0.1  (§22) The quotient topology

• Rmk. Injections gave us subspaces and embeddings. Surjections will give quotients — the dual construction, with finest in place of coarsest.

• Def. An equivalence relation $\sim$ on $X$: reflexive, symmetric, transitive. Classes $[x]=\{y\mid x\sim y\}$; they are nonempty, cover $X$, are mutually disjoint. $X/\sim =\{[x]\}$, canonical surjection $\pi (x)=[x]$.

• Lem. Any surjection $f:X\to Y$ arises this way: set $x\sim y\iff f(x)=f(y)$; then $h([x])=f(x)$ is a bijection $X/\sim \to Y$ with $f=h\circ \pi$.

• $\mathrm{★}$ Def. Quotient topology on $Y$ from a surjection $f:X\to Y$: $U\subset Y\text{ open}\iff f^{-1}(U)\text{ open in }X.$ A surjection carrying this topology is a quotient map.

  Proof. It is a topology because $f^{-1}$ commutes with $\cup$ and $\cap$. Quotient maps are continuous by construction.

• Lem. The quotient topology is the finest topology on $Y$ making $f$ continuous.

• Lem. $f$ surjective is a quotient map $\iff$ ($A\subset Y$ closed $\iff f^{-1}(A)$ closed).

• Lem. A bijective quotient map is a homeomorphism, and conversely.

• Def. $f$ is an open map if $f(U)$ is open for all open $U$; a closed map if $f(A)$ is closed for all closed $A$.

• Lem. $f$ open $\iff$ $f(B)$ open for every basis element $B$.

• $\mathrm{★}$ Lem. A surjective open map is a quotient map. So is a surjective closed map.

  Proof. $U=f(f^{-1}(U))$ by surjectivity; then apply openness (resp. closedness).

• $\mathrm{★}$ Ex. $f:[0,1]\to S^1$, $f(t)=(\cos 2\pi t,\sin 2\pi t)$: continuous, surjective, closed (compactness), hence a quotient map. Not open. With $0\sim 1$: $[0,1]/\sim \cong S^1$.

• $\mathrm{★}$ Ex. Torus. $g=f\times f:{[0,1]}^2\to S^1\times S^1$ is a quotient map; identify $(s,0)\sim (s,1)$ and $(0,t)\sim (1,t)$. Classes: the four corners; the paired edge points; the interior singletons. So $T^2\cong {[0,1]}^2/\sim$.

• Ex. $ℝ\to \{n,z,p\}$ by sign. Quotient topology $\{\mathrm{\varnothing },\{n\},\{p\},\{n,p\},Y\}$: not Hausdorff; $z$ is the only closed point.

• Ex. ${\pi }_1:{ℝ}^2\to ℝ$ is open (hence a quotient map) but not closed: the hyperbola $C=\{xy=1\}$ is closed, ${\pi }_1(C)=ℝ-\{0\}$ is not.

• Caution. Ex. Restricting a quotient map can destroy the property. With $A=C\cup \{(0,0)\}$, the map ${\pi }_1|A:A\to ℝ$ is a continuous surjection but not a quotient map: $\{0\}$ is not open in $ℝ$, yet its preimage $\{(0,0)\}$ is open in $A$.

• Thm. $f:X\to Y$ a quotient map, $B\subset Y$, $A=f^{-1}(B)$ saturated, $g=f|A:A\to B$. Then $g$ is a quotient map if either (1) $A$ is open, or $A$ is closed; or (2) $f$ is an open map, or a closed map.

  Proof. Two identities do the work: $g^{-1}(V)=f^{-1}(V)$ for $V\subset B$, and $g(A\cap U)=B\cap f(U)$. Both use $A=f^{-1}(B)$.

• Rmk. Composites of quotient maps are quotient maps. Caution. Products of quotient maps need not be; some local compactness hypothesis is required. Quotients of Hausdorff spaces need not be Hausdorff.

• $\mathrm{★}$ Thm. (Universal property.) $f:X\to Y$ a quotient map, $h:X\to Z$ constant on the fibres of $f$. Then $h$ factors uniquely as $h=g\circ f$; moreover $g$ is continuous $\iff h$ is, and $g$ is a quotient map $\iff h$ is.

• Cor. $h:X\to Z$ a continuous surjection, $x\sim y\iff h(x)=h(y)$. Then $h$ induces a continuous bijection $g:X/\sim \to Z$; $g$ is a homeomorphism $\iff h$ is a quotient map; and if $Z$ is Hausdorff so is $X/\sim$.

1  Connectedness and Compactness

1.1  (§23) Connected spaces

• Def. For disjoint spaces $C,D$: the disjoint union $C\bigsqcup D$ carries the topology consisting of those $W$ with $C\cap W$ open in $C$ and $D\cap W$ open in $D$; the finest making both inclusions continuous.

• Rmk. $C\bigsqcup D$ is also called the sum, or coproduct, of $C$ and $D$; it has a universal property dual to that of the product $C\times D$. Every $X$ is trivially $C\bigsqcup D$ with $C=\mathrm{\varnothing }$ or $D=\mathrm{\varnothing }$; if $X\cong C\bigsqcup D$ non-trivially, $X$ is disconnected.

• $\mathrm{★}$ Def. A separation of $X$: a pair $U,V$ of disjoint nonempty open sets with $U\cup V=X$. $X$ is connected if no separation exists.

• Rmk. Being connected is a topological property. The empty space needs care: some authors declare $\mathrm{\varnothing }$ not connected, much as $1$ has no proper factors yet is not counted as a prime.

• $\mathrm{★}$ Lem. $X$ is connected $\iff$ the only clopen subsets are $\mathrm{\varnothing }$ and $X$.

  Proof. In a separation $V=X-U$, so “$U,V$ both open” $=$ “$U$ clopen”, and “both nonempty” $=$ “$U\ne \mathrm{\varnothing },X$”. Cleanest working form of the definition.

• Lem. $U,V$ a separation $\Rightarrow X\cong U\bigsqcup V$; conversely $X=C\bigsqcup D$ with both nonempty gives a separation.

• Lem. Equivalent formulation: a separation is a pair of disjoint nonempty sets $A,B$ with $A\cup B=X$, neither containing a limit point of the other ($\overline{A}\cap B=\mathrm{\varnothing }=A\cap \overline{B}$).

• Ex. One-point spaces are connected. Sierpiński $\{a,b\}$ is connected ($\{a\}$ open not closed, $\{b\}$ closed not open). $[-1,0)\cup (0,1]$ is disconnected.

• Rmk. Coming in §24: $ℝ$ is connected, as is every interval $[a,b]$, $[a,b)$, $(a,b]$, $(a,b)$ for $-\mathrm{\infty }\le a\le b\le \mathrm{\infty }$.

• Ex. Every $X\subset \mathbb{Q}$ with $\ge 2$ points is disconnected: pick irrational $a$ between in $X$ and cut at $a$.

• $\mathrm{★}$ Lem. $U,V$ a separation of $X$, $A\subset X$ connected $\Rightarrow A\subset U$ or $A\subset V$.

  Proof. $A\cap U$ is clopen in $A$. The single most-used lemma of the section.

• $\mathrm{★}$ Thm. A union of connected subspaces with a point in common is connected.

• $\mathrm{★}$ Thm. $A\subset B\subset \overline{A}$, $A$ connected $\Rightarrow B$ connected. (Adding limit points cannot disconnect.)

• $\mathrm{★}$ Thm. The continuous image of a connected space is connected.

1.2  (§24) Connected subspaces of the real line

• Def. $C\subset ℝ$ is convex if in $C$ $\Rightarrow [a,b]\subset C$.

• Ex. The convex subsets of $ℝ$: $\mathrm{\varnothing }$; the intervals $(a,b),[a,b),(a,b],[a,b]$; the rays; and $ℝ$.

• $\mathrm{★}$ Thm. Every convex $C\subset ℝ$ is connected. In particular $ℝ$ is connected.

  Proof. Reduce to $[a,b]$. Given $[a,b]=U\bigsqcup V$ with $a\in U,b\in V$, let $c=supU$. Then $c\in \overline{U}=U$, so $c\ne b$; but $U$ open gives points $>c$ in $U$ — contradicting the supremum. Least upper bound property is doing all the work.

• $\mathrm{★}$ Thm. (Intermediate value theorem.) $f:X\to ℝ$ continuous, $X$ connected, $r$ between $f(a)$ and $f(b)$ $\Rightarrow \exists c$ with $f(c)=r$.

  Proof. Otherwise $f^{-1}(-\mathrm{\infty },r)$ and $f^{-1}(r,\mathrm{\infty })$ separate $X$.

• Def. A path from $x$ to $y$: a map $f:[a,b]\to X$, $f(a)=x$, $f(b)=y$. $X$ is path connected if any two points are joined by a path.

• $\mathrm{★}$ Lem. Path connected $\Rightarrow$ connected.

  Proof. A separation of $X$ would pull back to a separation of $[a,b]$.

• Lem. The continuous image of a path connected space is path connected.

• Def. $C$ in a real vector space is convex if $(1-t)x+ty\in C$ for $x,y\in C$, $t\in [0,1]$.

• Ex. Convex subsets of ${ℝ}^n$ are path connected — e.g. the unit ball $B^n=\{\Vert x\Vert \le 1\}$.

• Ex. ${ℝ}^n-\{0\}$ is path connected for $n\ge 2$ (not for $n=1$). Hence $S^{n-1}=\{\Vert x\Vert =1\}$ is path connected for $n\ge 2$, being the image of $x\mapsto x/\Vert x\Vert$. ($S^0=\{\pm 1\}$ is not.)

• $\mathrm{★}$ Ex. Topologist’s sine curve. ,   $\overline{S}=S\cup V$ with $V=\{0\}\times [-1,1]$. $S$ is connected (continuous image of $(0,1]$), hence so is $\overline{S}$. But $\overline{S}$ is not path connected.

  Proof. A path from $V$ into $S$: reparametrise so $f(0)\in V$, $f(t)\in S$ for $t>0$. Write $f=(x,y)$. Construct $t_n\to 0$ with $y(t_n)={(-1)}^n$ via the IVT applied to $x$. Then $y(t_n)$ does not converge — contradicting continuity.