bbb C$-scheme to a proper $\Bbb C$-scheme necessarily an isomorphism?

Let $$X$$ be a quasi-projective scheme over $$\mathbb{C}$$ and $$Y$$ be a proper scheme over $$\mathbb{C}$$, and $$f:X\to Y$$ be a bijective, etale morphism.

Is this necessary for $$f$$ to be an isomorphism? If the answer is not, furthermore assume $$X$$ is also proper(hence projective), one can show that $$f$$ is actually finite. Then is $$f$$ an isomorphism under this additional assumption?

When they are both varieties and $$X$$ is also proper(hence projective), one can show that $$f$$ is actually finite, and hence $$f$$ is an isomorphism (see for example, Harris' Algebraic Geometry: A First Course, p. 179.). But I'm a little bit confused about non-reduced case.

Thanks for any help.

Fact: Let $$f:Y\to X$$ be a bijective etale morphism of finite type $$\mathbf{C}$$-schemes. Then, $$f$$ is an isomorphism.
Proof: Note that $$\Delta:Y\to Y\times_X Y$$ is surjective. To see this, note that since $$\Delta(Y)$$ is a locally closed subset of $$Y\times_X Y$$its complement if non-empty must contain a $$\mathbf{C}$$-point. So, it suffices to prove that $$\Delta(Y)$$ contains all $$\mathbf{C}$$-points of $$Y\times_X Y$$. But, this is clear since if $$(y_1,y_2)$$ is a $$\mathbf{C}$$-point of $$Y\times_X Y$$ then $$f(y_1)=f(y_2)$$ and since $$f$$ is injective this implies that $$y_1=y_2$$ and so $$(y_1,y_2)\in\Delta(Y)$$. By Tag 01S4 this implies that $$f$$ is universally injective. Since $$f$$ is also etale, Tag 02LC implies that $$f$$ is an open embedding. Since $$f$$ is a bijection, it is thus an isomorphism. $$\blacksquare$$
Exercise: What properties of $$\mathbf{C}$$ did I implicitly use? What generality can one extend it to?