[Munkres Topology] 22. The Quotient Topology

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Definition. Let $X$ and $Y$ be topological spaces; let $\rho:X\rightarrow Y$ be a surjective map.(In homeomorphism, injective) The map $p$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.

Another way of describing a quotient map is as follows: We say that a subset $C$ of $X$ is saturated (with respect to the surjective map $p:X\rightarrow Y$) if $C$ contains every set $p^{-1}(\{y\})$ that is intersects. Thus $C$ us saturated if it equals the complete inverse image of a subset of $Y$. To say that $p$ is a quotient map is equivalent to saying that $p$ is continuous and $p$ maps saturated open sets of $X$ to open sets of $Y$ (or saturated closed sets of $X$ to closed sets of $Y$). (We need 'saturated' condition because of case $f(A_1)=B$, $f(A_2)=B$, when $A_1,B$ is open but $A_2$ is not. We have to consider whole connection.)

Two special kinds of quotient maps are the open maps and the closed maps. Recall that a map $f:X\rightarrow Y$ is said to be an open map if for each open set $U$ of $X$, the set $f(U)$ is open in $Y$. It is said to be a closed map if for each closed set $A$ of $X$, the set $f(A)$ is closed in $Y$. It follows immediately from the definition that if $p:X\rightarrow Y$ is a surjective continuous map that is either open or closed, then $p$ is a quotient map.

Now we show how the notion of quotient map can be used to construct a topology on a set.

Definition. If $X$ is a space and $A$ is a set and if $p:X\rightarrow A$ is a surjective map, then there exists exactly one topology $\mathcal{T}$ on $A$ relative to which $p$ is a quotient map; it is called the quotient topology induced by $p$.

Check.

The topology $\mathcal{T}$ is of course defined by letting it consist of those subsets $U$ of $A$ such that $p^{-1}(U)$ is open in $X$. It is easy to check that $\mathcal{T}$ is a topology. The sets $\varnothing$ and $A$ are open because $p^{-1}(\varnothing)=\varnothing$ and $p^{-1}(A)=X$. The other two conditions follows from the equations $$p^{-1}(\bigcup_{\alpha\in J}U_\alpha)=\bigcup_{\alpha\in J} p^{-1}(U_\alpha),$$ $$p^{-1}(\bigcap_{i=1}^n U_i)=\bigcap_{i=1}^n p^{-1}(U_i).$$

Definition. Let $X$ be a topological space, and let $X^*$ be a partition of $X$ into disjoint subsets whose union is $X$. Let $p:X\rightarrow X^*$ be the surjective map that carries each point of $X$ to the element of $X^*$ containing it. In the quotient topology induced by $p$, the space $X^*$ is called a quotient space of $X$.

Given $X^*$, there is an equivalence relation on $X$ of which the elements of $X^*$ are the equivalence classes. One can think of $X^*$ as having been obtained by 'identifying' each pair of equivalent points. For this reason, the quotient space $X^*$ is often called an identification space, or a decomposition space, of the space $X$.

We can describe the topology of $X^*$ in another way. A subset $U$ of $X^*$ is a collection of equivalence classes, and the set $p^{-1}(U)$ is just the union of the equivalence classes belonging to $U$. Thus the typical open set of $X^*$ is a collection of equivalence classes whose union is an open set of $X$.

Theorem 22.1. Let $p:X\rightarrow Y$ be a quotient map; let $A$ be a subspace of $X$ that is saturated with respect to $p$; let $q:A\rightarrow p(A)$ be the map obtained by restricting $p$.

(1) If $A$ is either open or closed in $X$, then $q$ is a quotient map.

(2) If $p$ is either an open map or a closed map, then $q$ is a quotient map.

Proof.

Step 1. We verify first the following two equations: $$q^{-1}(V)=p^{-1}(V)\quad \mbox{if }V\subset p(A);$$ $$p(U\cap A)=p(U)\cap p(A)\quad \mbox{if }U\subset X.$$ To check the first equation, we note that since $V\subset p(A)$ and $A$ is saturated, $p^{-1}(V)$ is contained in $A$. It follows that both $p^{-1}(V)$ and $q^{-1}(V)$ equals all points of $A$ that are mapped by $p$ into $V$. To check the second equation, we note that for any two subsets $U$ and $A$ of $X$, we have the inclusion $$p(U\cap A)\subset p(U)\cap p(A).$$ To prove the reverse inclusion, suppose $y=p(u)=p(a)$, for $u\in U$ and $a\in A$. Since $A$ is saturated, $A$ contains the set $p^{-1}(p(a))$, so that in particular $A$ contains $u$($u\in A$). Then $y=p(u)$($u\in U$), where $u\in U\cap A$.

Step 2. Now suppose $A$ is open or $p$ is open. Given the subset $V$ of $p(A)$, we assume that $q^{-1}(V)$ is open in $A$ and show that $V$ is open in $p(A)$.

Suppose first that $A$ is open. Since $q^{-1}(V)$ is open in $A$ and $A$ is open in $X$, the set $q^{-1}(V)$ is open in $X$. Since $q^{-1}(V)=p^{-1}(V)$, the latter set is open in $X$, so that $V$ is open in $Y$ because $p$ is a quotient map. In particular, $V$ is open in $p(A)$.

Now suppose $p$ is open. Since $q^{-1}(V)=p^{-1}(V)$ and $q^{-1}(V)$ is open in $A$, we have $p^{-1}(V)=U\cap A$ for some set $U$ open in $X$. Now $p(p^{-1}(V))=V$ because $p$ is surjective; then $$V=p(p^{-1}(V))=p(U\cap A)=p(U)\cap p(A).$$ The set $p(U)$ is open in $Y$ because $p$ is an open map; hence $V$ is open in $p(A)$.

Step 3. Closed case is just substitude open to close of Step 2.

Composites of maps behave well, but products of maps and Hausdorff condition not.

Perhaps the most important result in the study of quotient spaces has to do with the problem of constructing continuous functions on a quotient space. We consider that problem now. When we studied product spaces, we had a criterion for determining whether a map $f:Z\rightarrow \prod X_\alpha$ into a product space was continuous. Its counterpart in the theory of quotient space is continuous. One has the following theorem:

Theorem 22.2. Let $p:X\rightarrow Y$ be a quotient map. Let $Z$ be a space and let $g:X\rightarrow Z$ be a map that is constant on each set $p^{-1}(\{y\})$, for $y\in Y$. Then $g$ induces a map $f:Y\rightarrow Z$ such that $f\circ p=g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.

Proof.

For each $y\in Y$, the set $g(p^{-1}(\{y\}))$ is a one-point set in $Z$ (since $g$ is constant on $p^{-1}(\{y\})$). If we let $f(y)$ denote this point, then we have defined a map $f:Y\rightarrow Z$ such that for each $x\in X$, $f(p(x))=g(x)$. If $f$ is continuous, then $g=f\circ p$ is continuous. Conversely, suppose $g$ is continuous. Given an open set $V$ of $Z$, $g^{-1}(V)$ is open in $X$. But $g^{-1}(V)=p^{-1}(f^{-1}(V))$; because $p$ is a quotient map, it follows that $f^{-1}(V)$ is open in $Y$. Hence $f$ is continuous.

If $f$ is a quotient map, then $g$ is the composite of two quotient maps and is thus a quotient map. Conversely, suppose that $g$ is a quotient map. Since $g$ is surjective, so is $f$. Let $V$ be a subset of $Z$; we show that $V$ is open in $Z$ if $f^{-1}(V)$ is open in $Y$. Now the set $p^{-1}(f^{-1}(V))$ is open in $X$ because $p$ is continuous. Since this set equals $g^{-1}(V)$, the latter is open in $X$. Then because $g$ is a quotient map, $V$ is open in $Z$.

Corollary 22.3. Let  $g:X\rightarrow Z$ be a surjective continuous map. Let $X^*$ be the following collection of subsets of $X$: $$X^*=\{g^{-1}(\{z\})|z\in Z\}.$$ Give $X^*$ the quotient topology.

(a) The map $g$ induces a bijective continuous map $f:X^*\rightarrow Z$, which is a homeomorphism if and only if $g$ is a quotient map.

(b) If $Z$ is Hausdorff, so it $X^*$.

Proof. By the preceding theorem, $g$ induces a continuous map $f:X^*\rightarrow Z$; it is clear that $f$ is bijective. Suppose that $f$ is a homeomorphism. Then both $f$ and the projection map $p:X\rightarrow X^*$ are quotient maps, so that their composite $q$ is a quotient map. Conversely, suppose that $g$ is a quotient map. Then it follows from the preceding theorem that $f$ is a quotient map. Being bijective, $f$ is thus a homeomorphism.

Suppose $Z$ is Hausdorff. Given distinct points of $X^*$, their images under $f$ are distinct and ths possess disjoint neighborhoods $U$ and $V$. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint neighborhoods of the two given points of $X^*$.