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Recall that a morphism of schemes $X \to Y$ is called a universal homeomorphism if the underlying map of topological spaces is a homeomorphism, and for all morphisms $Z \to Y$, the lifted map $X \times_Y Z \to Z$ is again a homeomorphism.

A universal homeomorphism need not be an isomorphism. A counterexample is given by the normalization morphism of the cuspidal cubic. That is, the map $\operatorname{Spec} k[t] \to \operatorname{Spec} k[x,y]/(x^3-y^2)$ defined from the ring map $x \mapsto t^2$ and $y \mapsto t^3$ is (I think) a universal homeomorphism; for it is integral hence universally closed (Stacks, Tag 01WG), it is surjective hence universally surjective (Stacks, Tag 01RY), and it is universally injective because the diagonal map is surjective (Stacks, Tag 0CI0).

Question. Let $X$ and $Y$ be two schemes. If there exists a universal homeomorphism from $X$ to $Y$, and a universal homeomorphism from $Y$ to $X$, does that mean $X$ and $Y$ are isomorphic as schemes? If needed you may assume that the schemes considered are reasonable (e.g. locally Noetherian, quasi-separated).

I suspect the claim is false, and hopefully even in reasonable situations, because I see no reason for it to be true (in line with the general philosophy that schemes are not at all determined by their underlying topological spaces). However I can't quite come up with a counterexample.

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    $\begingroup$ I think if you allow non-reduced schemes, then the answer to your question is "no". Let $k$ be a field, and consider the maps $k \to k[x]/(x^2) \to k$. The corresponding scheme morphisms are universal homeomorphisms since they are integral, universally injective, and surjective [Stacks, 04DF] (for universal injectivity, it suffices to note that the maps on residue fields are just the identity on $k$ [Stacks, 01S4]). $\endgroup$ – Takumi Murayama Jun 6 '18 at 15:59
  • $\begingroup$ Whoops, off course, how could I not have considered that example. Do you think adding/removing reducedness is the only 'class' of counterexamples? $\endgroup$ – I I Jun 6 '18 at 16:26
  • $\begingroup$ In characteristic positive, Frobenius might be such a map. $\endgroup$ – Mohan Jun 6 '18 at 16:56
  • $\begingroup$ Dear Mohan, I only know of Frobenius endomorphisms, but domain and codomain are trivially isomorphic in that case. Or does Frobenius also mean something else? $\endgroup$ – I I Jun 6 '18 at 19:26
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Here is a more interesting example, following Mohan's suggestion. I would still like to see an example where both $X$ and $Y$ are reduced, though.

Let $X$ be a reduced scheme over an imperfect field $k$ of characteristic $p > 0$ that is not geometrically reduced (see, e.g., this MathOverflow answer). Consider the commutative diagram $$\require{AMScd} \begin{CD} X @>F_{X/k}>> X^{(p)} @>F'>> X\\ @. @VVV \lrcorner @VVV\\ @. \operatorname{Spec} k @>F>> \operatorname{Spec} k \end{CD} $$ where the square is cartesian. The morphism $F_{X/k}\colon X \to X^{(p)}$ is the relative Frobenius morphism of $X/k$, and is induced by the universal property of the fiber product with the composition $X \to X^{(p)} \to X$ being the absolute Frobenius.

Then, both $F'$ and $F_{X/k}$ are universal homeomorphisms [Stacks, Tags 0CC8 and 0CCB], but $X$ and $X^{(p)}$ are not isomorphic since $X$ is reduced while $X^{(p)}$ is not [Stacks, Tag 0CCE].

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