David Mumford gives in his book Algebraic Geometry I, Complex Projective Varieties on page 43 the definition of topologically unibranch points of affine variety and I have a lot of problems to extract from this a way how to think about unibranch points intuitively:
(3.9) Definition. Let $X$ be an affine variety (in this book always 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.
Then Mumford continues:
Note that smooth points are topologically unibranch: in fact if $x$ is smooth on $X$, $x$ has analytic coordinates in a neighborhood $U$ (according to Corollary 1.26):
$$\alpha: \text{ polycylinder } V_{\epsilon} := \{z \in \mathbb{C}^r \vert \ \vert z_i \vert < \epsilon \} \xrightarrow[\cong]{\alpha} U \subset X$$
$$ \text{ with } \ \ \ \ \ 0 \mapsto x$$
Essentially, $U$ becomes affine space. Every (alg closed subset) $Y \subsetneqq X$ is contaned in the set of zeroes of some polynomial $p$ which not vanish on $X$. Hence $\alpha^{-1}(Y)$ is contaned in the set of zeroes of some convergent power series $q(z_1,...,z_n)$ which doesn't vanish identically.
And then Mumford clamed that the set $V_{\epsilon} - (\text{ zeroes of } q)$ is connected. His "argument": If $x,y \in V_{\epsilon} - (\text{ zeroes of } q)$, look at line $tx+(1-t)y$ joining them (vector notation). Then $tx+(1-t)y \in V_{\epsilon} - (\text{ zeroes of } q)$ if $\vert t-\frac{1}{2} \vert \le \frac{1}{2}$ and $q(tx+(1-t)y)$ has only a finite set of zeroes on the disc $\vert t-\frac{1}{2} \vert < \frac{1}{2}$. Thus $x$ and $y$ can be joint by a path avoiding zeroes of $q$]. But what if $X$ is $2$-dimensional and $Y$ one-dimensional and contains $x$, for example if $Y$ is an affine line, see my conterexmple below in P1?
problems:
P1: where lives the parameter $t$ of the line $tx+(1-t)y$ introduced in the proof? Mumford called the definition domain of $t$ the disc $\vert t-\frac{1}{2} \vert < \frac{1}{2}$. Is it the complex ! disc $B_{\frac{1}{2}}(\frac{1}{2})= \{z \in \Bbb C \ \vert \ \vert z-\frac{1}{2} \vert < \frac{1}{2} \}$ or the "real" disc around $\frac{1}{2}$ which is in our case $[0,1]$? My guess is $B_{\frac{1}{2}}(\frac{1}{2})$ because for $t \in [0,1]$ the proof looks wrong. The zeroes of $q(tx+(1-t)y)$ cannot be avoided running $t$ in $[0,1]$.
as KReiser noticed in his comment my "conterexample" is wrong since X−Y={(x,y)|x≠0} is connected in C2 but not R2 and the later was the picture in had previously in mind. That is the next question P2 can be ignored:
P2: I think the proof is wrong. Take for example the affine plane $X= U = \mathbb{C}^2$ and $x=(0,0)$. Clearly $x$ is smooth in $X$. Take as $Y$ the vanishing set
of monomial $z_1 \in \mathbb{C}[z_1,z_2]$. Then $x \in Y$ and $Y$ "slices" $X$
in two parts along $y$-axis: $X-Y= \{(x,y) \vert x \neq 0 \}$
Then every open neighbourhood $U$ of $x$ in is divided by $Y$ in two parts as well and so $U - U \cap Y$ can never be connected. So according to this definition of topologically unibranchness even affine plane isn't unibranch, yeah? Therefore I want to clarify if the book's definition of topologically unibranch is indeed wrong or did I miss something.
P3: And moreover I would like to know how to think about topologically unibranchness intuitively. Is there any way to visualize geometrically what is going on there?