MATH31052 Topology
3.1 Remark. Given a topological space \(X\), is there a natural way of putting a topology on a subset \(X_1 \subset X\) ? One desirable property is the following since we do not expect the codomain of a map to affect whether or not it is continuous.
For all topological spaces \(Y\),
\(f\colon Y\to X_1\) is continuous \(\quad \Leftrightarrow \quad \) \(i\circ f\colon Y\to X\) is continous.
Here \(i\colon X_1\to X\) denotes the inclusion map \(i(x)=x\) for all \(x\in X_1\).
There is just one topology on \(X_1\) which has this property and this is known as the subspace topology. This property is known as the universal property of the subspace topology.
3.2 Definition. Given a topological space \((X,\tau )\) and a subset \(X_1\subset X\), then the subspace topology on \(X_1\) (induced by \(\tau \)) is given by \(\tau _1=\{\,U\cap X_1\mid U\in \tau \,\}\), i.e. \(V\subset X_1\) is open in \(X_1\) if and only if \(V=U\cap X_1\) where \(U\) is some open set in \(X\).
With this topology we say that \(X_1\) is a subspace of \(X\).
3.3 Proposition. Given a topological space \(X\) and a subset \(X_1\subset X\), Definition 3.2 defines a topology on \(X_1\). With this topology,
(a) the inclusion map \(i\colon X_1\rightarrow X\) is continuous;
(b) given a continuous function \(f\colon X\rightarrow Y\) (where \(Y\) is an topological space), the restriction \(f|X_1=f\circ i\colon X_1\rightarrow Y\) is continuous;
(c) (the universal property) a function \(f\colon Y\rightarrow X_1\) (where \(Y\) is any topological space) is continuous if and only if \(i\circ f\colon Y\rightarrow X\) is continuous.
Proof. To see that Definition 3.2 defines a topology we check the properties in Definition 2.11.
(i) \(\emptyset \) and \(X\) are open in \(X\) and so \(\emptyset \cap X_1 = X\) and \(X\cap X_1 = X_1\) are open in \(X_1\).
(ii) Given \(V_1\) and \(V_2\) open in \(X_1\) then \(V_i = U_i\cap X\) for \(U_i\) open in \(X\) (\(i=1\), 2). Hence \(V_1\cap V_2 = (U_1\cap X_1) \cap (U_2\cap X_1) = (U_1\cap U_2)\cap X_1\) is open in \(X_1\) since \(U_1\cap U_2\) is open in \(X\).
(iii) Given \(V_{\lambda }\) open in \(X_1\) for \(\lambda \in \Lambda \). Then \(V_{\lambda } = U_{\lambda }\cap X_1\) where \(U_{\lambda }\) is open in \(X\) (\(\lambda \in \Lambda \)). Hence \(\bigcup _{\lambda \in \Lambda }V_{\lambda } = \bigcup _{\lambda \in \Lambda }(U_{\lambda }\cap X_1) = \left (\bigcup _{\lambda \in \Lambda }U_\lambda \right )\cap X\) is open in \(X_1\) since \(\bigcup _{\lambda \in \Lambda }U_{\lambda }\) is open in \(X\).
(a) Given \(U\) open in \(X\) then \(i^{-1}(U) = U\cap X_1\) is open in \(X_1\) and so \(i\) is continuous.
(b) This follows from the fact the composition of continuous functions is continuous (Proposition 2.13).
(c) `\(\Rightarrow \)': This follows from the fact that the composition of continuous functions is continuous.
`\(\Leftarrow \)': Suppose that \(f\colon Y\to X_1\) is a function from a topological space \(Y\) such that \(i\circ f\colon Y\to X\) is continuous. Then given \(V\) open in \(X_1\), \(V=U\cap X_1= i^{-1}(U)\) for \(U\) open in \(X\). Thus
\(f^{-1}(V) = f^{-1}i^{-1}(U) = (i\circ f)^{-1}(U)\) is open in \(Y\) since \(i\circ f\) is continuous. Hence \(f\) is continuous.
□
3.4 Remark.
(a) The subspace topology on \(X\subset \R ^n\) induced by the usual topology on \(\R ^n\) is the usual topology on \(X\). [Exercise. Note that \(B_{\e }^X(x) = B_{\e }(x)\cap X\) for \(x\in X\) and \(\e >0\).]
(b) Given a subspace \(X_1\) of a topological space \(X\) it is not in general true that an open [closed] subset of \(X_1\) is open [closed] in \(X\). For example, \((1/2,1]\) is open in \([0,1]\) with the usual topology (since \((1/2,1] = (1/2,3/2)\cap [0,1]\)) but is not open in \(\R \).
3.5 Proposition. Given a subspace \(X_1\) of a topological space \(X\), a subset \(B \subset X_1\) is closed in \(X_1\) if and only if \(B = A \cap X_1\) where \(A\) is some closed set in \(X\).
Proof. Exercise. □ class="theoremendmark" >
3.6 Proposition. Suppose that \(X_1\) is a subspace of a topological space \(X\). Then all closed subsets of the subspace \(X_1\subset X\) are closed in \(X\) if and only if \(X_1\) is a closed subset of \(X\).
Proof. `\(\Rightarrow \)': If all closed subsets of \(X_1\) are closed in \(X\) then, since \(X_1\) is closed in \(X_1\), it is closed in \(X\).
`\(\Leftarrow \)': Suppose that \(X_1\) is a closed subset of \(X\). Then, given \(B\) closed in \(X_1\), \(B = A\cap X_1\) where \(A\) is closed in \(X\) (by Proposition 3.5) and so \(B\) is closed in \(X\) (the intersection of two closed
subsets). □
3.7 Theorem (Gluing Lemma). Suppose that \(X_1\) and \(X_2\) are closed subspaces of a topological space \(X\) such that \(X = X_1\cup X_2\). Suppose that \(f_1\colon X_1\rightarrow Y\) and \(f_2\colon X_2\rightarrow Y\) are continuous functions to a topological space \(Y\) such that, for all \(x\in X_1\cap X_2\), \(f_1(x)=f_2(x)\). Then the function \(f\colon X\rightarrow Y\) defined by \(f(x) = f_1(x)\) if \(x\in X_1\), \(f(x) = f_2(x)\) if \(x\in X_2\) is well-defined and continuous.
Proof. \(f\) is well-defined by the condition on \(f_1\) and \(f_2\) in the theorem. To see that \(f\) is continuous it is sufficient, by Problems 2, Question 8, to prove that the inverse image of a closed set in \(Y\) is closed in \(X\). Given \(A\) closed in \(Y\), \(f_j^{-1}(A)\) is closed in \(X_j\) (\(j=1\),2) since \(f_j\) is continuous and so, using Proposition 3.6, \(f_j^{-1}(A)\) is closed in \(X\) since \(X_j\) is closed in \(X\). If follows that \(f^{-1}(A) = f_1^{-1}(A) \cup f_2^{-1}(A)\) is closed in \(X\) and so \(f\) is continuous. □
3.8 Example. This result gives a justification for the continuity of the product of two paths \(\sigma _1 * \sigma _2\) in a topological space \(X\) (generalizing Definition 1.13(c) to topological spaces). For suppose that \(\sigma _1\colon [0,1]\to X\) and \(\sigma _2\colon [0,1]\to X\) are two paths in \(X\) so that \(\sigma _1(1) = \sigma _2(0)\). Then the product path \(\sigma _1\ast \sigma _2\colon [0,1]\to X\) is given by
\[\sigma _1\ast \sigma _2(s) = \left \{\begin {array}{l}\sigma _1(2s)\quad \mbox {for $0\leq s\leq 1/2$,}\\\sigma _2(2s-1)\quad \mbox {for $1/2\leq s\leq 1$.}\end {array}\right .\]
(generalizing Definition 1.13(c)). Define
\[f_1\colon [0,1/2]\mapsto X \mbox { to be the composition } [0,1/2] \stackrel {s\mapsto 2s}{\longrightarrow } [0,1] \stackrel {\sigma _1}{\longrightarrow } X, \mbox { and}\]
\[f_2\colon [1/2,1]\to X \mbox { to be the composition }[1/2,1]\stackrel {s\mapsto 2s-1}{\longrightarrow } [0,1] \stackrel {\sigma _2}{\longrightarrow } X.\]
Then \(f_1\) and \(f_2\) are compositions of continuous functions and so continuous. We can apply the Gluing Lemma to these two functions since \([0,1/2]\) and \([1/2,1]\) are closed in \([0,1]\), the intersection \([0/1/2]\cap [1/2,1]=\{1/2\}\) and \(f_1(1/2) = \sigma _1(1) = \sigma _2(0) = f_2(1/2)\). The well-defined continuous function given by the Lemma is \(\sigma _1\ast \sigma _2\).
3.9 Remark. Given topological spaces \(X_1\) and \(X_2\), is there a natural way of putting a topology on the cartesian product \(X_1\times X_2\) ? One desirable property is the following since it would generalize the familiar property that a function into \(\R ^n\) is continuous if and only if the coordinate functions are continuous (see Remarks 0.22(b)).
For all topological spaces \(Y\),
\(f\colon Y\to X_1\times X_2\mbox { is continuous }\)
\(\Leftrightarrow \)
\(p_i\circ f\colon Y\to X_i\mbox { is continous for $i=1,2$.}\)
Here \(p_i\colon X_1\times X_2\to X_i\) denotes the projection map \(p_i(x_1,x_2)=x_i\) for all \((x_1,x_2)\in X_1\times X_2\).
There is just one topology on \(X_1\times X_2\) which has this property and this is known as the product topology. This property is known as the universal property of the product topology.
3.10 Definition. Given topological spaces \(X_1\) and \(X_2\). The product topology on \(X_1\times X_2\) is the topology with a basis \(\{\,U_1\times U_2\mid U_i \mbox { open in }X_i\mbox { for $i=1$, 2}\,\}\). With this topology \(X_1\times X_2\) is called the product of the spaces \(X_1\) and \(X_2\).
3.11 Proposition. Given topological spaces \(X_1\) and \(X_2\), the set given above is the basis for a topology on \(X_1\times X_2\).
With this topology,
(a) the projection functions \(p_i\colon X_1\times X_2\rightarrow X_i\) are continuous;
(b) (the universal property) a function \(f\colon Y\rightarrow X_1\times X_2\) (for \(Y\) any topological space) is continuous if and only if the coordinate functions \(p_i\circ f\colon Y\rightarrow X_i\) are continuous for \(i=1\), 2.
Proof. To see that the collection of subsets in Definition 3.10 is a basis for a topology on \(X_1\times X_2\) we use the result of Problems 2, Question 11. Given two basic open sets \(U_1\times U_2\) and \(U_1'\times U_2'\) in \(X_1\times X_2\) (i.e. \(U_i\) and \(U_i'\) are open in \(X_i\) for \(i=1\), 2), then
\[(U_1\times U_2)\cap (U_1'\times U_2') = (U_1\cap U_1')\times (U_2\cap U_2')\]
(by Proposition 0.7(iii)) which is also a basic open set since \(U_i\cap U_i'\) is open in \(X_i\) for \(i=1\), 2.
(a) For \(U\) open in \(X_1\), \(p_1^{-1}(U) = U\times X_2\) which is open in \(X_1\times X_2\). Hence \(p_1\) is continuous. Similarly, \(p_2\) is continuous.
(b) `\(\Rightarrow \)': This follows from the continuity of a composition of continuous functions.
`\(\Leftarrow \)': To prove that \(f\colon Y\rightarrow X_1\times X_2\) is continuous it is sufficient to prove that \(f^{-1}(U_1\times U_2)\) is open in \(Y\) for basic open sets \(U_1\times U_2\) by Problems 2, Question 9.
Given such a basic open set and a function \(f\colon Y\to X_1\times X_2\) such that the coordinate functions \(p_1\circ f\) and \(p_2\circ f\) are continuous, \((p_1\circ f)^{-1}(U_1) = f^{-1}p_1^{-1}(U_1) = f^{-1}(U_1\times X_2)\) is
open in \(Y\) and, similarly, \((p_2\circ f)^{-1}(U_2) = f^{-1}(X_1\times U_2)\) is open in \(Y\). Hence, by taking the intersection of these open sets, \(f^{-1}(U_1\times X_2)\cap f^{-1}(X_1\times U_2) = f^{-1}(U_1\times X_2\cap
X_1\times U_2) = f^{-1}(U_1\times U_2)\) is open in \(Y\) and so \(f\) is continuous. □
3.12 Remark.
(a) In the same way we can define the product topology on any finite product \(X_1\times X_2\times \cdots \times X_n\) of topological spaces: a basis is given by subsets of the form \(U_1\times U_2 \times \cdots \times U_n\) where \(U_i\) is an open subset of \(X_i\).
(b) For each point \(x_2\in X_2\), the subspace \(X_1\times \{x_2\}\) of the product space \(X_1\times X_2\) is homeomorphic to \(X_1\).
To see this we prove that the obvious bijection \(f\colon X_1\to X_1\times \{x_2\}\) given by \(f(x) = (x,x_2)\) for \(x\in X_1\) is a homeomophism by using using the universal properties of the product topology and the subspace topology.
First of all, \(f\) is continuous if and only if \(i_1 = i\circ f\colon X_1 \to X_1\times {x_2} \to X_1\times X_2\) is continuous (by the universal property of the subspace topology) if and only if \(p_1\circ i_1 = \id _{X_1}\colon X_1\to X_1\) is continuous and \(p_2\circ i_1=c_{x_2}\colon X_1\to X_2\) is continuous (by the universal property of the product topology) and these maps are continuous by Examples 2.17(e) and (f). Hence \(f\) is continuous.
Secondly, the function \(f^{-1}\colon X_1\times \{x_2\}\to X_1\) is the restriction of the projection map \(p_1\colon X_1\times X_2 \to X_1\) which is continuous by Proposition 3.11(a) and so is continuous by Proposition 3.3(b).
(c) Given subsets \(Y_1\subset X_1\) and \(Y_2\subset X_2\) of topological spaces \(X_1\) and \(X_2\) then \(Y_1\times Y_2\) may be topologized as (i) a subspace of the product space \(X_1\times X_2\), and (ii) the product of the subspaces \(Y_1\) and \(Y_2\). These two topologies are the same. [Exercise. Use the universal properties to show that the identity map \(\id _{Y_1\times Y_2}\colon (Y_1\times Y_2, \tau _1) \to (Y_1\times Y_2,\tau _2)\) is a homeomorphism where \(\tau _1\) is the topology (i) and \(\tau _2\) is the topology \(\tau _2\).]
3.13 Example.
(a) Euclidean \(n\)-space \(\R ^n\) with the usual topology is homeomorphic to the product space \(\R ^{n-1}\times \R \) (with the usual topologies on \(\R \) and \(\R ^{n-1}\). [Exercise.]
(b) If \(X\) and \(Y\) have the discrete topology then the product topology on \(X\times Y\) is the discrete topology.
(c) The product space \([0,1]\times S^1\) is called the cylinder.
(d) The product space \(S^1\times S^1\) is called the torus.
(e) The product space \(D^2\times S^1\) is called the solid torus.
1 . Suppose that \(X_2\subset X_1\subset X\) where \(X\) is a topological space. Prove that the subspace topology on \(X_2\) induced by the topology on \(X\) is the same as the subspace topology on \(X_2\) induced by the subspace topology on \(X_1\).
2 . Suppose that \(X\subset \R ^n\). Prove that the subspace topology on \(X\) induced by the usual topology on \(\R ^n\) is the same as the usual topology on \(X\) (given by Definition 2.5). [Remark 3.4]
3 . Prove Proposition 3.5: For a subspace \(X_1\) of a topological space \(X\), the closed subsets of \(X_1\) are all subsets of the form \(A\cap X_1\) where \(A\) is a closed subset of \(X\).
4 . Prove that all open subsets of a subspace \(X_1\) of a topological space \(X\) are open in \(X\) if and only if \(X_1\) is open in \(X\).
Deduce the following version of the Gluing Lemma (cf. Theorem 3.7). Suppose that \(X_1\) and \(X_2\) are open subspaces of \(X\) such that \(X=X_1\cup X_2\) and \(f_i\colon X_i\to Y\) are continuous functions to a topological space \(Y\) (\(i=1\), 2) such that \(f_1(x)=f_2(x)\) for \(x\in X_1\cap X_2\). Then the function \(f\colon X\to Y\) defined by \(f(x) = f_i(x)\) for \(x\in X_i\) is well-defined and continuous.
5 . Find subspaces \(X_1\) and \(X_2\) of a topological space \(X\) and continuous functions \(f_i\colon X_i\to Y\), to a topological space \(Y\) such that \(f_1(x)=f_2(x)\) for all \(x\in X_1\cap X_2\) such that the function \(f\colon X\to Y\) given by \(f(x)=f_i(x)\) for \(x\in X_i\) is not continuous. [Hint: What is the simplest non-continuous function you can think of? Find some subspaces on which its restriction is continuous.]
6 . Prove that the product space \(\R \times S^1\) (the infinite cylinder) is homeomorphic to the punctured plane \(\R ^2-\{0\}\).
7
. Given subsets \(Y_1\subset X_1\) and \(Y_2\subset X_2\) of topological spaces \(X_1\) and \(X_2\) prove that the following two topologies on
\(Y_1\times Y_2\) are the same:
(i) \(Y_1\times Y_2\) is a subspace of the product space \(X_1\times X_2\);
(ii) \(Y_1\times Y_2\) is a product of the subspaces \(Y_1\) (of \(X_1\)) and \(Y_2\) (of \(X_2\)).
[Remark 3.12(c)]
[Hint: Consider the identity function \((Y_1\times Y_2, \tau _1)\rightarrow (Y_1\times Y_2,\tau _2)\) where \(\tau _1\) and \(\tau _2\) are the two topologies defined in the question. Use the universal properties of the product topology and
the subspace topology to show that this identity function and its inverse are both continuous.]
8 . Prove that the product space \(\R \times \{0,1\}\), where \(\R \) has the usual topology and \(\{0,1\}\) has the indiscrete topology, is path-connected.
9 . Suppose that \(X_1\) and \(X_2\) are disjoint topological spaces (i.e. \(X_1\cap X_2=\emptyset \)). Prove that we may define a topology on the union \(X_1\cup X_2\) by: \(U\subset X_1\cup X_2\) is open if and only if \(U\cap X_1\) is open in \(X_1\) and \(U\cap X_2\) is open in \(X_2\). With this topology \(X_1\cup X_2\) is called the disjoint union of the topological spaces (or sometimes the coproduct), often denoted \(X_1\sqcup X_2\).
Prove the following universal property of the disjoint union topology. A function \(f\colon X_1\sqcup X_2\rightarrow Y\) to a topological space \(Y\) is continuous if and only if the restricted functions \(f|X_1=f\circ i_1\colon X_1\rightarrow Y\) and \(f|X_2=f\circ i_2\colon X_2\rightarrow Y\) are continuous (where \(i_j\colon X_j\rightarrow X_1\sqcup X_2\) for \(j=1,2\) are the inclusion maps).