In David Mumford's book Algebraic Geometry I, Complex Projective Varieties
treating mainly complex varieties as objects of interest on page
43 he defines what is a topologically unibranch variety $X$ at $x \in X$.


(3.9) Definition. Let $X$ be an affine variety over $\mathbb{C}$
and $x\in X$. Then $X$ is
topologically unibranch at $x$ if for all closed algebraic subsets $Y \subsetneqq
X$, $x$ has fundamental system neighborhoods $U\_n$ in
classical topology of $X$ s.t. $U\_n-(U\_n\cap Y)$ is connected.


The main motivation behind this seems to provide tools to
analyze what happens around a singularity. Archetypical examples where this definition works fine
are the nodal and cuspidal curves with singularity in $(x,y)=(0,0)$.
In neighborhood in $(0,0)$ the nodal curve curve looks like the cross
$XY=0$. This means that two "branches" $X=0$ and $Y=0$ cross in $(0,0)$ transversally
and therefore nodal curve isn't unibranch by definition. (take as $Y$ of the two branches). The cuspidal curve is also singular in $(0,0)$ but it's "pick" is locally irreducible
and cannot be splitted in more that one "branches". Intuitively that's
the reason why the cusp is unibranch in $(0,0)$. Another
feature of unibranchness sounds more problematical too me. Mumford
claims that if $x \in X$ is a smooth point then $X$ is always unibranch
in $x$.


Mumford proved that smooth points are unibranch and possibly the main ingredient for his proof
was the fact that we consider a variety over $\mathbb{C}$.
More concretely he used that the unit disc in $\mathbb{C}$ is
pathconnected after removing a finite number of points. [Here I
asked a question](https://math.stackexchange.com/questions/3870088/topologically-unibranch-in-mumfords-book-complex-projective-varieties) which deals with this proof and it can also be considered as
starting point of this question.


Question: Can the *unibranchness* be defined in reasonable way for
more general varieties over arbitrary field $K$ in similar *topological* flavour like (3.9)
preserving the property that in "smooth" points the considered space
is always unibranch? The main point of my interest if it's possible
to do it not only for (projective) varieties over *complex* numbers?


And can it be garanteed that the definition preserves the property
that all smooth points of the variety are unibranch?


My motivation is a serious obstacle which I observed in the question I
already quoted before: Let consider the affine plane $X= U = K^2$ over
field $K$ and $x=(0,0)$. Clearly $x$ is smooth in $(0,0)$. Take as
$Y$ the vanishing set $V(z\_1)$ of monomial $z\_1 \in K[z\_1,z\_2]$.
Then $x \in Y$ and $Y$ "slices" $X$
in two parts along $y$-axis: $X-Y= \{(x,y) \vert x \neq 0 \}$.


Take an arbitrary open neighbourhood $U$ of $x$ and consider
$U- U \cap Y$. The question is when is it connected and when not?
The observation is that for $K=\mathbb{C}$ this set is connected, for
$K= \mathbb{R}$ not.


This shows that the definition in (3.9) of topologically unibranchness
seemingly only works for varieties of complex numbers. And I would
like to know it there exist a resonable definition of unibranchness
for varieties over arbitrary fields $K$? Primary interested in case $\mathbb{R}$. Until now I found only [pure algebraical definitions](https://stacks.math.columbia.edu/tag/0BQ1) and would like to know if it can be also defined pure topologically.



Unibranch points (definition for varieties over arbitrary field)
For a scheme $X$, say that $X$ is *topologically* unibranch at $x$ if $\mathop{Spec} O\_{X,x}$ is geometrically unibranch (meaning that $O\_{X,y}$ is geometrically unibranch at all generisations $y$ of $x$).


Assume $X$ is irreducible for simplicity.
Then by <https://stacks.math.columbia.edu/tag/0BQ4>, $X$ is topologically unibranch at $x$ if and only if for all closed $Y \subsetneqq X$, $x$ has fundamental system of *étale neighborhoods* $U\_n$ s.t. $U\_n-(U\_n\cap Y)$ is connected.


Indeed we may assume $Y$ irreducible, $y$ be its generic point. Then the punctured spectrum of $O^{\mathrm{sh}}\_{X,y}$ is connected by the Lemma above. But the strict henselization is the inductive limit of the étale neighboors, hence the equivalence with $U\_n-(U\_n\cap Y)$ connected for $U\_n$ sufficiently small.


Your example of $X/\mathbb{R}$ is topologically unibranch because $X\_\mathbb{C} \to X$ is an étale cover.


