In this question I'd like to examine some properties of universally closed morphisms.
The question is self-contained. It can also be seen as a follow-up to this [question](https://mathoverflow.net/q/397132).


Let $R$ be a discrete valuation ring, $X$ and $S$ two $R$-flat and universally closed $R$-schemes.


Let $f : X\to S$ be a universally closed morphism. Assume the special fiber of $f$ is an isomorphism.



> 
> * If $X$ and $S$ are reduced, is $f$ an isomorphism?
> * If $X$ and $S$ have reduced fibers, is $f$ an isomorphism?
> 
> 
> 


Without either of the reducedness assumptions, the answer is "no", as shown by the example provided [here](https://mathoverflow.net/q/397190).


My expectation is that the answer is still "no" to both questions, but an example seems subtle to produce.


**Remarks**
In these remarks we further assume $f$ is flat, to see what more can be said in this case. After all, should the questions have positive answer, $f$ would have to be flat. The remarks are listed increasing the strength of the additional assumptions, to see what can be said in each ever-more-special case.


* From flatness we deduce $f$ is surjective, since generizations lift along $f$ (by flatness), and every point in $S$ specializes to a point in the special fiber (by universal closedness of $S$, if $s\in S$ is contained in the generic fiber, it maps to the generic point in $\text{Spec}(R)$. This specializes to the closed point in $\text{Spec}(R)$, and specializations lift along $S\to \text{Spec}(R)$ by universal closedness, so there is some $s\_0\in S$ contained in the special fiber, to which $s$ specializes).
* If $f$ is, in addition, *separated* and *locally of finite type*, then it is also proper, since $f$ is quasi-compact by universal closedness. It is therefore an isomorphism (it is quasi-finite, hence finite, hence finite flat of degree $1$). The question is interesting only if $f$ is *not* locally of finite type.
* To summarize, the case when $f$ is flat and separated is already unanswered but one has access to more info on $f$. A non-separated example, flat or not, to answer the question in the negative, would be already what I'm looking for. A flat separated example, if any, would be the best way to settle this puzzle, and I expect it is very subtle to find.



On universally closed morphisms of reduced schemes
Let $R$ be $\mathbb Z\_p$ (or any other dvr).


Let $S$ be obtained by gluing two copies of $\mathbb P^1\_{\mathbb Z\_p}$ away from the $0$-point in the special fiber, i.e. away from the vanishing locus of the ideal $(p,x)$ in local coordinates away from $\infty$.


Let $X$ be obtained by gluing two copies of $\mathbb P^1\_{\mathbb Z\_p}$ away from the $0$-section, i.e. away from the vanishing locus of the ideal $x$.


These are obtained by gluing two flat and universally closed schemes along an open set, hence flat and universally closed.


Then we can map $X \to S$ by mapping each copy of $\mathbb P^1\_{\mathbb Z\_p}$ to a different copy of $\mathbb P^1\_{\mathbb Z\_p}$, which is compatible with the gluing since we glue along a larger open set in $S$ than in $X$.


This map is an isomorphism in the special fiber since the constructions of $X$ and $S$, restricted to the special fiber, are identical.


But it is manifestly not an isomorphism, e.g. because it makes a non-separated generic fiber to a separated one.


