← Metric Spaces → Regular and Normal Spaces
What should you be acquainted with? Set Theory.

Basics of Topological Spaces

Topologies and continuous functions

A topology ${\cal T}$ on a set $X$ is a collection of subsets of $X$, i.e. ${\cal T}\sbe{\cal P}(X)$ with the following properties: 1. $\emptyset,X\in{\cal T}$. 2. For all $U,V\in{\cal T}$ we have: $U\cap V\in{\cal T}$. 3. For any family $U_\a\in{\cal T}$ we have: $\bigcup U_\a\in{\cal T}$. $(X,{\cal T})$ is said to be a topological space and every subset $U\in{\cal T}$ is called an open subset of $X$. $(X,{\cal T})$ is said to be metrizable, if there exists a metric $d$ on $X$, such that ${\cal T}$ coincides with the metric topology of $(X,d)$.
Let ${\cal T}$ be a topology on $X$. A collection of subsets ${\cal B}\sbe{\cal T}$ is said to be a basis of the topology ${\cal T}$, if every set $U\in{\cal T}$ is the union of sets in ${\cal B}$.
A collection of subsets $\B\sbe{\cal P}(X)$ is a basis of a topology ${\cal T}$ on $X$, if and only if the following conditions hold: 1. For $x\in X$ there exists $B\in\B$ such that $x\in B$. 2. For all $A,B\in\B$ and all $x\in A\cap B$ there exists $C\in\B$ such that $x\in C\sbe A\cap B$.
A collection of subsets ${\cal S}\sbe{\cal T}$ is called a subbasis of the topology ${\cal T}$, if the collection, which consists of all finite intersections of sets of ${\cal S}$, is a basis for ${\cal T}$.
Let $(X,{\cal T})$ be a topological space. A subset $A$ of $X$ is called a neighborhood of a point $x\in X$, if there exists an open subset $U$ such that $x\in U\sbe A$. The set of all neighborhoods of $x$ and all open neighborhoods respectively, will be benoted by ${\cal U}(x)$ and ${\cal U}(x)^\circ$ respectivley. A subset $A$ of $X$ is said to be closed, if $A^c$ is open. The interior $A^\circ$ of a subset $A$ of $X$ is the union of all open subsets contained in $A$ and the closure of $A$ is the intersection of all closed subsets containing $A$, i.e.: $$ A^\circ\colon=\bigcup\{U\sbe A:U\in{\cal T}\}\quad \cl A\colon=\bigcap\{C\spe A:C^c\in{\cal T}\}~. $$ The set $\pa A\colon=\cl{A}\sm A^\circ$ is called the boundary of $A$.
A mapping $f:X\rar Y$ from the topological space $(X,{\cal T}_X)$ into the topological space $(Y,{\cal T}_Y)$ is said to be continuous at $x\in X$, if for all $V\in{\cal U}(f(x))$ we can find some $U\in {\cal U}(x)$ such that $f(U)\sbe V$ i.e. $f^{-1}(V)\spe U$. This is equivalent to: $f^{-1}({\cal U}(f(x))\sbe{\cal U}(x)$. $f$ is said to be continuous on $X$, if $f$ is continuous at all points of $X$. A bijection $f:X\rar Y$ is called a homeomorphism, if both $f$ and $f^{-1}$ are continuous.
Let $(X,{\cal T}_X)$ and $(Y,{\cal T}_Y)$ be topological spaces and suppose that ${\cal B}_Y$ and ${\cal S}_Y$ respectively is a basis and a subbasis of ${\cal T}_Y$ respectively. For a mapping $f:X\rar Y$ the following assertions are equivalent: 1. $f$ is continuous on $X$. 2. The preimage of any open subset of $Y$ is open in $X$. 3. The preimage of any closed subset of $Y$ is closed in $X$. 4. $f^{-1}({\cal B}_Y)\sbe{\cal T}_X$. 5. $f^{-1}({\cal S}_Y)\sbe{\cal T}_X$. 6. For all $A\sbe X$: $f(\cl A)\sbe\cl{f(A)}$.
A function $f:X\rar\R$ is continuous if and only if for all $t\in\R$ the sets $[f\leq t]\colon=\{x\in X:f(x)\leq t\}$ and $[f\geq t]$ are closed.

Semi continuous functions
A function $f:X\rar\cl{\R}$ is said to be lower semi continuous (l.s.c), if for all $t\in\cl{\R}$ the set $[f\leq t]$ is closed. $f$ is said to be upper semi continuous (u.s.c), if $-f$ is l.s.c.
Suppose $f_\a:X\rar\cl{\R}$ is an arbitrary family of l.s.c functions. Then $f\colon=\sup_\a f_\a$ is l.s.c.

Open and closed mappings
A mapping $f:X\rar Y$ is said to be open (closed) if the image of any open (closed) subset of $X$ is open (closed) in $Y$.
A mapping $f:X\rar Y$ is closed if and only if for all $y\in Y$ and all open subsets $U\sbe X$, containing $f^{-1}(y)$ we can find some open neighborhood $V$ of $y$ such that $f^{-1}(V)\sbe U$.
A continuous and injective mapping $f:X\rar Y$ is said to be an embedding if $f:X\rar f(X)$ is a homeomorphism.
A continuous and injective mapping $f:X\rar Y$ is an embedding if and only if for all closed subsets $A$ of $X$ and all $x\notin A$: $f(x)\notin\cl{f(A)}$.
Let $Y$ be a subset of a topological space $X$ and $f:X\rar Y$ a continuous mapping. $f$ is a retraction, if $f$ is onto and $f\circ f=f$. Given that, $Y$ is said to be a retract of $X$.

Countable bases and coverings
A subset $A$ of a topological space $(X,{\cal T})$ is said to be dense if $\cl A=X$. $X$ is called separable, if there exists a countable dense subset. We say that $X$ has a countable basis, if the topology ${\cal T}$ has a countable basis.
1. Every topological space which has a countable base is separable. 2. Every separable metric space has a countable base. 3. Suppose $X$ has a countable base and $U_\a$, $\a\in I$, is another basis for the topology of $X$, then there exists a countable subset $J$ of $I$, such that $U_\a$, $\a\in J$, is a basis for the topology of $X$.
Every subspace of a separable metric space is separable $X$ and the countable product of separable metric spaces is again a separable metric space. Moreover the Borel $\s$-algebra of a separable metric space is countably generated.
A covering of a set $X$ is a collection of subsets ${\cal U}\sbe{\cal P}(X)$, such that: $X=\bigcup{\cal U}\colon=\bigcup\{U:U\in{\cal U}\}$. A subcollection ${\cal V}\sbe{\cal U}$, which is a covering of $X$ is said to be a subcovering of $X$.
A covering ${\cal U}$ of a topological space $X$ is called an open covering if all sets of ${\cal U}$ are open. A covering ${\cal U}$ of $X$ is locally finite, if for any point $x\in X$ we can find some open neighborhood $U$ such that the set $\{V\in{\cal U}:U\cap V\neq\emptyset\}$ is finite.
Suppose $D_\a$, $\a\in I$, is a locally finite covering of a topological space $X$. Then for every subset $J\sbe I$ the set $\bigcup\{\cl D_\a:\a\in J\}$ is closed and $\{\cl D_\a:\a\in I\}$ is locally finite.

Filters
A collection $\F$ of subsets of a set $X$ is said to be a filter, if 1. $\emptyset\notin\F$. 2. For all $F,G\in\F$: $F\cap G\in\F$.3. If $G$ is a subset of $F\in\F$, then $G\in\F$.
A collection $\B$ of subsets of $X$ is said to be a filter basis if $\emptyset\notin\F$ and for all $F,G\in\F$ there exists some $H\in\B$ such that $H\sbe F\cap G$.
A collection $\F\sbe{\cal P}(X)$ is called a subbasis of a filter, if the collection of all finite intersections of sets in $\F$ is a filter basis.
If $\F$ is a filter basis, than $\wt{\cal F}\colon=\{G\in{\cal P}(X):\exists F\in\F: F\sbe G\}$ is a filter - the filter generated by $\F$.
The collection ${\cal U}(x)$ of all neighborhoods of a point $x$ is a filter!
For every sequence $x_n$ in $X$, the sets $F_n\colon=\{x_m:m\geq n\}$ form a filter basis and the filter generated by these sets is called the Frechetfilter of the sequence.
The sequence $x_n$ converges to $x\in X$, if for every neighborhood $U$ of $x$ there exists some index $n_0$, such that for all $n\geq n_0$: $x_n\in U$. This holds if and only if the Frechetfilter of the sequence $x_n$ is finer than the neighborhood filter of $x$.
A filter $\F$ on a topological space $X$ converges to $x\in X$ if ${\cal U}(x)\sbe\F$ - we will denote this by $\F\rar x$. Accordingly, a filter basis $\B$ converges to $x$, if the filter generated by $\B$ converges to $x$.
A topological space $X$ is called a Hausdorff space, if for all pairs $x,y\in X$ of distinct points we can find neighborhoods $U$ and $V$ of $x$ and $y$ respectively, such that $U\cap V=\emptyset$.
In a Hausdorff space $X$ a convergent filter $\F$ converges to exactly one point; moreover every finite set is closed and every subspace of $X$ is again Hausdorff.
Let $A$ be a subset of a topological space $X$. Then $x\in X$ is in the closure $\cl A$, if there exists a filter basis $\B$ on $A$, which converges in $X$ to $x$.
A mapping $f:X\rar Y$ is continuous at $x$, if for every filter $\F$ on $X$, which converges to $x$ the filter basis $f(\F)$ converges to $f(x)$.
An accumulation point of a filter basis $\B$ on a topological space $X$ is any point in the set $$ x\in\bigcap\{\cl F:F\in\B\}~. $$
Let $\B$ be a filter basis on the topological space $X$. $x$ is an accumulation point of $\B$ if and only if there exists some filter basis ${\cal G}\spe\B$, such that ${\cal G}\rar x$.
The set of all filters on a given set $X$ is inductively ordered by inklusion, thus, by Zorn's lemma we can find for every filter $\F$ on $X$ a maximal filter $\F_0$ containing $\F$ - $\F_0$ is in general not unique. Every maximal filter on $X$ is called an ultra filter on $X$.
A filter $\F$ on a set $X$ is an ultra filter, if and only if for any subset $A$ of $X$ we have: $A\in\F$ or $A^c\in\F$. Moreover if $\F$ is an ultra filter on $X$ and $f:X\rar Y$ any mapping, then $f(\F)$ is basis of an ultra filter on $Y$.

Final and initial topologies

 For the following let $X$ be a set, $Y_\a$ topological spaces, $f_\a:X\rar Y_\a$ and $g_\a:Y_\a\rar X$ arbitary mappings.
The inital topology ${\cal T}_i$ on $X$ with respect to the familly of mappings $f_\a$ is the coarserst topology on $X$, which makes all the mappings $f_\a:(X,{\cal T})\rar Y_\a$ continuous. The final topology on $X$ with respect to the family of mappings $g_\a$ is the finest topology ${\cal T}_f$ on $X$, such that all mappings $g_\a:Y_\a\rar(X,{\cal T})$ are continuous.
Let ${\cal T}_\a$ be the topology on $Y_\a$. Then $\bigcup_{\a\in I}f_\a^{-1}({\cal T}_\a)$ is a subbasis of the initial topology on $X$ and $\bigcap_{\a\in I}\{U\sbe X:g_\a^{-1}(U)\in{\cal T}_\a\}$ is the final topology on $X$.
If $X$ carries the initial Topology with repect to the mappings $f_\a:X\rar Y_\a$, then $h:Z\rar X$ is continuous if and only if all mappings $f_\a\circ h:Z\rar Y_\a$ are continuous.
If $X$ carries the final topology with respect to the mappings $g_\a:Y_\a\rar X$, the $h:X\rar Z$ is continuous if and only if all mappings $h\circ g_\a:Y_\a\rar Z$ are continuous.
The product topology on the cartesian product $X=\prod X_\a$ of a family of topological spaces $X_\a$ is the initial topology on $X$ with respect to the canonical projections $\Prn_\a:X\rar X_\a$. Suppose $U\sbe X$ is an open subset of $X$ and $x=(x_\a)\in U$, then by definition there is a finite subset $J$ von $I$ and open neighborhoods $U_\a$ of $x_\a$, such that for all $\a\notin J$: $U_\a=X_\a$ and $\prod U_\a\sbe U$.
A topological space $X$ is Hausdorff if and only if the diagonal $\D_X\colon=\{(x,x):x\in X\}$ is closed in $X\times X$.
Suppose $f,g:X\rar Y$ are continuous, $Y$ Hausdorff and $D$ a dense subset of $X$. If for all $x\in D$: $f(x)=g(x)$, then $f=g$.
If $X$ carries the initial topology with respect to the mappings $f_\a:X\rar Y_\a$, then a filter basis $\F$ on $X$ converges to $x$, if all $f_\a(\F)\to f_\a(x)$.
Suppose all spaces $X_\a$ are Hausdorff and $X$ carries the initial topology with respect to the mappings $f_\a:X\rar X_\a$. Then $X$ is Hausdorff if and only if for each pair $x\neq y\in X$ there exists some $\a\in I$, such that $f_\a(x)\neq f_\a(y)$. A family of mappings with the later property is said to separate points. 2. If $f_\a$ separates points, then $X$ is homeomorphic to a subset of $\prod X_\a$. In particular $X$ is metrizable (and separable) if $I$ is at most countable and all spaces $X_\a$ are metrizable (and separabel).
The topology on the disjoint union $Z\colon=\coprod X_\a$ is the final topology on $Z$ with respect to the canonical inclusions $j_\a:X_\a\rar Z$. A subset $B\sbe Z$ is open (closed), if and only if for all $\a$: $j_\a^{-1}(B)$ is open (closed) in $X_\a$.
Let $R\sbe X\times X$ be an equivalence relation on the topological space $X$ and denote by $\pi:X\rar X/R$ the quotient map. The space $X/R$ equipped with the final topology with respect to the quotient mapping is called the quotient space $\wh X\colon=X/R$ of $X$. We put: $R(x)\colon=\pi^{-1}(\pi(x))=\{y\in X:\,xRy\}$. $\pi$ is open (closed), if and only if for all open (closed) subsets $B$ of $X$ the following set is open (closed) in $X$: $$ R(B)\colon=\pi^{-1}(\pi(B))=\{y\in X:\,\exists x\in B:\ xRy\}~. $$
Suppose $f:X\rar Y$ is continuous, open (closed) and onto. Then $Y$ carries the final topology with respect to $f$. Let $R$ be the equivalence relation $x_1Rx_2$: $\Lrar f(x_1)=f(x_2)$, then $\wh f:X/R\rar Y$ is a homeomorphism.
A surjection $f:X\rar Y$ is called a quotient map if $Y$ carries the final topology with respect to $f$. Hence the previous proposition states that an open (or closed) continuous surjection $f:X\rar Y$ is a quotient map.
Let $R\sbe X\times X$ be an equivalence relation on $X$. 1. If $\wh X\colon=X/R$ is Hausdorff, then $R$ is closed in $X\times X$. 2. If $R\sbe X\times X$ is closed and $\pi:X\rar X/R$ open, then $X/R$ is Hausdorff.
Let $R$ be an equivalence relation on $X$. Then the quotient mapping $\pi:X\rar X/R$ is closed, if and only if for all $x\in X$ and all open supersets $U$ of $R(x)$ there exists another open set $V$, such that $R(x)\sbe V\sbe U$ and $R(V)=V$.

Collapsing

For a closed subset $A$ of a topological space $X$ we define the equivalence relation $R_A\colon=\D_X\cup(A\times A)$. $X/R_A$ is called the $A$-collaps $X/A$ of $X$. $R_A\sbe X\times X$ is closed and for all subsets $C$ von $X$: $R(C)=C$ or $R(C)=C\cup A$. The quotient mapping $X\rar X/A$ is always closed (but not necessarily open). $X/A$ need not be Hausdorff!

Suspension

Put $I=[0,1]$ and $Z\colon=X\times I$. Define an equivalence relation by $R(x,s)=\{(x,s)\}$ if $0< s< 1$ and $R(x,0)=X\times\{0\}$, $R(x,1)=X\times\{1\}$. The quotient space $X\times I/R$ is called the suspension $SX$ of $X$. $R$ is closed and for every subset $D$ of $X\times I$ the subset $R(D)$ coincides with one of the following sets: $$ D, \quad D\cup(X\times\{0\}), \quad D\cup(X\times\{1\}), \quad D\cup(X\times\{0,1\}), $$ Therefore the quotient mapping $X\times I\rar SX$ is closed but it general it fails to be open.

Attaching

Suppose $X,Y$ are topological spaces, $A\sbe X$ closed, $f:A\rar Y$ continuous and $Z$ the disjoint union of $X$ and $Y$. On $Z$ define an equivalence relation $R$ by $$ zRw\colon\Lrar z=w \quad\mbox{or}\quad z\in A\mbox{ and }w=f(z) \quad\mbox{or}\quad w\in A\mbox{ and }z=f(w)~. $$ $Z/R$ and $f$ respectively are called the attaching space $X\cup_f Y$ of $X$ and $Y$ and the attaching map respectively - $X$ and $Y$ will be attached along $A$ by $f$; if $f:A\rar Y$ is an embedding, then we say that $X\cup_f Y$ are attached along $A$. Collapsing is a special case of attaching! If both $X$ and $Y$ are Hausdorff, then $R$ is a closed subset and the quotient map $Z\rar X\cup_f Y$ is closed if and only if $f$ is closed. Now suppose that $g:X\rar T$ and $h:Y\rar T$ are continuous. These mappings can be attached to a mapping $G:X\cup_f Y\rar T$ if and only if $h\circ f=g|A$ and by the definition of the quotient topology $G$ ist continuous.

Join

Suppose $X_1,\ldots,X_{n+1}$ are Hausdorff and let $\D^n\colon=\{s\in[0,1]^{n+1}:s_1+\cdots+s_{n+1}=1\}$ be the $n$-dimensional simplex. The join $X_1*\cdots*X_{n+1}$ of the spaces $X_1,\ldots,X_{n+1}$ is the quotient space $\prod X_j)\times\D^n/R$ where $$ R(x_1,\ldots,x_{n+1},s_1,\ldots,s_{n+1}) =s_1(x_1)\times\cdots\times s_{n+1}(x_{n+1}) \times\{(s_1,\ldots,s_n)\} $$ and $s_j(x_j)=X_j$ if $s_j=0$ and $s_j(x_j)=\{x_j\}$ otherwise. Any point in $X_1*\cdots*X_{n+1}$ will be represented by a formal sum: $\sum_js_jx_j$ skipping any term of the form $0x_j$. $$ \sum_js_jx_j=\sum t_jy_j \colon\Lrar \forall j:\ s_j=t_j \quad\mbox{and}\quad s_j(x)=t_j(y)~. $$ The quotient map is closed but fails to be open in general.

Projective Space

For what is going to follow put $\bK=\R,\C$ or $\H$. On $\bK^{n+1*}\colon=\bK^{n+1}\sm\{0\}$ we define an equivalence relation $xRy\colon\Lrar\exists\l\in\bK:y=\l x$ and denote by $\pi$ the quotient map. $P^n(\bK)\colon=\bK^{n+1*}/R$ is called the $n$-dimensional real or complex or hypercomplex projective space -- $P^n(\bK)$ is the space of all $1$-dimensional subspaces of $\bK^{n+1}$. By Definition a set $V\sbe P^n(\bK)$ is open if and only if $\pi^{-1}(V)$ is open in $\bK^{n+1*}$. Obviously $P^n(\bK)$ is Hausdorff and $\pi$ is open.
Homogeneous coordinates: Let $a\in\bK^{n+1}$ be normed and denote by $a^\perp$ the space orthogonal to $a$, i.e. $E\colon=a^\perp=\{x\in\bK^{n+1}:\la x,a\ra=0\}$. Now put $$ \Phi_a(x)\colon=\la x,a\ra^{-1}(x-\la x,a\ra a) $$ then $\Phi_a:\bK^{n+1}\sm E\rar E$ is continuous, open and onto. Geometrically $\Phi_a(x)+a$ is the intersection of the space generated by $x$ and the hyperplane $E+a$. Since for all $\l\in\bK\sm\{0\}$: $\Phi_a(\l x)=\Phi_a(x)$, there is exactly one map $\vp_a:\pi(E)^c\rar E$, such that $\Phi_a=\vp_a\circ\pi$; it follows that $\vp_a:\pi(E)^c\rar E$ is a homeomorphism. Now put in particular $a=e_j$, $j=1,\ldots,n+1$, e.g.: $$ \Phi_j(x)\colon=\Phi_{e_j}(x) =(x_j^{-1}x_1,\ldots,x_j^{-1}x_{j-1},x_j^{-1}x_{j+1},\ldots,x_j^{-1}x_{n+1}) $$ These $n$-tuple of numbers are called the homogeneous coordinates of $\pi(x)\in\pi(E)^c$.

Quotient spaces with respect to discrete groups

Suppose $\G$ is a discrete group, $X$ a Hausdorff space and $(x,g)\mapsto xg$ a continuous map from $X\times\G$ into $X$, such that for all $x\in X$ and all $g,h\in\G$: $xe=x$ and $x(gh)=(xg)h$ -- we say: the group $\G$ operates (from the right) on $X$. Putting $R_\G(x)\colon=x\G$ we get again an equivalence relation; the quotient $X/R_\G$ is usually denoted by $X/\G$ and is called the orbit space, for it's elements are the orbits $\{xg:g\in\G\}$ of the ponints $x\in X$. The quotient map $\pi:X\rar X/\G$ is always open but need not be closed. $\pi$ is closed if and only if for every open superset $U$ of $x\G$ we can find an open neighborhood $V$ of $x$, such that $V\G\sbe U$ -- for finite groups this is always true. $X/\G$ is Hausdorff, if and only if $R_\G$ is closed in $X\times X$ and this holds if and only if for all $x\in X$ and all $y\notin x\G$ there are neighborhoods $U$ of $x$ and $V$ of $y$ such that $V\cap U\G=\emptyset$. If moreover for all $x\in X$ there exists some open neighborhood $U$ such that $\{g\in\G:U\cap Ug\neq\emptyset\}=\{e\}$, then $\G$ is said to operate properly on $X$.

Quotient spaces of normed spaces

Let $X$ be a normed space, $Y$ a closed subspace and define an equivalence relation: $xRy\colon\Lrar -x+y\in Y$. $R$ is closed and the quotient map $\pi:X\rar X/R$ is open, thus $X/R$ is Hausdorff; $X/R$ will be denoted by $X/Y$. On the quotient space $X/Y$ we put $$ \norm{\wh x}\colon=\inf\{\norm{x+y}:y\in Y\}~. $$ Since $Y$ is closed this is actually a norm on $X/Y$. Let $\wh X$ be the space $X/Y$ furnished with this norm. By definition we get for the open unit balls: $\pi(B_X)\sbe B_{\wh X}$. Conversely, for any $\wh x\in B_{\wh X}$ we have $\norm{\wh x}<1$ and thus there is some $x\in X$ such that $\pi(x)=\wh x$ and $\norm x<1$. It follows that $B_{\wh X}\sbe\pi(B_X)$, i.e. we also have: $B_{\wh X}\sbe\pi(B_X)$. The quotient map $\pi:X\rar\wh X$ is thus both continuous and open, hence the identity map $X/Y\rar\wh X$ is a homeomorphism: in particular $X/Y$ is normable.

Surfaces

 Let $P$ a closed (convex) polygon in $\R^2$ with $2n$ vertices. Each of the edges will be assign both a letter $\{a,b,\ldots\}$ and a direction $\rar$ (clockwise) or $\lar$ (anti-clockwise). For an edge with letter $a$ and anti-clockwise direction we will write $a$ and for an edge with symbol $b$ and clockwise direction: $b^{-1}$. This way we get for each polygon a word of length $2n$ consisting of symbols of the form $ab^{-1}c^{-1}d\ldots$, more specifically $$ \TT^2:bab^{-1}a^{-1},\quad S^2:abb^{-1}a^{-1},\quad P^2(\R):ba^{-1}ba^{-1}b,\quad K:ba^{-1}b^{-1}a^{-1}~. $$ These words are subjected to a single condition: each letter must occure exactly twice - $a$ and $a^{-1}$ are interpreted as the same letter. Now we define an equivalence relation $\sim$ on the set of edeges as follows: two edges are identified, if both are assigned the same letter and this identification will be achieved by an affine mapping which maps the starting point of the first edge onto the starting point of the second and the endpoint of the first onto the endpoint of the second. Each word $A$ thus represents a topological space $X=P/\sim$. Which words represent homeomorphic spaces? Obviously, we can permute the letters of a word cyclically without changing the topology. If $A=a\ldots b$, then we put $A^{-1}=b^{-1}\ldots a^{-1}$, where, as customary $(a^{-1})^{-1}\colon=a$. Apparently $A^{-1}$ and $A$ represent homeomorphic spaces. Instead of straight lines we can also use arclike curves to represent edges graphically. In this manner we see, that both the sphere $S^2$ and the projective space $P^2(\R)$ can be represented by words of length $2$: $aa^{-1}$ and $aa$ respectively. Moreover, if $A$ is a word consisting of mutually distict letters, then the word $AA^{-1}$ represents the sphere $S^2$.
Attaching two tori: remove from each of the tori a 'simple' disc-like area $D$ and attach the remaining parts along the boundary $e\colon=\pa D$ of $D$. Thus $\TT\sharp\TT$ is going to be represented by the word $d^{-1}c^{-1}dca^{-1}b^{-1}ab$. More generally, let $A=a\ldots b$ be a word representing the topological space $X$ and let $e$ be a character not occuring in $A$. Then the word $Ae$ represents the space $X\sm D$; similarly $B=c\ldots de$ represents another space $Y\sm D$ - $A$ and $B$ must not have any letter in common! the attaching of $X$ and $Y$ along the boundary $e$ of $D$ is thus represented by $a\ldots bd^{-1}\ldots c^{-1}$, which is $AB^{-1}$. Here are a few examples: $P^2(\R)\sharp P^2(\R)$ is represented by: $aabb$, $\TT^2\sharp P^2(\R)$ is represented by: $a^{-1}b^{-1}abcc$ and $S^2\sharp S^2$ is represented by: $aa^{-1}b^{-1}b=a^{-1}b^{-1}ba=(ba)^{-1}ba$, which represents again a sphere.
The Klein bottle $K=ba^{-1}b^{-1}a^{-1}$ is homeomorphic to the attaching of two projective spaces. Indeed, after cutting $K$ along the diagonal $x$, we mirror the upper triangle about the second axis and attach both triangles along the edge $a$. By means of our letter algebra this comes down to: $$ K=(ba^{-1}x^{-1})(xb^{-1}a^{-1}) =(x^{-1}ba^{-1})(abx^{-1}) =x^{-1}bbx^{-1} =x^{-1}x^{-1}bb $$ where the equality sign indicates that both sides represent homeomorphic spaces.
Suppose $A,B$ are words, such that neither contains the letters $a$ or $x$. Then the words $aaBA$ and $xAxB^{-1}$ represent homeomorphic spaces.

Connected spaces
A topological $X$ space is said to be connected if for any pair $U,V$ of open subsets satisfying: $U\cup V=X$ and $U\cap V=\emptyset$ we have: $U=X$ or $V=X$. $X$ is said to be locally connected if every point has a basis consisting of connected neighborhoods.
$X$ is connected if and only if every continuous function $f:X\rar\{0,1\}$ is constant.
1. If $A$ is a connected subset of $X$, then $\cl A$ is connected. 2. Suppose $A,B$ are connected subsets of $X$, such that $\cl A\cap B\neq\emptyset$. Then $A\cup B$ is connected. 3. Let $Z_\a$ be connected subsets of $X$, such that $\bigcap Z_\a\neq\emptyset$. Then $\bigcup Z_\a$ is connected.
A subset $A$ of $\R$ is connected if and only if $A$ is an interval.
If $X$ is connected and $f:X\rar Y$ continuous, then $f(X)$ is connected.
In a normed space the distance of any point $x$ to a closed subset $A$ equals the distance of $x$ to the boundary $\pa A$ of $A$; we are going to show, that this property is essentially equivalent to the connectedness of the unit ball: So let's consider a metric space $X$ and a closed subset $A$ of $X$. Assume that for some point $x\in A^c$: $d_A(x) < r < d_{\pa A}(x)$. Then: $B_r(x)\cap A\neq\emptyset$ and $$ \emptyset=B_r(x)\cap\pa A =B_r(x)\cap(A\sm A^\circ) \quad\mbox{i.e.}\quad B_r(x)\cap A=B_r(x)\cap A^\circ $$ This precisely means that $B_r(x)=(B_r(x)\cap A^c)\cup(B_r(x)\cap A^\circ)$, i.e. $B_r(x)$ is the disjoint union of two none empty open subsets, i.e. $B_r(x)$ is not connected. Thus if every open ball $B_r(x)$ is connected, then for any closed subset $A$ of $X$ we have: $d_A=d_{\pa A}$.
A topological space $X$ is said to be path-connected, if for each pair $x,y$ of points in $X$ there exists a continuous curve $c:[0,1]\rar X$, such that $c(0)=x$ and $c(1)=y$. $X$ is said to be locally path-connected, if every point has a basis consisting of path-connected neighborhoods.
If $X$ is connected and locally path-connected, then $X$ is path-connected.
An open and connected subset of a topological space $X$ is commonly called a domain. A domain $U$ in a normed space is always path-connected. Moreover in this case any two points can be joined both by a smooth curve and a polygon.
Suppose $X_\a$, $\a\in I$ are connected spaces. Then $\prod X_\a$ is connected.
For any point $x$ in a topological space $X$ the connected component of $x$ is the union of all connected subsets of $X$ containing $x$.
Let $X$ be a separable locally connected space. Then there exists an at most countable family of pairwise disjoint connected subsets $Z_n$, which are both open and closed (so called clopen sets), such that $X=\bigcup Z_n$.
Suppose $U$ is an open subset of the reals, then $U$ is the union of a sequence of pairwise disjoint open intervals.
Let $X$ be a connected space and let $A$ be a closed subset of $X$ such that $A^\circ=\emptyset$. If for every point $x\in X$ we can find an open and connected neighborhood $V$, such that $V\sm A$ is connected, then $X\sm A$ is connected .

Manifolds

A metrizable, separable space $M$ is said to be a topological manifold, if every point $x\in M$ has a neighborhood homeomorphic to $\R^n$ - $n=n(x)$ may depend on $x$.
If $M$ is a connected topological manifold, then the function $x\mapsto n(x)$ must be constant - this is a consequence of the invariance of domains theorem in algebraic topology: two open subsets of $\R^n$ and $\R^m$ respectively can only be homeomorphic if $n=m$. In this case we call the uniquely defined number $n$ the dimension of the manifold - sometimes expressed by adding a superscript: $M^n$.
Let $U_k$ be an open cover of the $n$-dimensional topological manifold $M$ and $\vp_k:U_k\rar\R^n$ homeomorphisms. Then the set of pairs $(U_k,\vp_k)$ is called an atlas of $M$ and each pair $(U_k,\vp_k)$ is said to be a chart. The projective spaces $P^n(\R)$, $P^n(\C)$ and $P^n(\H)$ are topological manifolds of dimension $n$, $2n$ and $4n$ respectively.
A smooth manifold $M$ is a topological manifold equipped with an atlas $(U_k,\vp_k)$ such that the following property holds: for all $j,k$ the so called transition map $\vp_j\circ\vp_k^{-1}$ is smooth. If all the transition maps are holomorphic, $M$ is said to be a complex manifold.
Suppose $M,N$ are smooth (complex) manifolds. A smooth (holomorphic) map $F:M\rar N$ is a mapping such that for every $x\in M$ there exists a chart $(U,\vp)$ about $x$ and a chart $(V,\psi)$ about $y=F(x)$, such that the mapping $\psi\circ F\circ\vp^{-1}$ is smooth (holomorphic).
← Metric Spaces → Regular and Normal Spaces
<Home>
Last modified: Tue Sep 12 18:42:24 CEST 2023