# The underlying space of a scheme remembers its affineness?

Let $$f:X\rightarrow Y$$ be a morphism of schemes. We know that if $$Y$$ is affine and $$f$$ induces homeomorphism on the underlying spaces then $$X$$ is affine. Is it true that if $$X$$ is affine and $$f$$ induces a homeomorphism on the underlying spaces then $$Y$$ is affine?

More generally, is it true that a scheme whose underlying space is homeomorphic (possibly via a homeomorphism that is not induced by a morphism of schemes) to the underlying space of an affine scheme is affine? EDIT: actually, the answer to the last question is a very strong "NO" as the underlying space of any scheme is sober and Noetherian sober spaces are spectral (i.e. homeomorphic to the underlying space of an affine scheme).

• The answer to the second question is no: over an infinite field, $\mathbb{A}^1$ and $\mathbb{P}^1$ are homeomorphic. Apr 13, 2019 at 2:40
• The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $\{U_i\}$ such that $U_i\cap U_j$ is affine. It then follows from the quoted result that $f^{-1}(U_i)$ and $f^{-1}(U_i\cap U_j)$ are affine. So $X$ is defined by the same patching. Apr 13, 2019 at 3:04
• An idea that I don't have time to pursue: Let $E$ be a supersingular elliptic curve over $\mathbb{F}_p$. For each $\alpha \in H^1(E, \mathcal{O})$, there is an affine bundle $X_{\alpha}$ over $E$, whose corresponding line bundle is trivial. There is a Frobenius map $E_{\alpha} \to E_{\alpha^p}$ which is a homeomorphism on the topological spaces. Since $E$ is supersingular, $\alpha^p$ is $0$ in $H^1(E, \mathcal{O})$, so we get a map $A_{\alpha} \to A_0$. I know that $A_0 \cong A \times \mathbb{A}^1$ is not affine; is $A_{\alpha}$ affine? Apr 17, 2019 at 16:31
• As others have said, curves are hopeless in the Zariski topology. But in higher dimensions you might be interested in the recent Lieblich-Olsson reconstruction theorem (arxiv.org/pdf/1902.04668.pdf). This is very much in the normal projective variety setting, but in that case they show that the Zariski topological space together with the data of the rational equivalence relation on divisors is enough to determine the underlying abstract scheme. This suggests that your question probably requires additional hypotheses. Apr 20, 2019 at 19:12
• @TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $\mathbb{C}$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
– user138661
Apr 21, 2019 at 18:19

Here is a counterexample. Fix a field $$k$$, and let $$Y$$ be built from two copies of the affine nodal curve $$y^2=x^3+x^2$$, glued together on the complement of the singular point. In other words $$Y$$ is a nodal curve with doubled singular point. Then $$Y$$ is not affine because it is not separated. However, there is a homeomorphism $$\mathbb{A}^1\to Y$$, which is built from the usual parameterization of the nodal curve by $$\mathbb{A}^1$$ (which passes through the singular point twice) by choosing one of version of the singular point the first time through and the other version the second time through.
• do you think there is a counter-example with $Y$ separated?
Take a nodal cubic in $$\mathbb{P}^2$$, this is an irreducible scheme proper over $$\mathbb{C}$$, take its normalization and throw out a point from the inverse image of the node. We get a $$\mathbb{C}$$-morphism from the affine line to our cubic, which is bijective on the underlying spaces. Since a non-empty proper closed subspace of either the affine line or the cubic is a finite union of closed points, this morphism is also closed, thus it induces a homeomorphism on the underlying spaces. I find it somewhat amusing that this example was not pointed out earlier (indeed, it seems to be even simpler than Julian Rosen's much upvoted example).