Product and Quotient Topology

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1. Basis of Topology

Definition. Basis of a topology
If $X$ is a set, a basis for a topology on $X$ is a collection $\B$ of subsets of $X$ (called basis elements) such that

• $\displaystyle\bigcup_{B\in\B}B=X$
• $\forall B_1,B_2\in\B.(x\in B_1\cap B_2\implies \exists B_3\in \B.x\in B_3\subset B_1\cap B_2)$

If the collection $\B$ satisfies such conditions, we define the topology $\T$ generated by $\B$ as follows. A subset $U$ of $X$ is said to be open (i.e. be an element of $\T$) if

$$\forall x\in U.\exists B\in\B.x\in B\subset U$$

and with open set defined (using the language of basis), $\T$ is defined to be the union of all such open sets.

By definition, each basis element itself is open, thus an element of $\T$.

However we only claimed (without proof) the “topology” spanned from the basis is a topology. How do we know if the generated topology is well-defined? The following lemma shows this.

Lemma. What a basis generates is indeed a topology!
Let $X$ be a set, let $\B$ be a basis which generates $\T$. Then $\T$ is indeed a topology on $X$.
Pf. Only need to verify if the generated “topology” aligns with the definition of topology on $X$

1. $X=\bigcup_{B\in \B}B\in \T$. Empty set $\emptyset$ satisfies the definition of openness (from basis) vacuously, so $\emptyset\in\T$
2. Take an indexed family $\{U_α\}_{α∈J}$ of elements of $\T$. $\forall x\in U_\alpha$, $\exists B\in\B.x\in B\subset U_\alpha$, so $x\in B\subset \bigcup_{\alpha\in J}$, which means that the union of $\{U_\alpha\}_{\alpha\in J}$ is open.
3. Take any two elements $U,V\subset \T$. Given $x\in U\cap V$, $\exists B_1\in\B$ such that $x\in B_1\subset U$, $\exists B_2\in\B$ such that $x\in B_2\subset V$. By the second requirement of a basis element we know $\exists B_3\in\B$ such that $x\in\B_3\subset B_1\cap B_2\subset U\cap V$. Since the choice of $x$ is arbitrary, we know $U\cap V$ is open.

Note. In the definition of basis, why do we say $B_3\subset B_1\cap B_2$ instead of $B_3=B_1\cap B_2$ ? Consider the fact that the standard topology (the metric topology of Euclidean metric) on $\R^2$, which is generated by the basis as all open balls on $\R^2$. Given two balls $B_1$ and $B_2$, $B_1\cap B_2$ will be obviously open, but $B_1\cap B_2$ is not a ball (under most cases).

Lemma. Another way of defining “generate”.
If $X$ is a set, $\B$ is a basis for a topology $\T$ on $X$, then $\T$ equals the collection of arbitrary union of elements in $\B$. i.e.

$$U\in\T\iff U=\bigcup_{\alpha\in\Lambda}B_{\alpha}$$

where $\{B_\alpha\}_\index$ is a subcollection of $\B$.
Pf. We want to show that the “generated topology” defined above is equivalent to this definition.

• Given a collection of elements of $\B$, they are also elements of $\T$. Because $\T$ is a topology, their union is in $\T$.
• Conversely, given $U\in \T$, choose for each $x\in U$ an element $B_x$ of $\B$ such that $x\in B_x\subset U$ . Then $\displaystyle U = \bigcup_{x\in U} B_x$ , so $U$ equals a union of elements of $\B$.

Example.

1. In $\Rn$, the following are all valid basis
   • $\B_1$: the collection of all open balls
   • $\B_2$: the collection of all open cubes
2. In a metric space $(x,d)$, the collection $\B$ of all open metric balls is a basis.
3. $X\neq \emptyset$, in a discrete topological space $(X, 2^{X})$, the basis is a collection $\B$ of all singletons in $X$ (note that the second condition is vacuously true now)

Lemma. Basis of a topology is not unique
Given a topological space $(X, \T)$. Let $\D$ be a collection of open sets such that $\forall U\in \T.\forall x\in U$

$$\exists D\in \D.x\in D\subset U$$

Then $\D$ is also a basis for the same topology $\T$
Pf. First show that $\D$ is a basis for a topology, and then show that $\T$ is same as the space spanned from $\D$

• $X=\bigcup_{x\in X}D_x$ where $D_x\subset X$ so we have $\bigcup_{D\in \D}D\supset X$
• $D_1,D_2\in \D$, $D_1\cap D_2$ is open thus $\exists D\in\D.D\subset D_1\cap D_2$
  So $\D$ is indeed a basis.

By the above definitions and lemmas, we know how to build a topology if the basis is given. The following lemmas shows how to obtain the basis from a collection of open sets.

Lemma. Obtain basis from a collection of open sets
Let $X$ be a topological space. Suppose $\B$ is a collection of open sets of $X$ with respect to a topology $\T$, such that for each open set $U$ of $X$, there exists $B\subset \B$ such that $x\in B\subset U$. Then $\B$ is the basis for the topology $\T$ of $X$.
Pf. It could be checked from definition that $\B$ is indeed a basis for some topology of $X$. Furthermore we need to check that specific topology $\T'$ equals $\T$, by showing that $\T'\subset \T$ and $\T\subset \T'$.

Lemma. (Munkres Lemma 13.3) Criterion for finer topologies.
Let $\B_1$ and $\B_2$ be basis for the topologies $\T_1$, $\T_2$ of $X$, respectively, then $\T_1$ is finer than $\T_2$ if and only if $\forall B_1\in\B_1$, $\forall x\in B_1$, $\exists B_2\in\B_2$, s.t. $x\in B_1\subset B_2$.

If $\B_1$ is finer than $\B_2$, just image $\B_1$ to be a truckload of dust while $\B_2$ is a truckload of pebbles.

Example. Basis of $\R$

1. By default, when we say a topology of $\R$, we mean the standard topology, which is generated by the basis $\{(a,b)\mid a<b, a,b\in \R\}$
2. The lower limit topology $\R_{\mathscr l}$ is generated by the basis $\{[a,b)\mid a<b, a,b\in \R\}$
3. the standard topology of $\R$ can also be generated by $\{(a,b)\mid a,b\in\Q\}$, but cannot be generated by $\{[a,b)\mid a,b\in \Q\}$

Definition. Subbasis
A subbasis $\S$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis is defined to be the collection $\T$ of all unions of finite intersections of elements of $\S$, that is

$$\B:=\{\bigcap_{i=1}^mS_i\mid S_i\in\S\}$$

is a basis for the topology $\T$

Note. The difference between basis and subbasis
Given a topology $\T$ on a set $X$,

• A basis of $\T$ is always a subbasis of $\T$ (by definition)
• A subbasis of $\T$ is not necessarily a basis of $\T$
• A basis of $\T$ can generate $\T$ using only union operations.
• A subbasis of $\T$ must generate $\T$ using both finite intersections and union operations.
• Every subbasis element is also open, just like basis elements. (there exists a basis $\B$ of $X$ such that $\S\subset\B$)

Example. $\S:=\{(-\infty,a)\}\cup\{(b,+\infty)\}$ is a subbasis of $\R$.

Lemma. Basis of subspace topology
If $\B$ is a basis for the topology of $X$ then the collection $\B_Y = \{B \cap Y\mid B\in \B\}$ is a basis for the subspace topology on $Y$.

Just check the definition.

Theorem. continuity criterion using basis and subbasis
$f:X\to Y$ where $(X,\T_X)$ and $(Y,\T_Y)$ are two topological spaces.

1. If $Y$ is given by a basis $\B$, then $f$ is continuous iff: $\forall B\in\B.f^{-1}(B)\in\T_X$
2. If $Y$ is given by a subbasis $\S$, then $f$ is continuous iff: $\forall S\in\S.f^{-1}(S)\in\T_X$.
   Pf.
3. Given any open set $U\subset Y$, $U=\cup_{\alpha\in\Lambda}{B_\alpha}$ where $\{B_\alpha\}$ is a collection of basis elements. If $f$ is continuous on $X$, at least $f^{-1}(B)$ should be open for all $B\in\B$ because every basis element is open. Since $f^{-1}(U)=\cup_{\alpha\in\Lambda}f^{-1}(B_\alpha)$, when the preimages of all basis elements are open, the preimage of any open set should be open.
4. Definition of subbasis tells us every finite intersection of subbasis elements is a basis element. This means that there exists a basis $\B$ such that $\S\subset\B$. Let $B=\cap_{i=1}^n S_i\in\B$ be a basis element of $Y$, $f^{-1}(B)=\cap_{i=1}^nf^{-1}(S_i)$. If the preimage of any subbasis element is open then $f$ is continuous by definition. if $f$ is continuous, the preimage of any subbasis element should be open because every subbasis element is open.

Since $\{(a,+\infty)\}\cup\{(-\infty,b)\}$ is a subbasis of $\R$, it’s natural to make the following definition according to continuity criterion from subbasis of the codomain.

Definition. Semicontinuous functions on $\R$
Let $U$ be an open subset of $\R$. A function $f:U\to\R$ is said to be

1. upper semicontinuous: if $\forall t\in\R$. $f^{-1}((-\infty,t))$ is open
2. lower semicontinuous: if $\forall t\in\R$. $f^{-1}((t,+\infty))$ is open
   By the continuity criterion, $f$ is continuous if $f$ is both upper semicontinuous and lower semicontinuous.

2. Product Topology

With basis defined, we can now define the topology created from the product of given topologies. This is a natural idea.

Definition. box topology
$\{X_\alpha\}_{\index}$ is a collection of topological spaces. $X:=\prod_{\index}X_\alpha$ is a space equipped with the topology generated by the basis

$$\B_\square:=\{\prod_\index U_\alpha\mid U_\alpha\in\T_{X_\alpha}\}$$

which is called the box topology of $X$. $\B_\square$ is indeed A BASIS because of the following lemma.

Lemma. Product of basis elements is a basis for box topology

• $X=\prod_\index X_\alpha\in\B$ so $\bigcup_{B\in\B}B=X$
• Suppose $U_\alpha$ and $V_\alpha$ are open in $X_\alpha$, then

$$(\prod_\index U_\alpha)\cap(\prod_\index V_\alpha)=\prod_\index (U_\alpha \cap V_\alpha)\in\B$$

by definition $\B$ is a basis for the topology on $X$.

Definition. Projection mapping
Now we generalize the subbasis formulation of the definition. Let

$$π_\beta :\prod_\index X\alpha \to X_\beta$$

be the function assigning to each element of the product space its $β$-th coordinate,

$$\pi_\beta ((x_α)_\index) = x_\beta ;$$

it is called the projection mapping associated with the index $β$.

Lemma. Properties of finite-dimensional projection mapping
Let $X=\prod_{i=1}^nX_i$ where $\pi_i:X\to X_i$ is defined as $\pi_i(x_1,x_2,\ldots,x_n)\mapsto x_i$. Then the following are true.

• $\pi_i$ is surjection
• If $U_i\subset X_i$, then $\pi_i^{-1}(U_i)=X_1\times X_2\times \ldots\times U_i\times \ldots\times X_n$
• If $U_i\subset X_i$, then $\bigcap_{i=1}^n\pi_{i}^{-1}(U_i)=\prod_{i=1}^n U_i$

Definition. product topology (Tychonoff’s Topology)
Let $\{X_\alpha\}_{\alpha\in\Lambda}$ be a collection of topological spaces. $S_\beta:=\{\pi_\beta^{-1}(U_\beta)\mid U_\beta\in\T_{X_\beta}\}$ and let $\S:=\bigcup_\index S_\alpha$, The topology generated by this subbasis $\S$ is called a product topology. With this topology $X:=\prod_\index X_\alpha$ is called a product space. $\S$ is indeed a subbasis of $X$ due two property (2) of the above lemma.

Note. When we say product spaces, we always mean spaces equipped with product topologies by default.

Note. Consider the following questions. Given a map $f:X\to Y$, how can we

• find the finest topology $\T_Y$ with fixed $\T_X$; or
• find the coarsest topology $\T_X$ with fixed $\T_Y$
  such that $f$ is continuous. It’s quite natural because given an open set $O\in\T_Y$, we are not sure if $f^{-1}(O)$ is open in $X$, but if $\T_X$ is fine enough, then we are likely to make it. The second question is quite similar.

This problem could be answered when $X$ is a product space and $Y$ is one of its dimensions. the product topology is the smallest (coarsest) topology that makes every projection mapping $\pi_\alpha$ continuous.

Lemma. The product topology of $X=\prod_\index X_\alpha$ is the coarsest topology that makes each projection mapping $\pi_\alpha$ continuous.
Pf. Suppose $\pi_\alpha:X\to X_\alpha$ is continuous. Take $U_\alpha\subset X_\alpha$, we have

$$\pi^{-1}_\alpha(U_\alpha)=(\prod_{\beta<\alpha}X_\beta)\times U_\alpha\times (\prod_{\beta>\alpha}X_{\beta})$$

should be open in $X$. Since any topology of $X$ must contain

• arbitrary unions of preimages
• finite intersections of preimages
  By definition we know these are exactly what the product topology contains.

Corollary. The projection mapping is continuous with respect to product topology.

Theorem. (Munkres Thm 19.1) Box topology is finer than the product topology.
Pf. The box topology on $X=\prod_{\alpha}X_\alpha$ has a basis all sets of the form $\prod U_\alpha$ where $U_\alpha$ is open in $X_\alpha$. The product topology on $X$ has a basis all sets of the form $\prod U_\alpha$ where $U_\alpha$ is open in $X_\alpha$, and $U_\alpha=X_\alpha$ for all but finitely many $\alpha$’s .
Pf. The claim about box topology is just a restatement of its definition. Let $\B$ be the basis for the product topology that the subbasis $\S$ generates.

Another thing is clear from the theorem above: In finite-dimensional product spaces, the box topology is same as the product topology.

Theorem. (Munkres Thm 19.6)
Let $f:A\to \prod_{\alpha\in J} X_\alpha$. Let $X=\prod_{\alpha\in J} X_\alpha$ have the product topology. $f$ is continuous if and only if for every $\alpha\in J$, $f_\alpha=\pi_\alpha\circ f$ is continuous on $A$.
Pf. When $f$ is continuous, $f_\alpha$ is the composite of two continuous functions, thus continuous. When each $f_\alpha$ is continuous on $A$, by the continuity criterion, $f$ is continuous if and only if for every subbasis element $S\in\S$ of $X$, $f^{-1}(S)$ is open in $A$ (why subbasis? because in product topology it’s easy for us to find a subbasis!). Consider the subbasis $\{ \pi_\alpha^{-1}(U_\alpha)\mid U_\alpha$ is open in $X_\alpha\}$ of $X$, we know $f^{-1}\circ\pi_\alpha^{-1}(U_\alpha)=f^{-1}_{\alpha}(U_\alpha)$, which is open in $A$. Since the choice of $U_\alpha$ is arbitrary, $f$ is continuous on $A$.

Note. This property doesn’t hold for the box topology, because it allows too many sets to be open, but they shouldn’t be open. See the counterexample below.

Example. Let $X_\alpha=\R$, $X=\R^\omega$, $U=\prod_{n\in\N}(-\frac 1 n, \frac 1 n)$, $f_\alpha=\Id_\R$, then $f^{-1}(U)=\bigcap_{n\in\N}(-\frac 1 n, \frac 1 n)=\{0\}$ . $U$ is open (w/ box topology) in $X$ but $f^{-1}(U)$ is not open in $\R$.

Theorem. (Munkres Thm 19.5) subspace property and closure property
Let $\{X_\alpha\}_{\alpha\in\Lambda}$ be a family of topological spaces. $X:=\prod_{\alpha\in\Lambda}X_\alpha$. For each $\alpha$, $A_\alpha$ is a subspace of $X_\alpha$. Then the following properties hold with respect to either the box topology and the product topology

1. $\prod_{\alpha\in\Lambda}A_\alpha$ is a subspace of $\prod_{\alpha\in\Lambda}X_\alpha$.
2. $\prod_{\alpha\in\Lambda}\clo A_\alpha=\clo{\prod_{\alpha\in\Lambda} A_\alpha}$.

3. Quotient Topology

Definition. Open map / closed map
Given two topological spaces $X$ and $Y$, a map $f:X\to Y$ is said to be a

• open map: if for any open set $U\subset X$, $f(U)\subset Y$ is open in $Y$
• closed map: if for any closed set $U\subset X$, $f(U)\subset Y$ is closed in $Y$

Definition. Quotient map
Let $X$ and $Y$ be two topological spaces. A surjective map $p:X\to Y$ is said to be a quotient map if given any subset $U\subset Y$, $U$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.

Note that this definition seems a bit different from what we usually see in algebra. By definition a quotient map is continuous. Sometimes a quotient map is also called a strong continuous function.

Definition. Saturated
Let $X$ and $Y$ be two topological spaces. Given a surjective map $p:X\to Y$. A subset $C\subset X$ is said to be saturated if $C$ contains every set $p^{-1}(y)$ that it intersects. i.e.

$$\forall y\in Y.(C\cap f^{-1}(y)\neq \emptyset)\implies (f^{-1}(y)\subset C)$$

To say that $p$ is a quotient map, it’s equivalent to say $p$ is continuous and $p$ maps every saturated open sets of $X$ to open sets of $Y$.

Definition. Quotient topology
If $X$ is a topological space, $A$ is a set. If there is a surjective map $p:X\to A$, then there exists exactly one topology $\T$ on $A$ relative to which $p$ is a quotient map; $\T$ is called the quotient topology of $A$ introduced by $p$.
Pf. Why does this topology exist? Define the openness in $A$ this way

• $U\subset A$ is open if $p^{-1}(U)\subset X$ is open in $X$
  The three requirements of topology is then automatically satisfied
• $p^{-1}(A)=X, p^{-1}(\emptyset)=\emptyset$
• $\displaystyle{p^{-1}(\bigcap_{i=1}^n U_i)=\bigcap_{i=1}^np^{-1}(U_i)}$
• $\displaystyle{p^{-1}(\bigcup_{\alpha\in\Lambda} U_\alpha)=\bigcup_{\alpha\in\Lambda}p^{-1}(U_\alpha)}$
  So such a topology $\T$ exists. Why is $\T$ unique? If there’s another topology $\T'$ such that $p$ is a quotient map. $U\in\T'\iff p^{-1}(U)$ is open in $X\iff U\in\T$.

Definition. Quotient space / Identification space / decomposition space
Let $X$ be a topological space, and let $X^*$ is a partition of $X$ into disjoint subsets of $X$ whose union is $X$. Let $p:X\to X^*$ be the surjective map that carries each point of $X$ to the element of $X^*$ containing it. In the quotient topology introduced by $p$, the space $X^*$ is called quotient space of $X$. Given two topological spaces $X$ and $Z$, and a continuous surjection $p:X\to Z$, we can even more naturally define $X^*:=\{p^{-1}(z)\mid z\in Z\}$.

Definition. Another way of defining quotient map
Let $X$ and $Y$ be two topological spaces. A surjective map $p:X\to Y$ is said to be a quotient map if $X$ is a topological space with an equivalence relation $\sim$ defined on it. $\pi:X\to (X/\sim)=Y$ is the quotient map.

Theorem.
$h:X\to Z$ induces a bijection $f:X^*\to Z$ such that $h=f\circ \pi$. Moreover, $f$ is a homeomorphism iff $h$ is a quotient map. If $X$ is Hausdorff, so is $X^*$

Lemma. continuous maps on a compact metric space are quotient
$X$ is a compact metric space, and $Y$ is a metric space. If $f:X\to Y$ is continuous and surjective, then $f$ is a quotient map.

This should be obvious by definition.

// universal property of quotient topology