We next need to discuss several ways of constructing new topological spaces from spaces that we already know about.
Let $Y$ and $Z$ be disjoint sets, and put $X=Y\cup Z$. Suppose we are given topologies on $Y$ and $Z$, which we use to regard them as topological spaces. We then declare that a subset $U\subseteq X$ is open iff $U\cap Y$ is open in the given topology on $Y$, and $U\cap Z$ is open in the given topology on $Z$. This is easily seen to give a topology on $X$, which we call the coproduct topology.
We now have a topology on $X$, and $Y$ is a subset of $X$, so we can use Definition 3.25 to define a subspace topology on $Y$. It is easy to see that this is just the the same as the original topology on $Y$. Similarly, if we regard $Z$ as a subspace of $X$, then the subspace topology is just the same as the original topology. In particular, the inclusion maps $Y\stackrel{\mathit{i}}{\to}X\stackrel{\mathit{j}}{\leftarrow}Z$ are continuous.
The category-theoretic viewpoint encourages us to ask the following kind of question: whenever we construct a new object $X$ in a category $\mathcal{C}$, we should try to prove theorems describing the set $\mathcal{C}(T,X)$ of morphisms into $X$, and/or the set $\mathcal{C}(X,T)$ of morphisms out of $X$, for an arbitrary object $T\in \mathcal{C}$.
Let $X$, $Y$ and $Z$ be as above, and let $T$ be another topological space. To describe a function $f\mathrm{:}X\mathrm{\to}T$, it is enough to specify the restricted functions $g\mathrm{=}{f\mathrm{|}}_{Y}\mathrm{:}Y\mathrm{\to}T$ and $h\mathrm{=}{f\mathrm{|}}_{Z}\mathrm{:}Z\mathrm{\to}T$. Moreover, the combined map $f$ is continuous (with respect to the coproduct topology) iff $g$ and $h$ are continuous (with respect to the originally given topologies on $Y$ and $Z$). Thus, we have a bijection between $\mathrm{Top}\mathit{}\mathrm{(}X\mathrm{,}T\mathrm{)}$ and $\mathrm{Top}\mathit{}\mathrm{(}Y\mathrm{,}T\mathrm{)}\mathrm{\times}\mathrm{Top}\mathit{}\mathrm{(}Z\mathrm{,}T\mathrm{)}$.
Consider an open set $A\subseteq T$. We then find that ${f}^{-1}(A)\cap Y={g}^{-1}(A)$ and ${f}^{-1}(A)\cap Z={h}^{-1}(A)$. Thus, ${f}^{-1}(A)$ is open in $X$ iff ${g}^{-1}(A)$ is open in $Y$ and ${h}^{-1}(A)$ is open in $Z$. Thus, $f$ is continuous iff this condition holds for all $A$ iff $g$ and $h$ are continuous. Thus, to give a continuous map $f\in \mathrm{Top}(X,T)$ is the same as to give a pair of continuous maps $g\in \mathrm{Top}(Y,T)$ and $h\in \mathrm{Top}(Z,T)$, or in other words a pair $(g,h)\in \mathrm{Top}(Y,T)\times \mathrm{Top}(Z,T)$. ∎
There is a general notion of coproduct objects in category theory. The above proposition can be interpreted as saying that $X$ (equipped with the coproduct topology) is the coproduct of $Y$ and $Z$ in this general categorical sense.
Let $X$, $Y$, $Z$ and $T$ be as above, and consider a function $p\mathrm{:}T\mathrm{\to}X$. Then $p$ is continuous iff
The sets $A={p}^{-1}(Y)$ and $B={p}^{-1}(Z)$ are open in $T$.
The restricted map $q={p|}_{A}:A\to Y$ is continuous with respect to the subspace topology on $A\subseteq T$.
The restricted map $r={p|}_{B}:B\to Z$ is continuous with respect to the subspace topology on $B\subseteq T$.
Left as an exercise. ∎
Let $Y$ and $Z$ be topological spaces, and consider the product
$$X=Y\times Z=\{(y,z)|y\in Y\text{and}z\in Z\}.$$ |
For a point $a=(b,c)\in Y\times Z$, a box around $a$ means a set of the form $V\times W$, where $b\in V$ and $c\in W$ and $V$ is open in $Y$ and $W$ is open in $Z$. We declare that a set $U\subseteq X$ is open iff for all $a\in U$, there is a box around $a$ that is contained in $U$.
The picture shows the case where $Y=[0,2]$ and $Z=[0,1]$. We have indicated a subset $U\subset Y\times Z$, two points $a,{a}^{\prime}\in U$, a box around $a$ contained in $U$, and a box around ${a}^{\prime}$ contained in $U$.
Suppose that $V$ is an open subset of $Y$ and $W$ is an open subset of $Z$. Then $V\mathrm{\times}W$ is open in $Y\mathrm{\times}Z$.
Consider a point $a=(b,c)\in V\times W$, so $b\in V$ and $c\in W$. We need to show that there is a box $B$ around $a$ such that $B\subseteq V\times W$. We can just take $B=V\times W$. ∎
Definition 7.6 specifies a topology on $Y\mathrm{\times}Z$ (which we call the product topology).
First suppose we have a family of open sets ${U}_{i}\subseteq Y\times Z$; we must show that the union ${U}^{*}={\bigcup}_{i}{U}_{i}$ is open. Consider a point $a\in {U}^{*}$. By the definition of the union, this means that $a\in {U}_{i}$ for some $i$. As ${U}_{i}$ is open, we can choose a box $B=V\times W$ around $a$ such that $B\subseteq {U}_{i}$. Now ${U}_{i}\subseteq {U}^{*}$, so we also have $B\subseteq {U}^{*}$. Thus, every point in ${U}^{*}$ has a box that is contained in ${U}^{*}$, so ${U}^{*}$ is open as required.
Now suppose we have a finite list of open sets ${U}_{1},\mathrm{\dots},{U}_{n}$; we must show that the intersection ${U}^{\mathrm{\#}}={U}_{1}\cap \mathrm{\dots}\cap {U}_{n}$ is open. Consider a point $a=(b,c)\in {U}^{\mathrm{\#}}$. By the definition of the intersection, we have $a\in {U}_{i}$ for all $i$. As ${U}_{i}$ is open, we can choose a box ${B}_{i}={V}_{i}\times {W}_{i}$ around $a$ such that ${B}_{i}\subseteq {U}_{i}$. Put ${V}^{\mathrm{\#}}={V}_{1}\cap \mathrm{\dots}\cap {V}_{n}$ and ${W}^{\mathrm{\#}}={W}_{1}\cap \mathrm{\cdots}\cap {W}_{n}$ and ${B}^{\mathrm{\#}}={B}_{1}\cap \mathrm{\cdots}\cap {B}_{n}={V}^{\mathrm{\#}}\times {W}^{\mathrm{\#}}$. For all $i$ we have $b\in {V}_{i}$ and $c\in {W}_{i}$, so $b\in {V}^{\mathrm{\#}}$ and $c\in {W}^{\mathrm{\#}}$. By the topology axioms for $Y$, the set ${V}^{\mathrm{\#}}$ is open in $Y$. By the topology axioms for $Z$, the set ${W}^{\mathrm{\#}}$ is open in $Z$. It follows that ${B}^{\mathrm{\#}}$ is a box around $a$ that is contained in ${U}^{\mathrm{\#}}$, as required.
We also see that $\mathrm{\varnothing}$ is open (because there are no points $a\in \mathrm{\varnothing}$ to check, so the definition is vacuously satisfied). Also, the set $X=Y\times Z$ itself is a box around each of its points, so $X$ is open. This completes the proof that we have defined a topology. ∎
Again, the categorical viewpoint encourages us to try to analyse the continuous maps to or from $Y\times Z$. For maps out of $Y\times Z$ there is no simple answer, but maps into $Y\times Z$ are quite easy.
The projection maps $Y\stackrel{\mathit{p}}{\mathrm{\leftarrow}}Y\mathrm{\times}Z\stackrel{\mathit{q}}{\mathrm{\to}}Z$ (given by $p\mathit{}\mathrm{(}y\mathrm{,}z\mathrm{)}\mathrm{=}y$ and $q\mathit{}\mathrm{(}y\mathrm{,}z\mathrm{)}\mathrm{=}z$) are continuous.
Let $V\subseteq Y$ be an open set. We must show that ${p}^{-1}(V)$ is open in $Y\times Z$, but ${p}^{-1}(V)$ is just the same as $V\times Z$, which is open by Lemma 7.8. The proof for $q$ is essentially the same. ∎
Now consider a topological space $T$ and a function $f:T\to Y\times Z$. This must have the form $f(t)=(g(t),h(t))$ for some functions $g:T\to Y$ and $h:T\to Z$. We denote this by $f=\u27e8g,h\u27e9$. We can also express the component functions $g$ and $h$ as $g=p\circ f$ and $h=q\circ f$.
The combined map $f\mathrm{=}\mathrm{\u27e8}g\mathrm{,}h\mathrm{\u27e9}\mathrm{:}T\mathrm{\to}Y\mathrm{\times}Z$ is continuous (with respect to the product topology) iff the component functions $g\mathrm{=}p\mathrm{\circ}f\mathrm{:}T\mathrm{\to}Y$ and $h\mathrm{=}q\mathrm{\circ}f\mathrm{:}T\mathrm{\to}Z$ are continuous. Thus, there is a bijection between the sets $\mathrm{Top}\mathit{}\mathrm{(}T\mathrm{,}Y\mathrm{\times}Z\mathrm{)}$ and $\mathrm{Top}\mathit{}\mathrm{(}T\mathrm{,}Y\mathrm{)}\mathrm{\times}\mathrm{Top}\mathit{}\mathrm{(}T\mathrm{,}Z\mathrm{)}$, in which the element $f\mathrm{\in}\mathrm{Top}\mathit{}\mathrm{(}T\mathrm{,}Y\mathrm{\times}Z\mathrm{)}$ corresponds to the pair $\mathrm{(}g\mathrm{,}h\mathrm{)}\mathrm{\in}\mathrm{Top}\mathit{}\mathrm{(}T\mathrm{,}Y\mathrm{)}\mathrm{\times}\mathrm{Top}\mathit{}\mathrm{(}T\mathrm{,}Z\mathrm{)}$.
If $f$ is continuous, then the composites $g=p\circ f$ and $h=q\circ f$ are continuous by Lemma 7.10 and Proposition 3.24.
Suppose instead that we start from the assumption that $g$ and $h$ are both continuous. Consider an open set $U\subseteq Y\times Z$; we must show that ${f}^{-1}(U)$ is open. The simplest case is where $U$ is a box, say $U=V\times W$, where $V$ is open in $Y$ and $W$ is open in $Z$. We then have
$$t\in {f}^{-1}(V\times W)\iff (g(t),h(t))\in V\times W\iff \left(g(t)\in V\text{and}h(t)\in W\right)\iff t\in {g}^{-1}(V)\cap {h}^{-1}(W).$$ |
This shows that ${f}^{-1}(V\times W)={g}^{-1}(V)\cap {h}^{-1}(W)$, and this set is open because $g$ and $h$ are continuous.
Now return to the general case of an arbitrary open set $U\subseteq Y\times Z$, which need not be a box. For each point $x\in {f}^{-1}(U)$, we have $f(x)\in U$. As $U$ is open with respect to the product topology, there is a box ${B}_{x}$ with $f(x)\in {B}_{x}\subseteq U$. We put ${A}_{x}={f}^{-1}({B}_{x})$, so $x\in {A}_{x}\subseteq {f}^{-1}(U)$. The previous paragraph shows that ${A}_{x}$ is open. Moreover, we find that ${f}^{-1}(U)$ is the union of all the sets ${A}_{x}$. The union of any family of open sets is open, so ${f}^{-1}(U)$ is open as required. ∎
There is a general notion of product objects in category theory. The above proposition can be interpreted as saying that $Y\times Z$ (equipped with the product topology) is the product of $Y$ and $Z$ in this general categorical sense.
We now turn our attention to quotient constructions. We recall the basic definitions.
Let $X$ be a set, and let $E$ be a relation on $X$ (so for each pair $(x,y)\in X\times X$ we have a statement $xEy$, which may or may not be satisfied).
We say that the relation is reflexive if $xEx$ for all $x\in X$.
We say that the relation is symmetric if whenever $xEy$, we also have $yEx$.
We say that the relation is transitive if whenever $xEy$ and $yEz$, we also have $xEz$.
We say that the relation is an equivalence relation if it is reflexive, symmetric and transitive.
If we have an equivalence relation, we write $[x]=\{y|xEy\}\subseteq X$, and call this the equivalence class of $x$. We note that
The relation $xEy$ holds iff $[x]=[y]$
For each $x\in X$ we have $x\in [x]$
If $x\overline{)\phantom{\rule{-1.66666666666667pt}{0ex}}}Ey$, then the subsets $[x],[y]\subseteq X$ are disjoint, i.e. $[x]\cap [y]=\mathrm{\varnothing}$.
Although the equivalence classes are defined as subsets of $X$, we will often deemphasise that fact, and just treat them as abstract symbols satisfying property (i).
We write $X/E$ for the set of all equivalence classes, so $X/E=\{[x]|x\in X\}$. We define a map $\pi :X\to X/E$ by $\pi (x)=[x]$, so $\pi $ is surjective and $\pi (x)=\pi (y)$ iff $xEy$. Informally, we say that $X/E$ is obtained from $X$ by identifying points $x$ and $y$ whenever $xEy$. We may also say “gluing together” instead of “identifying”.
Suppose we have sets $X$ and $Y$, and an equivalence relation $E$ on $X$. We say that a function $f:X\to Y$ is $E$-saturated if whenever $xE{x}^{\prime}$ in $X$, we have $f(x)=f({x}^{\prime})$ in $Y$.
The quotient map $\pi :X\to X/E$ is $E$-saturated.
For any function $\overline{f}:X/E\to Y$, the composite $\overline{f}\circ \pi $ is $E$-saturated.
For any $E$-saturated function $f:X\to Y$, there is a well-defined function $\overline{f}:X/E\to Y$ given by $\overline{f}([x])=f(x)$. This is the unique $\overline{f}:X/E\to Y$ such that $\overline{f}\circ \pi =f$. (We call it the function induced by $f$.)
Claim (a) follows from Definition 7.13(e)(i), and claim (b) follows immediately from (a). For claim (c), we need to define $\overline{f}(u)\in Y$ for each equivalence class $u\in X/E$. To do this, we pick any $x$ with $u=[x]$, and take $\overline{f}(u)=f(x)$. There could in principle be a problem with this. If we choose a different ${x}^{\prime}$ such that $u$ is also equal to $[{x}^{\prime}]$, then we should also have $\overline{f}(u)=f({x}^{\prime})$, and this would be inconsistent if $f({x}^{\prime})$ was different from $f(x)$. However, in this situation we have $[x]=[{x}^{\prime}]$ so $xE{x}^{\prime}$ so $f(x)=f({x}^{\prime})$ by the $E$-saturation condition. Thus, no inconsistency can arise, and we have a well-defined function as claimed. The equation $\overline{f}\circ \pi =f$ is just another way of writing the condition $\overline{f}([x])=f(x)$, so it is clear that there is a unique function $\overline{f}$ with this property. ∎
There is a one-to-one correspondence between functions $X\mathrm{/}E\mathrm{\to}Y$, and $E$-saturated functions $X\mathrm{\to}Y$. ∎
Now suppose we have a topological space $X$, together with an equivalence relation $E$ on $X$. We declare that a subset $V\subseteq X/E$ is open iff the preimage ${\pi}^{-1}(V)\subseteq X$ is open in $X$.
The above definition gives a topology on $X\mathrm{/}E$ (which we call the quotient topology). Moreover, the quotient map $\pi \mathrm{:}X\mathrm{\to}X\mathrm{/}E$ is continuous with respect to this topology.
Suppose we have a family of subsets ${V}_{i}\subseteq X/E$ that are open with respect to the quotient topology. We must show that the union ${V}^{*}={\bigcup}_{i}{V}_{i}$ is also open. By the definition of the quotient topology, we see that the sets ${U}_{i}={\pi}^{-1}({V}_{i})$ are open in $X$, and we must show that ${\pi}^{-1}({V}^{*})$ is also open in $X$. As the sets ${U}_{i}$ are all open, the set ${U}^{*}={\bigcup}_{i}{U}_{i}$ is also open, by the topology axioms for $X$. Moreover, we have ${\pi}^{-1}({V}^{*})={U}^{*}$ by Lemma 3.11(a), so ${\pi}^{-1}({V}^{*})$ is open as required. The proof for finite intersections is similar. We also have ${\pi}^{-1}(X/E)=X$, and $X$ is open in $X$ by the topology axioms for $X$, so $X/E$ is open in $X/E$ by our definition of open sets in $X/E$. Similarly, we have ${\pi}^{-1}(\mathrm{\varnothing})=\mathrm{\varnothing}$, which is open in $X$, so $\mathrm{\varnothing}$ is open in $X/E$. This completes the proof that we have a topology on $X/E$.
We also want to prove that $\pi :X\to X/E$ is continuous. In other words, for every set $V\subseteq X/E$ that is open with respect to the quotient topology, we must show that ${\pi}^{-1}(V)$ is open in $X$. But this is true by the very definition of the quotient topology. ∎
Let $X$ and $E$ be as above, and let $\overline{f}$ be a function from $X\mathrm{/}E$ to another topological space $Y$. Then $\overline{f}$ is continuous (with respect to the quotient topology on $X\mathrm{/}E$) iff the composite $\overline{f}\mathrm{\circ}\pi \mathrm{:}X\mathrm{\to}Y$ is continuous.
We have seen that $\pi $ is continuous and that composites of continuous functions are continuous. Thus, if $\overline{f}$ is continuous, then the composite $\overline{f}\circ \pi $ is also continuous as claimed.
Suppose instead that we start from the assumption that $\overline{f}\circ \pi :X\to Y$ is continuous; we must show that $\overline{f}:X/E\to Y$ is continuous. Let $W\subseteq Y$ be open; we must show that the preimage $V={\overline{f}}^{-1}(W)$ is open in $X/E$. By the definition of the quotient topology, it is equivalent to prove that the set ${\pi}^{-1}(V)$ is open in $X$. However, we have ${\pi}^{-1}(V)={\pi}^{-1}({\overline{f}}^{-1}(W))={(\overline{f}\circ \pi )}^{-1}(W)$, and this is open as required because $\overline{f}\circ \pi $ is assumed to be continuous. ∎
Let $f\mathrm{:}X\mathrm{\to}Y$ be an $E$-saturated function. Then the induced function $\overline{f}\mathrm{:}X\mathrm{/}E\mathrm{\to}Y$ is continuous iff $f$ is continuous. Thus, we have a one-to-one correspondence between continuous functions $X\mathrm{/}E\mathrm{\to}Y$, and $E$-saturated continuous functions $X\mathrm{\to}Y$.
As $\overline{f}\circ \pi =f$, this is just a restatement of the proposition. ∎
This corollary could be put in a common framework with the first isomorphism theorem for groups, if we took the time to develop the relevant categorical notion of coequalisers.
We can define an equivalence relation $E$ on ${S}^{n}$ by $xEy$ iff $y=\pm x$. The real projective space $\mathbb{R}{P}^{n}$ is defined to be the quotient space ${S}^{n}/E$. Similarly, we can regard ${S}^{2n+1}$ as the unit sphere in the space ${\u2102}^{n+1}={\mathbb{R}}^{2n+2}$. We can then define another equivalence relation $F$ on ${S}^{2n+1}$ by declaring that $xFy$ iff $y={e}^{i\theta}x$ for some $\theta \in \mathbb{R}$. The complex projective space $\u2102{P}^{n}$ is defined to be ${S}^{2n+1}/F$.
We will prove that the real projective space $\mathbb{R}{P}^{1}$ is homeomorphic to ${S}^{1}$. Recall that $\mathbb{R}{P}^{1}={S}^{1}/E$, where $uEv$ iff $v=\pm u$. We will identify ${S}^{1}$ with $\{z\in \u2102||z|=1\}$ by the usual Argand corrspondence $(x,y)\leftrightarrow x+iy$. We define $f:{S}^{1}\to {S}^{1}$ by $f(z)={z}^{2}$. This clearly satisfies $f(-z)=f(z)$, so it is $E$-saturated and induces a map $\overline{f}:\mathbb{R}{P}^{1}\to {S}^{1}$ with $\overline{f}(\pi (z))={z}^{2}$ for all $z$. Now suppose we have a point $w={e}^{i\theta}\in {S}^{1}$. Then the set $g(w)=\{{e}^{i\theta /2},-{e}^{i\theta /2}\}$ is an $E$-equivalence class, or in other words a point of the space $\mathbb{R}{P}^{1}$, so we have a map $g:{S}^{1}\to \mathbb{R}{P}^{1}$. It is easy to see that this is inverse to $\overline{f}$, so $\overline{f}$ is a continuous bijection. We just need to check that $g$ is also continuous. Suppose that $U\subseteq \mathbb{R}{P}^{1}$ is open, and contains $g(w)$ for some point $w={e}^{i\theta}\in {S}^{1}$. This means that ${e}^{i\theta /2}\in {\pi}^{-1}(U)$, and ${\pi}^{-1}(U)$ is open in ${S}^{1}$ by the definition of the quotient topology, so there is some $\u03f5>0$ such that ${e}^{i\varphi}\in {\pi}^{-1}(U)$ whenever $$. It follows that ${e}^{i{\theta}^{\prime}}\in {g}^{-1}(U)$ whenever $$. From this we can deduce that ${g}^{-1}(U)$ is open as required.
Put
$$X=\{(x,y,z)\in {\mathbb{R}}^{3}|{x}^{2}+{y}^{2}\le 1\text{and}z=\pm 1\}.$$ |
This consists of two closed discs, as shown on the left below.
If we glue together the boundary circles of the two discs, we get a sphere, as shown on the right. We can make this more formal and rigorous as follows. We introduce an equivalence relation $E$ on $X$ by declaring that $(x,y,z)E({x}^{\prime},{y}^{\prime},{z}^{\prime})$ if $({x}^{\prime},{y}^{\prime},{z}^{\prime})=(x,y,z)$, or if $({x}^{\prime},{y}^{\prime},{z}^{\prime})=(x,y,-z)$ with ${x}^{2}+{y}^{2}=1$. We define $f:X\to {S}^{2}$ by $f(x,y,z)=(x,y,z\sqrt{1-{x}^{2}-{y}^{2}})$, as illustrated below.
This is clearly continuous. Also, if ${x}^{2}+{y}^{2}=1$ then we have $f(x,y,1)=f(x,y,-1)=(x,y,0)$, so $f$ is $E$-saturated. We therefore have an induced map $\overline{f}:X/E\to {S}^{2}$, and Corollary 7.20 tells us that this is continuous. It is not hard to check that $\overline{f}$ is a bijection. However, we saw in Example 4.10 that not every continuous bijection has a continuous inverse, so we cannot immediately conclude that $\overline{f}$ is a homeomorphism. It would be possible but fiddly to prove this directly. However, we will see an efficient general method for this in Section 8, so we will defer further comment until then.
From the Knots and Surfaces course you should remember pictures like this:
We take a filled octagon $X$, with edges marked $a$, $b$, $c$ or $d$ as shown. There are two edges marked $a$, and we glue them together in the direction indicated by the arrows. We do the same for the labels $b$, $c$ and $d$. To make this more formal and rigorous, we introduce a relation $E$ as follows: we declare that $xEy$ if
$x=y$; or
for some $t\in [0,1]$ we have $x=(1-t){v}_{1}+t{v}_{2}$ and $y=(1-t){v}_{4}+t{v}_{3}$; or
for some $t\in [0,1]$ we have $x=(1-t){v}_{2}+t{v}_{3}$ and $y=(1-t){v}_{4}+t{v}_{3}$; or
for some $t\in [0,1]$ we have $x=(1-t){v}_{5}+t{v}_{6}$ and $y=(1-t){v}_{8}+t{v}_{7}$; or
for some $t\in [0,1]$ we have $x=(1-t){v}_{6}+t{v}_{7}$ and $y=(1-t){v}_{1}+t{v}_{8}$; or
for some $i,j$ we have $x={v}_{i}$ and $y={v}_{j}$.
One can check that this is an equivalence relation, so we can form the quotient space $X/E$. One can then show that $X/E$ is homeomorphic to the usual kind of double torus embedded in ${\mathbb{R}}^{3}$.
Consider the spaces $X={\mathbb{R}}^{2}$ and
$$Y={S}^{1}\times {S}^{1}=\{(w,x,y,z)\in {\mathbb{R}}^{4}|{w}^{2}+{x}^{2}={y}^{2}+{z}^{2}=1\}$$ |
(so $Y$ is a version of the torus). We introduce an equivalence relation on $X$ by declaring that $uEv$ iff $u-v\in {\mathbb{Z}}^{2}$. We then define $f:X\to Y$ by
$$f(x,y)=(\mathrm{cos}(2\pi x),\mathrm{sin}(2\pi x),\mathrm{cos}(2\pi y),\mathrm{sin}(2\pi y)).$$ |
It is easy to see that this is continuous and $E$-saturated, so it induces a continuous map $\overline{f}:X/E\to Y$. It is also easy to see that $\overline{f}$ is a bijection. It is again true that ${\overline{f}}^{-1}$ is also continuous, so that $\overline{f}$ is a homeomorphism, but we will again defer the proof.
Here is a more exotic example. We start with the space $X=\mathbb{R}\times \{1,-1\}$, and we declare that $({x}^{\prime},{y}^{\prime})E(x,y)$ if either $({x}^{\prime},{y}^{\prime})=(x,y)$, or $({x}^{\prime},{y}^{\prime})=(x,-y)$ with $x\ne 0$. In other words, we start with the two lines where $y=1$ and $y=-1$ and then we glue them together, except that we leave the points $(0,1)$ and $(0,-1)$ unglued. The quotient space $X/E$ is called the line with doubled origin. It has various unpleasant properties, and we mostly choose not to study spaces with those properties; that is the point of the Hausdorff condition to be introduced in Section 8.
There is one more construction that we would like to treat, but will not. Let $X$ be any metric space. From the Metric Spaces course, you should remember that the set $C(X,\mathbb{R})=\mathrm{Top}(X,\mathbb{R})$ (of continuous functions from $X$ to $\mathbb{R}$) has a metric of its own, which is useful for many purposes. It would also be useful to define a similar topology on $\mathrm{Top}(X,Y)$ for arbitrary topological spaces $X$ and $Y$. This seems natural, because it is easy to imagine what we mean by saying that two functions are close to each other, which is the basic idea that we need when defining a topology. Unfortunately this leads to a host of technical difficulties and subtle distinctions. A huge detour is necessary before one can set up a clear and coherent theory. Thus, we will motivate various constructions using the idea of treating $\mathrm{Top}(X,Y)$ as a topological space, but we will not use this idea in our formal definitions.
Let $Y$ and $Z$ be disjoint spaces, and take $X\mathrm{=}Y\mathrm{\cup}Z$, with the coproduct topology. Then ${\pi}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}$ is the disjoint union of ${\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Y\mathrm{)}$ and ${\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Z\mathrm{)}$.
We can define $f:X\to \{1,-1\}\subset \mathbb{R}$ by $f(y)=1$ for $y\in Y$, and $f(z)=-1$ for $z\in Z$. This is clearly continuous when restricted to $Y$ or $Z$, so it is continuous on all of $X$ by Proposition 7.3. If $u:[0,1]\to X$ is a continuous path, then $f\circ u:[0,1]\to \{-1,1\}$ is also continuous, so it clearly must be constant. (Formally, the proof uses the Intermediate Value Theorem.) Thus, the path lies wholly in $Y$ or wholly in $Z$. Given this, the claim is clear from the definitions. ∎
Let $Y$ and $Z$ be topological spaces. Then ${\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Y\mathrm{\times}Z\mathrm{)}$ can be identified with ${\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Y\mathrm{)}\mathrm{\times}{\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Z\mathrm{)}$. More precisely, let $Y\stackrel{\mathit{p}}{\mathrm{\leftarrow}}Y\mathrm{\times}Z\stackrel{\mathit{q}}{\mathrm{\to}}Z$ be the projections, which give rise to maps ${\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Y\mathrm{)}\stackrel{{p}_{\mathrm{*}}}{\mathrm{\leftarrow}}{\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Y\mathrm{\times}Z\mathrm{)}\stackrel{{q}_{\mathrm{*}}}{\mathrm{\to}}{\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Z\mathrm{)}$ as in Proposition 5.20. We can therefore define a map
$$\varphi :{\pi}_{0}(Y\times Z)\to {\pi}_{0}(Y)\times {\pi}_{0}(Z)$$ |
by $\varphi \mathit{}\mathrm{(}u\mathrm{)}\mathrm{=}\mathrm{(}{p}_{\mathrm{*}}\mathit{}\mathrm{(}u\mathrm{)}\mathrm{,}{q}_{\mathrm{*}}\mathit{}\mathrm{(}u\mathrm{)}\mathrm{)}$, and this is a bijection.
We would like to define $\psi :{\pi}_{0}(Y)\times {\pi}_{0}(Z)\to {\pi}_{0}(Y\times Z)$ as follows: an element of ${\pi}_{0}(Y)\times {\pi}_{0}(Z)$ can be written as $([a],[b])$ for some $a\in Y$ and $b\in Z$, and these give a point $(a,b)\in Y\times Z$ and a path component $[(a,b)]\in {\pi}_{0}(Y,Z)$, and we want to define $\psi ([a],[b])=[(a,b)]$. We must check that this is well-defined. Suppose that $([a],[b])=([{a}^{\prime}],[{b}^{\prime}])$ in ${\pi}_{0}(Y)\times {\pi}_{0}(Z)$. This means that $[a]=[{a}^{\prime}]$ in ${\pi}_{0}(Y)$, so there is a continuous path $u:a\rightsquigarrow {a}^{\prime}$ in $Y$. It also means that $[b]=[{b}^{\prime}]$ in ${\pi}_{0}(Z)$, so there is a continuous path $v:b\rightsquigarrow {b}^{\prime}$ in $Z$. We define $w:[0,1]\to Y\times Z$ by $w(t)=(u(t),v(t))$ (which is continuous by Proposition 7.11). This gives a path $(a,b)\rightsquigarrow ({a}^{\prime},{b}^{\prime})$ in $Y\times Z$, proving that $[(a,b)]=[({a}^{\prime},{b}^{\prime})]$, as required. Thus, our definition of $\psi $ is valid. We can describe $\varphi $ in similar terms by $\varphi ([a,b])=([a],[b])$. (We do not need to check separately that this is well-defined, because that was done already in Proposition 5.20 which we used to define ${p}_{*}$ and ${q}_{*}$.) It is now clear that $\varphi $ and $\psi $ are inverse to each other, so $\varphi $ is a bijection as claimed. ∎