MATH31052 Topology
3.14 Definition . Suppose that \(q\colon X\rightarrow Y\) is a surjection from a topological space \(X\) to a set \(Y\). Then the quotient topology (or the identification topology) on \(Y\) determined by \(q\) is given by the condition \(V\subset Y\) is open in \(Y\) if and only if \(q^{-1}(V)\) is open in \(X\). With this topology we call \(Y\) a quotient space of \(X\).
3.15 Proposition. Given a surjection \(q\colon X\rightarrow Y\) from a topological space
\(X\) to a set \(Y\), the above definition gives a topology on \(Y\). With this topology,
(a) the function \(q\colon X\rightarrow Y\) is continuous;
(b) (the universal property) a function \(f\colon Y\rightarrow Z\) to a topological space \(Z\) is continuous if and only if the composition \(f\circ q\colon X\rightarrow Z\) is continuous.
Proof. Exercise. You might use the proof of Proposition 3.3 as a model. □
3.16 Remark . Given an equivalence relation \(\sim \) on a topological space \(X\) there is a surjection \(q\colon X\rightarrow X/\!\!\sim \) to the set of equivalence classes given by sending each element of \(X\) to its equivalence class: \(q(x)=[x] = \{\,x'\in X\mid x'\sim x\,\}\) (see Definition 0.18). We can give \(X/\!\!\sim \) the quotient topology determined by \(q\). We call such a quotient space an identification space of \(X\).
3.17 Definition . Suppose that \(A\) is a non-empty subset of a topological space \(X\). Then we may define an equivalence relation on \(X\) by
\[x \sim x' \Leftrightarrow x=x' \mbox { or both } x \mbox { and }x'\in A.\]
In this case we write the set of equivalence classes \(X/\!\!\sim \) as \(X/A\). With the quotient topology this is called the identification space obtained from \(X\) by collapsing \(A\) to a point. Notice that, as a set, \(X/A\) has one point for each point \(x\in X \setminus A\) (since \([x]=\{x\}\)) and one point corresponding to all of \(A\) (since if \(a\in A\), \([a]=A\)).
3.18 Theorem . Suppose that \(f\colon X\rightarrow Y\) is a continuous surjection of topological spaces. Then we may define an equivalence relation on \(X\) by
\[x\sim x' \Leftrightarrow f(x)=f(x')\]
and then the bijection \(F\colon X/\!\!\sim \;\rightarrow Y\) induced by \(f\), i.e. \(F([x])=f(x)\) for \(x\in X\) (see Theorem 0.20), is a continuous bijection of topological spaces.
Proof. The proof that \(F\) is a bijection is the proof of Theorem 0.20.
The continuity of \(F\) follows from the universal property of the quotient topology since \(F\circ q(x) = F\bigl (q(x)\bigr ) = F([x]) = f(x)\) so that \(F\circ q=f\colon X \to Y\) which is continuous. □
3.19 Example . \([0,1]/\{0,1\}\cong S^1\).
The homeomorphism is induced by the continuous function \(f\colon [0,1]\rightarrow S^1\) given by \(f(t)=(\cos 2\pi t,\sin 2\pi t)\).
To see this notice that \(f\) is continuous (since the component functions are continuous) and
\begin{equation} f(t)=f(t') \Leftrightarrow t=t'\mbox { or }t,t'\in \{0,1\}. \end{equation}
Then, by Theorem 3.18, \(f\) induces a continuous bijection \(F\colon [0,1]/\{0,1\}\;\to S^1\) by \(F[t]=f(t)\). It will follow from a result in 5 that this is a homeomorphism but this may be shown directly as follows.
The inverse \(F^{-1}\colon S^1\to [0,1]/\{0,1\}\) is given by
\[F^{-1}(\x ) = \left \{\begin {array}{ll}q(\cos ^{-1}(x_1)/2\pi )&\mbox {for $x_2\geq 0$,}\\ q(1 - \cos ^{-1}(x_1)/2\pi )&\mbox {for $x_2\leq 0$,}\end {array}\right .\]
writing \(\x = (x_1,x_2)\).This uses the principal value of the inverse cosine, \(\cos ^{-1}\colon [-1,1]\to [0,\pi ]\) which is a continuous function, and the continuous quotient map \(q\colon [0,1]\to [0,1]/\{0,1\}\). This function is continuous on the two closed subsets \(\{\,\x \in S^1\mid x_2\geq 0\,\}\) and \(\{\,\x \in S^1\mid x_2\leq 0\,\}\), and agrees on their intersection \(\{(\pm 1,0)\}\). Hence, by the Gluing Lemma, \(F^{-1}\colon S^1 \to [0,1]/\{0,1\}\) is continuous and so \(F\) is a homeomorphism.
3.20 Example .
(a) There is a continuous bijection \(D^n/S^{n-1}\to S^n\) (which is in fact a homeomorphism). Define \(f\colon D^n \to S^n\) by
\[f(\x ) = \biggl (\frac {2\sqrt {|\x |-|\x |^2}}{|\x |}\,\x , 1 - 2|\x |\biggr ).\]
This is continuous since the component functions are continuous (actually this formula doesn't make sense if \(\x =\0\) but, as \(\x \to \0\), \(f(\x )\to (\0,1)\) and so if we put \(f(\0)=(\0,1)\) we get a continuous function). We can check that \(|f(\x )|=1\) so that \(f(\x )\in S^n\). Given \(\y \in S^n\), \(f(\x )=\y \Leftrightarrow 1 - 2|\x | = y_{n+1}\mbox { and } 2(\sqrt {|\x |-|\x |^2}/|\x |)\,\x = (y_1,\ldots ,y_n)\). This means that for \(\y \in S^n\) there is a unique \(\x \in D^n\) such that \(f(\x )=\y \) so long as \(|y_{n+1}|<1\). For \(y_{n+1}=1\), \(f(\x )=(\0,1)\Leftrightarrow \x =\0\). For \(y_{n+1}=-1\), \(f(\x )=(\0,-1)\Leftrightarrow |\x |=1\). Hence \(f\) is a continuous surjection and
\( \seteqnumber {2} \)
\begin{equation} f(\x )=f(\x ')\Leftrightarrow \x =\x '\mbox { or }\x ,\x '\in S^{n-1}. \end{equation}
Define \(F\colon D^n/S^{n-1}\to S^n\) by \(F([\x ]) = f(\x )\). This is well-defined and an injection by (2). It is a surjection since \(f\) is a surjection. Thus \(F\) is a continuous bijection. It will follow from a result in 5 that \(F\) is a homeomorphism (a direct proof is a little awkward in this case).
(b) There is an equivalence relation on the unit square \(I^2=I\times I\) (where \(I=[0,1]\) with the usual topology) such that there is a continuous bijection \(I^2/\!\!\sim \; \to I\times S^1\), the cylinder (which is in fact a homeomorphism). To see this, define a surjection \(f\colon I^2\to I\times S^1\) by \(f(x,y) = \bigl (x,\exp (2\pi iy)\bigr )\) where we think of \(S^1\) as \(\{\,z\in \C \mid |z|<1\,\}\) using the standard identification \(\C \cong \R ^2\). This function is continuous by the universal property of the product topology since the component functions are continuous and so Theorem 3.18 applies giving a continuous bijection \(F\colon I^2/\!\!\sim \;\to I\times S^1\). It will follow from a result in 5 that this is a homeomorphism. It is possible to prove directly that \(F^{-1}\) is continuous using the Gluing Lemma using a argument like of that at the end of Example 3.19.
The equivalence relation on \(I^2\) can be described explicitly by
\[ (x,y) \sim (x',y') \Leftrightarrow \left \{ \begin {array}{l}(x,y) = (x',y')\mbox { or }\\ x=x',\;y=0\mbox { and }y=1\mbox { or }\\ x=x',\; y=1\mbox { and }y'=0.\end {array}\right .\]
We say that this equivalence relation is generated by the relation \((x,0)\sim (x,1)\) for all \(x\in I\) (since the other relations are forced by reflexivity and symmetry).
(c) There is an equivalence relation on the unit square \(I^2\) such that there is a continuous bijection \(I^2/\!\!\sim \; \to S^1\times S^1\) (and again this is in fact a homeomorphism).
This is left as an exercise. The equivalence relation is generated by \((x,0)\sim (x,1)\) and \((0,y)\sim (1,y)\) for all \(x,\,y\in I\)
(d) We may generate an equivalence relation on \(I^2\) by \((x,0)\sim (1-x,1)\) for all \(x\in I\). The identification space \(I^2/\!\!\sim \) is called the Möbius band.
(e) We may generate an equivalence relation on \(I^2\) by \((x,0)\sim (x,1)\) and \((0,y)\sim (1,1-y)\) for all \(x,\,y\in I\). The identification space \(I^2/\!\!\sim \) is called the Klein bottle.
(f) We may define an equivalence relation on \(I^2\) by \((x,0)\sim (1-x,1)\) and \((0,y)\sim (1,1-y)\) for all \(x,\,y\in I\). The identification space \(I^2/\!\!\sim \) is homeomorphic to a space called the projective plane, denoted \(P^2\), usually defined as follows.
3.21 Definition . Define an equivalence relation on \(S^n\) by \(\x \sim \pm \x \) for all \(\x \in S^n\). Then the identification space \(S^n/\!\!\sim \) is called (real) projective \(n\)-space and is denoted \(P^n\) (or sometimes \(\R P^n\)).
3.22 . Remark The formal proof that the identification space of Example 3.20(f) is homeomorphic to the projective plane \(P^2\) is omitted.
1. Prove Proposition 3.15
2. Let \(X = \{\,x\in \R ^2 \mid 1\leq |x|\leq 2\,\}\) be an annulus in \(\R ^2\) with the usual topology. Define a continuous bijection from the
quotient space \(X/S^1\) to unit disc \(D^2\) in \(\R ^2\) with the usual topology.
[It follows from a general result in 5 that such a map is a homeomorphism.]
3. Suppose that \(\sim \) is the equivalence relation on \(I^2\) generated by \((x,0)\sim (x,1)\) and \((0,y)\sim (1,y)\) for all \(x\), \(y\in I=[0,1]\). Prove that there is a continuous bijection \(I^2/\!\!\sim \;\to S^1\times S^1\). [Example 3.20(c). It will follow from a general result in 5 that such a map is a homeomorphism.]
4. Define an equivalence relation on the product space \(S^1\times [-1,1]\) by \((x,t)\sim (x',t')\) if and only if \((\x ,t) = (\x ',t')\) or \(t = t' = 1\) or
\(t = t' = -1\). Define a continuous bijection from the quotient space \((S^1\times [-1,1])/\!\!\sim \) to \(S^2\) with the usual topology.
[It follows from a general result in 5 that such a map is a homeomorphism.]
5. Define an equivalence relation on \(\R ^{n+1}\setminus \{0\}\) by \(x\sim \lambda x\) for all non-zero real numbers \(\lambda \). Prove that the
quotient space \((\R ^{n+1}\setminus \{0\})/\!\!\sim \) is homeomorphic to \(P^n\), real projective \(n\)-space (as defined in 3.21 in the notes).
[This is the classical definition of projective \(n\)-space: as the set of lines through the origin in \(\R ^{n+1}\).]
6. Let \(f\colon S^2\rightarrow \R ^4\) be defined by
\[f(x_1,x_2,x_3) = (x_1^2-x_2^2,x_1x_2,x_1x_3,x_2x_3).\]
Prove that \(f\) induces a continuous bijection \(F\colon P^2 \rightarrow F(P^2)\subset \R ^4\) where \(F(P^2)\) has the usual topology as a subset of \(\R ^4\).
[It is a little tricky to show that \(f(\x )=f(\x ')\Rightarrow \x =\pm \x '\) so if this causes problems I suggest that you don't spend too much time on it.
It follows from a general result in 5 that this map is a homeomorphism. Such a homeomorphism is called an embedding of the projective plane in \(\R ^4\). It can be shown that there is no embedding of the projective plane in \(\R ^3\).]
7. Let \(D^n_r(\a ) = \{\,\x \in \R ^n\mid |\x -\a |\leq r\,\}\). Define an equivalence relation on the subset \(X=D^2_1(2,0)\cup D^2_1(-2,0)\) of
\(\R ^2\) with the usual topology by \((2,0)+\x \sim (-2,0)+\x \) for \(|\x |=1\). Prove that there is a homeomorphism from the quotient space \(X/\!\!\sim \) to \(S^2\) with the usual topology.
[This shows that gluing two discs together by their boundary circles gives a sphere. You can get a continuous function \(X\rightarrow S^2\) which induces the homeomorphism by mapping one disc to the upper hemisphere (using the map of \(D^2
\to \{\,\x \in S^2\mid x_2\geq 0\,\}\) asked for in Problems 1, Question 4) and the other to the lower hemisphere. You can prove that the inverse map is continuous using the Gluing Lemma.]
8. Define an equivalence relation on \(D^2\) by \(\x \sim \x ' \Leftrightarrow \x = \x ' \mbox { or } (\x ,\;\x '\in S^1 \mbox { and }\x ' = - \x \).
Prove that the there is a continuous bijection from \(D^2/\!\!\sim \) to \(P^2\).
[It follows from a general result in 5 that such a map is a homeomorphism. To construct the bijection you may find it useful to make use of the map \(D^2 \to \{\,\x \in S^2\mid x_2\geq 0\,\}\) asked for in Problems 1, Question 4.]