# Surjective maps of affine spaces are closed

Let $$f: \mathbb{A}^n \to \mathbb{A}^n$$ be a quasi-finite surjective morphism.

Question: Is $$f$$ closed ?

• Suggested tag: counterexamples. (I don't know about the downvote, but I do feel I have answered a very similar question here at some point...) Nov 11, 2022 at 19:10
• I fixed a bit of spelling, I hope you don't mind.
– M.G.
Nov 11, 2022 at 22:10

Example. Let $$f \colon \mathbf A^2 \to \mathbf A^2$$ be given by $$(x,y) \mapsto \big(x^2y^2,y(xy-1)+x\big).$$ Then $$f$$ is the composition of the maps $$\begin{array}{ccccc}\mathbf A^2 & \stackrel g\longrightarrow & \mathbf A^2 & \stackrel h\longrightarrow & \mathbf A^2 \\ (x,y) & \longmapsto & (xy,y(xy-1)+x),\! & & \\ & & (x,y) & \longmapsto & (x^2,y),\! \end{array}$$ where $$h$$ is finite surjective and $$g$$ is quasi-finite missing only $$(1,0)$$:
• If $$(a,b) \in \mathbf A^2(k)$$ has $$a = 1$$, then the preimage consists of those $$(x,y) \in \mathbf A^2(k)$$ with $$xy=1$$ and $$x=b$$. This is only $$(b,b^{-1})$$ for $$b \neq 0$$, and there is no solution if $$b = 0$$.
• If $$(a,b) \in \mathbf A^2(k)$$ has $$a \neq 1$$, then the preimage consists of those $$(x,y) \in \mathbf A^2(k)$$ with $$xy=a$$ and $$(a-1)y+x=b$$. Substituting $$y=(b-x)/(a-1)$$ in $$xy=a$$ and multiplying by $$(a-1)$$ gives the quadratic equation $$x^2-bx+a^2-a=0$$, which has one or two solutions.
Since $$g$$ and $$h$$ are quasi-finite, so is $$f$$, and since $$h(1,0) = h(-1,0)$$ we see that $$f$$ is surjective. But $$f$$ is not closed: the restriction to the hyperbola $$xy=1$$ is given by $$(x,y) \mapsto (1,x)$$, which misses the point $$(1,1)$$.