chap-manifold.tex

\chapter{Manifolds and surfaces}

\section{Sides of a surface}

\begin{defn}
Let $\mu$ be an endofuncoid on a set~$U$.
\emph{Surface side} of a set~$T\subseteq\Ob\mu$ is a connected component
(regarding $\mu$) of the filter $(\rsupfun{\mu}T)\setminus T$.
\fxnote{$\mu$ is used twice in this definition. We may generalize
for two different funcoids instead.}
\end{defn}

Keep in mind that the above definition may work nicely if~$\mu$ is
a complete funcoid induced by a topological space.

\begin{example}
For an $\mathbb{R}^{n-1}$ subspace~$T$ of a~$\mathbb{R}^n$ ($n\geq 1$)
euclidean space and the complete funcoid~$\mu$ induced by the usual topology:
\begin{enumerate}
\item $T$~has exactly two surface sides.
\item The filter $\rsupfun{\mu}@\{a\} \setminus T$ (for every $a\in T$)
  has exactly two connected components.
\end{enumerate}
\end{example}

\begin{proof}
Without loss of generality assume that
\[ T=\setcond{(x_0,x_1,\dots,x_{n-2},0)}{
x_0,x_1,\dots,x_{n-2}\in\mathbb{R}};
\quad
a = (0,\dots,0). \]

We have
\[ \rsupfun{\mu}@\{a\} =
\left(\uparrow\setcond{v\in\mathbb{R}^n}{v_{n-1}>0}\sqcap\rsupfun{\mu}@\{a\}\right)
\sqcup
\left(\uparrow\setcond{v\in\mathbb{R}^n}{v_{n-1}<0}\sqcap\rsupfun{\mu}@\{a\}\right). \]

Let us prove that
$\uparrow\setcond{v\in\mathbb{R}^n}{v_{n-1}>0}\sqcap\rsupfun{\mu}@\{a\}$ and
$\uparrow\setcond{v\in\mathbb{R}^n}{v_{n-1}<0}\sqcap\rsupfun{\mu}@\{a\}$ are
connected components.

??
\end{proof}

\subsection{Special points}

We will start from the example of open $T=\setcond{(x,y,0)}{x^2+y^2<1}$
and closed $T=\setcond{(x,y,0)}{x^2+y^2\leq 1}$ disks in~$\mathbb{R}^3$.

\begin{xca}
Prove that open disk (in a usual 3-dimensional space) has two surface sides
and closed disk has one surface side.
\end{xca}

\section{Special points}

\begin{defn}
\emph{Surface cardinality} of a point~$a$ (an element of the set $\Ob\mu$) is
the cardinality of the set of connected components of the filter
$\rsupfun{\mu}\{a\}\setminus T$.
\end{defn}

\begin{defn}
\emph{Cardinality regular point} is a point~$a$, which has a neighborhood
($X\in\up\rsupfun{\mu}\{a\}$) such that all points~$x\in X\cap T$
are of the same surface cardinality as the point~$a$.

\emph{Cardinality special point} is a point which is not cardinality regular.
\end{defn}

\begin{defn}
\emph{Isomorphism regular point} is a point~$a$, which has a neighborhood
($X\in\up\rsupfun{\mu}\{a\}$) such that for all points~$x\in X\cap T$
the filter $\rsupfun{\mu}\{a\}$ is isomorphic
to~$\rsupfun{\mu}\{x\}$.

\emph{Isomorphism special point} is a point which is not isomorphism regular.
\end{defn}

\fxnote{Try to replace isomorphism~$f$ with some kind of filter embedding.}

Consider the dihedral angle~$T$ produced by two half-planes. Are the points of
intersection of the half-planes isomorphism-special? (They should not
be considered special. If they are special, this is a probably flaw in
the definition of isomorphism special.)

Consider union~$T$ of two intersecting lines on a plane. The intersection
may be considered as a special point, because it has more connected
components that the rest. We don't want to consider it special, however.
We can restrict to consider special only points which have less connected
components (rather than more) to correct this trouble. Also try to define
it with some kind of morphisms of filters instead of isomorphism as in
isomorphism-special.

\begin{xca}
Excluding special points (either cardinality or isomorphism) from closed disk
produces open disk.
\end{xca}

Let us note that special points of closed disk have surface cardinality
$1$ which is less than surface cardinality ($2$) of regular points.
So, it is a conceivable idea to consider special points which have
lesser surface cardinality than nearby points.

Consider the following two subsets of a plane (the lines are the
set~$T$, the small black blob is the point~$a$, and the cyan
blob symbolizes the filter $(\rsupfun{\mu}\{a\})\setminus T$):

\begin{figure}
  \begin{subfigure}[c]{0.3\textwidth}
    \centering
    \begin{tikzpicture}
        \node at (0,0) [circle,draw,fill=cyan] (a) {};
        \fill (0,0) circle [radius=2pt,fill=black];
        \draw [ultra thick] (0,0) -- +(0:1cm);
        \draw [ultra thick] (0,0) -- +(120:1cm);
        \draw [ultra thick] (0,0) -- +(240:1cm);
    \end{tikzpicture}
    \caption{Three surface sides}
  \end{subfigure}
  \begin{subfigure}[c]{0.3\textwidth}
    \centering
    \begin{tikzpicture}
        \node at (0,0) [circle,draw,fill=cyan] (a) {};
        \fill (0,0) circle [radius=2pt,fill=black];
        \draw [ultra thick] (0,0) -- +(0:1cm);
        \draw [ultra thick] (0,0) -- +(180:1cm);
    \end{tikzpicture}
    \caption{Two surface sides}
  \end{subfigure}
  \caption{Examples of surface cardinality}
\end{figure}

For one of the sets surface cardinality of~$a$ is $3$ and for
another it is~$2$.

Now define \emph{shift special points}.

Let $I$ be an interval on~$\mathbb{R}$ (containing zero?)

A point~$a$ is \emph{shift special} if there exists a transformation
(that is a continuous function $f:I\times\mu\to\mu$ such that:
\begin{enumerate}
  \item $f(0)$ is identity. \fxwarning{Is this condition needed?}
  \item for every sufficiently small~$\epsilon>0$ we have $f(\epsilon,a)\in T$;
  \item there is $\epsilon>0$ such that for every $0<\epsilon'<\epsilon$ we have
    $f(\epsilon')$ being not continuous at~$a$ regarding complete funcoid
    defined by the function $x\mapsto\rsupfun{\mu}\{x\}\setminus T$.
\end{enumerate}

We may consider to additonally require that every~$f(\epsilon)$ is isomorphism
of funcoids.

\begin{example}
$T$~is disk $\setcond{(x,y,0)}{x^2+y^2\leq 1}$. $f$~is the contraction
$(\epsilon,v)\mapsto\frac{1}{1+\epsilon}v$. $a=(1,0,0)$.

In the usual topology~$f$ is continuous. In
$x\mapsto\rsupfun{\mu}\{x\}\setminus T$ we have the function
$\epsilon\mapsto f(\epsilon)$ not continuous at zero.
So~$a$ is a shift special point.
\end{example}

\begin{proof}
$f (0) (v) = v$. Thus $\langle f (0) \rangle (\rsupfun{\mu} \{ a
\} \setminus T) = \rsupfun{\mu} \{ a \} \setminus T$ intersects
the plane $Z = 0$. But $f (0, a)$

??
\end{proof}

\begin{question}
Can we exclude real numbers from the play?
\end{question}

\begin{question}
How cardinality special points, isomorphism special points and shift
special points are related with each others?
\end{question}

\begin{question}
How the number of surface sides is related with usual surface sides for
manifolds?
\url{https://en.wikipedia.org/wiki/Orientability#Orientability_of_manifolds}
\end{question}

\begin{rem}
Manifolds have no special points. (Prove!)
\end{rem}

Prove that $2$-manifold image which special points removed has the same number
of sides as the defined above.

Another way to define special points: A special point is a point
such that $T\sqcap\supfun{\mu}\{a\}$ is not isomorphic to
$T\sqcap\supfun{\mu}\{x\}$ for nearby points~$x$. Consider replacement
of isomorphism with injection, surjection, etc. here and above.

How many sides has in $\mathbb{R}^3$ a plane without one point?

Easy way to spot special points: They are boundary points in the
topology (or funcoid) induced on~$T$. Alternatively we can consider
points whose neighborhood in~$T$ is different (as non-isomorphic or
maybe non-injective or non-surjective or like this) than of nearby
points. Thus another way to remove special points: use interior funcoid.

\url{https://math.stackexchange.com/q/2836833/4876}