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Section 2.4 Quotient spaces

Let \(X\) be a set. Recall that an equivalence relation \(\sim\) on \(X\) is a relation that satisfies the following conditions:

  • For every \(x \in X\text{,}\) one has \(x \sim x\text{.}\)

  • For every \(x,y \in X\text{,}\) one has \(x \sim y\) if and only if \(y \sim x\text{.}\)

  • For every \(x,y,z \in X\text{,}\) if \(x \sim y\) and \(y \sim z\text{,}\) then \(x \sim z\text{.}\)

If \(x \in X\text{,}\) one can form the equivalence class \([x] \coloneq \{ y \in X : y \sim x \}\text{.}\) Now the set \(X/\!\sim \colon \{ [x] : x \in X \}\) of equivalence classes relative to \(\sim\) is called the quotient of \(X\) relative to \(\sim\text{.}\) The assignment \(x \mapsto [x]\) is a map \(q \colon X \to X/\!\sim\text{,}\) which is called the quotient map; The map \(q\) is a surjection, and it enjoys the following pleasant property: for every set \(Y\) and every map \(f \colon X \to Y\) with the property that for every \(x, y \in X\text{,}\) if \(x \sim y\text{,}\) then \(f(x) = f(y)\text{,}\) there exists a unique map \(f' \colon X/\!\sim \to Y\) such that \(f' \circ q = f\text{.}\)

Every surjection is secretly a quotient map: if \(p \colon X \to T\) is a surjection, then we may define an equivalence relation \(\sim\) on \(X\) in which \(x \sim y\) if and only if \(p(x) = p(y)\text{.}\) We may call this the equivalence relation induced by \(p\text{.}\) The unique map \(p' \colon X/\!\sim \to T\) through which \(p\) factors is then a bijection.

Finally, if \(R\) is any relation on \(X\text{,}\) then it generates an equivalence relation. That is, recall that \(\sim\) is really a subset

of \(X \times X\text{.}\) If we look at the collection of all subsets \(Q \in \PP(X \times X)\) that are equivalence relations that contain \(R\text{,}\) then the intersection of all these \(Q\) is itself an equivalence relation that contains \(R\text{.}\) This intersection is the equivalence relation generated by \(R\text{.}\)

Definition 2.4.1.

Let \(X\) be a topological space, let \(Y\) be a set, and let \(p \colon X \to Y\) be a surjection. The final topology with respect to \(p\) is the quotient topology on \(Y\text{;}\) this is the finest topology on \(Y\) such that the map \(p\) is continuous. We call \(p\) the quotient map. We also say that \(Y\) is the quotient space of \(X\) under the equivalence relation induced by \(p\text{.}\)

The ability to form quotients of topological spaces is really the most powerful motivation for defining abstract topological spaces in the first place.

Consider the real line \(\RR\text{,}\) and consider the equivalence relation \(\sim\) in which \(x \sim y\) if and only if \(y-x\) is an integer. Now consider the map \(t \mapsto \exp(2\pi t)\text{;}\) this is a continuous map \(e \colon \RR \to S^1\text{.}\) By the discussion above, there exists a unique map \(f \colon \RR/\!\sim \to S^1\) such that if \(q \colon \RR \to \RR/\!\sim\) is the quotient map, then \(f \circ q = e\text{.}\) This map is continuous with respect to the quotient topology on \(\RR/\sim\text{.}\) In fact, \(f\) is a homeomorphism.

Similarly, consider the closed interval \([0,1]\text{,}\) and consider the equivalence relation \(\sim\) generated by the requirement that \(0\sim 1\text{.}\) Then the inclusion map \([0,1] \inclusion \RR\) descends to a continuous map \([0,1]/\sim \to \RR/\!\sim\text{,}\) and this map is a homeomorphism as well. This justifies our intuition that the circle is obtained from the interval by identifying the endpoints.

Let \(X \subseteq \RR^n\) be a subspace, and let \(k \in \NN\) be a natural number. Then we have the subspace \(X^k \subseteq \RR^{nk}\text{.}\) Furthermore, we can define the subsspace

\begin{equation*} Y \coloneq \{(x_1,\dots,x_k) \in X^k : (\forall i \neq j)(x_i \neq x_j) \} \subseteq X^k\text{.} \end{equation*}

Now conisder the equivalence relation \(\sim\) on \(Y\) in which \((x_1, \dots, x_k) \sim (y_1,\dots, y_k)\) if and only if there exists a permutation

\(\sigma \in \Sigma_k\) such that for each \(i\text{,}\) one has \(y_i = x_{\sigma(i)}\text{.}\) Now we can think of the points of the quotient space \(C_k(X) \coloneq Y/\!\sim\) as unordered collections of \(k\) points on \(X\text{.}\) This space \(C_k(X)\) is called the configuration space of \(k\) marked points on \(X\text{.}\)

The space \(C_0(X)\) has exactly one point.

The space \(C_1(X)\) is \(X\) itself.

The space \(C_2(X)\) is much more interesting. For example, \(C_2(\RR^2) \cong \RR^3 \times S^1\text{.}\) The space \(C_2(S^1)\) is the Möbius band.

As \(k\) goes higher, these spaces become even more involved.

The spaces \(C_k(X)\) are extremely important in physics, since we may think of them as the space of possible states of \(k\) noninteracting particles on \(X\text{.}\)

Let \(V\) be a finite dimensional real vector space. Since \(V\) is a finite dimensional, it is isomorphic to \(\RR^n\text{.}\) Let us transport the topology from \(\RR^n\) to \(V\) along such an isomorphism.

Thus \(V\) is homeomorphic to \(\RR^n\text{.}\)

For every \(k \in \NN^{\ast}\text{,}\) the set \(V^k\) of \(k\)-tuples \((v_1, \dots, v_k)\) of vectors in \(V\) is also a vector space, and thus also a topological space that is homeomorphic to \(\RR^{nk}\text{.}\) Now we let \(L_{\RR}(k,V) \subseteq V^k\) be the set of \(k\)-tuples \((v_1, \dots, v_k)\) of vectors in \(V\) that are linearly independent, so that they span a \(k\)-dimensional subspace of \(V\text{.}\)

Now we define an equivalence relation \(\sim\) on \(L_{\RR}(k, V)\) by the rule that \((v_1, \dots, v_k) \sim (w_1, \dots, w_k)\) if and only if their spans are equal (as subspaces of \(V\)). The quotient space \(G_{\RR}(k,V) \coloneq L_{\RR}(k,V)/\!\sim\) is called the Grassmannian of \(k\)-dimensional subspaces of \(V\text{.}\) The points of \(G_{\RR}(k,V)\) are \(k\)-dimensional real linear subspaces of \(V\text{.}\) One often writes \(G_{\RR}(k,n)\) as a shorthand for \(G_{\RR}(k, \RR^n)\text{.}\)

In the particular case in which \(k=1\text{,}\) the space \(G_{\RR}(1,n)\) is the space of lines through the origin of \(\RR^n\text{;}\) this is also called the (real) projective space \(\PP^{n-1}_{\RR}\text{.}\)

We can do this same thing with \(\CC\) in place

of \(\RR\text{.}\)

In particular, \(G(k,n)\) is the space of complex lines through the origin in \(\CC^n\text{;}\) this is called the (complex) projective space \(\PP^{n-1}\text{.}\)

Let \(X\) and \(Y\) be topological spaces, and let \(p \colon X \to Y\) be a continuous surjection. Then \(p\) need not be a quotient map. For example, the map \(e \colon \left]0,1\right] \to S^1\) given by \(t \mapsto \exp(2\pi t)\) is not a quotient map: even though \(e^{-1}(e(\left]1/2,1/2\right])) = \left]1/2,1/2\right]\) is open, the subset \(e(\left]1/2,1/2\right]) \subseteq S^1\) is not. If \(p\) is an open map — that is, if it carries open sets to open sets — then it is a quotient map.

Let \(\{X_a\}_{a\in A}\) be an indexed family of sets. Then the coproduct of sets is by definition the union

\begin{equation*} \coprod_{a\in A}X_a\coloneq\bigcup_{a\in A}X_a\times\{a\}\text{.} \end{equation*}

It is engineered precisely so that a map out of the coproduct is determined by each of its components. More precisely, the coproduct is equipped with inclusion maps

\begin{equation*} \iota_b\colon\fromto{X_b}{\coprod_{a\in A}X_a}\text{,} \end{equation*}

one for every \(b\in A\text{,}\) given by \(\iota_b(x)=(x,b)\text{.}\) For any set \(Y\text{,}\) and for any collection \(\{ f_a \colon X_a \to Y\}_{a\in A}\) of maps, there exists a unique map

\begin{equation*} f \colon \coprod_{a \in A} X_a \to Y \end{equation*}

such that for each \(a\in A\text{,}\) one has \(f \circ \iota_a = f_a\text{.}\) Thus composition with the inclusions induces a bijection

\begin{equation*} \prod_{a\in A}-\circ \iota_{a} \colon\fromto{\Map\left(\coprod_{a\in A}X_a,Y\right)}{\prod_{a\in A}\Map(X_a,Y)}\text{.} \end{equation*}
Definition 2.4.5.

Let \(X\) be a topological space, and let \(A\subseteq X\) a subspace. Consider the map \(q\colon\fromto{X}{(X-A)\sqcup\{X-A\}}\) defined by

\begin{equation*} q(x)\coloneq \begin{cases}x\amp \text{ if } x\notin A;\\ X-A\amp \text{ if } x\in A. \end{cases} \end{equation*}

We define \(X/A\) as the set \((X-A)\sqcup\{X-A\}\) equipped with the final topology with respect to \(q\text{.}\)

Consider the subspace

\begin{equation*} D^{n}\coloneq\{x\in\RR^{n} : \|x\|\leq 1\}\subset\RR^{n} \end{equation*}

and the subspace

\begin{equation*} S^{n-1}\coloneq\{x\in D^{n} : \|x\|=1\}\subset D^{n}\text{.} \end{equation*}

Then the space \(D^n/S^{n-1}\) is homeomorphic to \((\RR^n)^+\text{,}\) which is in turn homeomorphic to \(S^n\text{.}\)

Definition 2.4.7.

Let \(\{(X_a,\tau_a)\}_{a\in A}\) be an indexed family of topological spaces. Then the coproduct topology \(\coprod_{a\in A}\tau_a\) is the final topology with respect to the set \(\{\iota_a\}_{\alpha\in A}\text{.}\)

Let \(S=\coprod_{a\in A}S_a\subseteq\coprod_{a\in A}X_a\) (so that \(S_a\subseteq X_a\)). If \(\tau\) is the coproduct topology, then

\begin{equation*} \tau(S)=\coprod_{a\in A}\tau_a(S_a)\text{.} \end{equation*}

If \(X\) is topological space, then \(X_+\) denotes the coproduct \(X\sqcup\{X\}\) with the coproduct topology.

Let \(A = (A,\delta)\) be a discrete topological space. Suppose \(f\colon\fromto{X}{A}\) a continuous map. Then we have a bijection \(G \colon \coprod_{a\in A}f^{-1}\{a\} \to X\text{.}\) Let us show that this is a homeomorphism, where the right hand side is given the coproduct topology (and the fibers \(f^{-1}\{a\} \subseteq X\) are given the subspace topology). By construction, the map \(G\) is continuous. It remains to show that \(G^{-1}\) is continuous; let \(U \subseteq \coprod_{a\in A}f^{-1}\{a\}\) be an open subset. Then \(U \cap f^{-1}\{a\}\) is open in \(f^{-1}\{a\}\text{,}\) and so since \(U\) is the union of these intersections, it suffices to prove that any open subset of \(f^{-1}\{a\}\) is open in \(X\text{.}\) Indeed, this is true, since \(f^{-1}\{a\}\) is itself open.

Conversely, if \(\{X_a\}_{a\in A}\) is an indexed family of topological spaces, then define the map

\begin{equation*} g\colon\fromto{\coprod_{a\in A}X_a}{A} \end{equation*}

such that \(g(x)=a\) if and only if \(x\in X_a\text{.}\) The map \(g\) is continuous.

In other words, a decomposition of a toploogical space into a coproduct of topological spaces is equivalent to a continuous map into a discrete topological space.

Much of the story of topology is the story of “gluing topological spaces together” to get new ones. We are now ready to say exactly what this means.

Let \(U\text{,}\) \(V\text{,}\) and \(X\) be three topological spaces, and let \(f\colon\fromto{X}{U}\) and \(g\colon\fromto{X}{V}\) be two continuous maps. Consider the equivalence relation \(\sim\) on \(U\sqcup V\) generated by declaring that: for every element \(x\in X\text{,}\) \(f(x) \sim g(x)\text{.}\) Write \(U \cup^X V\) for the quotient \((U\sqcup V)/\sim\) with the quotient topology.

Consider the subset

\begin{equation*} D^{n}\coloneq\{x\in\RR^{n} : \|x\|\leq 1\}\subset\RR^{n} \end{equation*}

and the subset

\begin{equation*} S^{n-1}\coloneq\{x\in D^{n} : \|x\|=1\}\subset D^{n}\text{.} \end{equation*}

Here's a standard way to build a new space from an old one. Suppose \(X\) a topological space, and suppose \(f\colon\fromto{S^{n-1}}{X}\) a continuous map. Then we can form the cell attachment for attaching map \(f\)

\begin{equation*} X\cup^{S^{n-1}}D^n\text{.} \end{equation*}

A cell complex is a topological space \(Y\) that is built via cell attachments from discrete spaces; that is, a cell complex is a topological space \(Y\) along with a sequence of subspaces

\begin{equation*} Y_0\subseteq Y_1\subseteq \cdots\subseteq Y \end{equation*}

such that:

  • \(Y=\bigcup_{n\geq 0}Y_n\text{;}\)

  • \(Y_0\) is discrete;

  • for any \(n\geq 1\text{,}\) there exist an indexed family \(\left\{f_{\alpha}\colon\fromto{S^{n-1}}{Y_{n-1}}\right\}_{\alpha\in A_n}\) of attaching maps such that

    \begin{equation*} Y_n=Y_{n-1}\cup^{\coprod_{\alpha\in A_n}S^{n-1}}\coprod_{\alpha\in A_n}D^{n}\text{.} \end{equation*}