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MATH31052 Topology

1 Topological Equivalence and Path-Connectedness

  • 1.1 Definition.   Suppose that \(X\) and \(Y\) are subsets of Euclidean spaces. A function \(f\colon X\rightarrow Y\) is a topological equivalence or a homeomorphism if it is a continuous bijection such that the inverse \(f^{-1}\colon Y\rightarrow X\) is also continuous. If such a homeomorphism exists then \(X\) and \(Y\) are topologically equivalent or homeomorphic, written \(X\cong Y\).

  • 1.2 Example.  

    • (a) The real line \(\R \) and the open half line \((0,\infty )=\{\,x \in \R \mid x>0\,\}\) are homeomorphic. A homeomorphism is given by \(\exp \colon \R \to (0,\infty )\) with inverse \(\log _e\colon (0,\infty )\to \R \).

    • (b) \(X=\R ^2\setminus \{\0\}\), the punctured plane, and \(Y=\{\,\x =(x_1,x_2,x_3)\in \R ^3\mid x_1^2+x_2^2=1\,\}\), the infinite cylinder, are homeomorphic. A homeomorphism \(f\colon X\to Y\) is given by

      \[ f(x_1,x_2) = \bigl (x_1/|\x |,x_2/|\x |,\log _e(|\x |)\bigr )\]

      with inverse \(g\colon Y\to X\) given by

      \[ g(y_1,y_2,y_3) = e^{y_3}(y_1,y_2).\]

  • 1.3 Exercise.  

    • (a) The punctured plane, \(X=\R ^2\setminus \{\0\}\), is homeomorphic to the complement of the unit disc, \(Z=\{\,\x \in \R ^2\mid |\x |>1\,\} = \R ^2\setminus D^2\) where \(D^2 = \{\,\x \in \R ^2\mid |\x |\leq 1\,\}\).

    • (b) \(S^1=\{\,\x \in \R ^2\mid |\x |=1\,\}\), the unit circle, and \(T=\{\,\x =(x_1,x_2)\in \R ^2\mid |x_1|+|x_2|=1\,\}\), the diagonal square, are homeomorphic.

  • 1.4 Problem.   We prove that two subsets are homeomorphic by writing down a homeomorphism. How can we prove that two subsets are not homeomorphic?

  • 1.5 Definition.   A property \(P\) of subsets of Euclidean spaces is a topological property when, if \(X\) and \(Y\) are homeomorphic subsets, then \(X\) has property \(P\) if and only if \(Y\) has property \(P\).

Thus, if \(X\) has property \(P\) and \(Y\) does not have property \(P\) then \(X\) and \(Y\) are not homeomorphic.

Exercises
  • 1.  Prove that the punctured plane, \(X=\R ^2\setminus \{\0\}\), is homeomorphic to the complement of the unit disc, \(Z=\{\,\x \in \R ^2\mid |\x |>1\,\} = \R ^2\setminus \D ^2\) where \(\D ^2 = \{\,\x \in \R ^2\mid |\x |\leq 1\,\}\).

  • 2.  Prove that \(S^1=\{\,\x \in \R ^2\mid |\x |=1\,\}\), the unit circle, and \(T=\{\,\x =(x_1,x_2)\in \R ^2\mid |x_1|+|x_2|=1\,\}\), the diagonal square, are homeomorphic.

  • 3

    • (a) Prove that any two closed intervals \([a_1,b_1]=\{\,x\in \R \mid a_1\leq x \leq b_1\,\}\) and \([a_2,b_2]\) (where \(a_i<b_i\)) are homeomorphic by writing down the formula for a homeomorphism.

    • (b) Prove that the open intervals

      \[(0,1)=\{\,x\in \R \mid 0<x<1\,\},\;(0,\infty )=\{\,x\in \R \mid x>0\,\},\;(-\infty ,\infty )=\R \]

      are all homeomorphic.

  • 4.  The (closed) ball in \(\R ^n\) of radius \(r\geq 0\) with centre \(\a \in \R ^n\) is defined to be the subset

    \[\D _r^n(\a ) = \{\,\x \in \R ^n\mid |\x -\a |\leq r\,\}.\]

    Prove that any two balls of positive radius in \(\R ^n\) are homeomorphic.

  • 5.  The cylinder of unit length may be defined as the following subset of \(\R ^3\)

    \[\{\,(x_1,x_2,x_3)\in \R ^3\mid x_1^2+x_2^2=1\mbox { and } 0\leq x_3\leq 1\,\}.\]

    Prove that this cylinder is homeomorphic to the annulus \(\{\,\x \in \R ^2\mid 1\leq |\x |\leq 2\,\}\).

  • 6.  Let \(S^n\) denote the unit sphere in \(\R ^{n+1}\) with centre the origin

    \[S^n = \{\,\x \in \R ^{n+1}\mid |\x |=1\,\}\]

    and \(D^n\) denote the closed unit ball in \(\R ^n\) centre the origin

    \[\D ^n = \{\,\x \in \R ^n\mid |\x |\leq 1\,\}.\]

    Prove that the closed semicircle \(\{\,\x \in S^1\mid x_2\geq 0\,\}\) is homeomorphic to the closed interval \([-1,1]\). More generally, prove that the closed hemisphere \(\{\,\x \in S^n\mid x_{n+1}\geq 0\,\}\) is homeomorphic to the closed ball \(\D ^n\).

Path-connected subsets of Euclidean space
  • 1.6 Definition.  

    • (a) Let \(X\) be a subset of some Euclidean space. A path in \(X\) is a continuous function \(\sigma \colon [0,1]\rightarrow X\) where \([0,1]=\{\,t\in \R \mid 0\leq t\leq 1\,\}\). The point \(\sigma (0)\) is the beginning point of the path and the point \(\sigma (1)\) is the terminal point of the path. We say that \(\sigma \) is a path in \(X\) from \(\sigma (0)\) to \(\sigma (1)\).

    • (b) The subset \(X\) is said to be path-connected if, for each pair of points \(\x \), \(\x '\in X\), there is a path in \(X\) from \(\x \) to \(\x '\).

  • 1.7 Proposition.   The closed unit ball (or disc) \(D^n=\{\,\x \in \R ^n\mid |\x |\leq 1\,\}\) in \(\R ^n\) is path-connected.

  • Proof. Given \(\x \),\(\x ' \in D^n\) define \(\sigma \colon [0,1]\to \R ^n\) by

    \[\sigma (s) = \x + s(\x '-\x ) = (1-s)\x + s\x '\]

    for \(s\in [0,1]\). Then \(\sigma \) is continuous, \(\sigma (0)=\x \) and \(\sigma (1)=\x '\) so \(\sigma \) is a path in \(\R ^n\) from \(\x \) to \(\x '\).

    However, for \(0\leq s\leq 1\), \(|\sigma (s)| = |(1-s)\x +sx'| \leq |(1-s)\x | + |s\x '|\) (by the triangle inequality) \(= (1-s)|\x | + s|\x '|\) (since \(s\geq 0\) and \(1-s\geq 0\)) \(\leq (1-s) + s\) (since \(x\), \(\x '\in D^n\)) \(= 1\), i.e. \(|\sigma (s)|\leq 1\). Hence \(\sigma (s)\in D^n\) and so \(\sigma \colon [0,1]\to D^n\) is a path in \(D^n\) from \(\x \) to \(\x '\).

    Hence \(D^n\) is path-connected.  □

  • 1.8 Exercise.   The unit circle \(S^1\) in \(\R ^2\) is path-connected.

  • 1.9 Theorem.   Let \(f\colon X\to Y\) be a continuous surjection where \(X\) and \(Y\) are subsets of Euclidean spaces. Then, if \(X\) is path-connected, so is \(Y\).

  • Proof. Exercise.  □

  • 1.10 Corollary.   Path-connectedness is a topological property.

  • Proof. Suppose that \(X\) and \(Y\) are homeomorphic subsets of Euclidean spaces. Then there is a homeomorphism \(f\colon X\to Y\). Then if \(X\) is path-connected so is \(Y\) by the Theorem since \(f\) is a continuous surjection. Conversely, if \(Y\) is path-connected then so is \(X\) since \(f^{-1}\colon Y\to X\) is a continuous surjection. Thus, \(X\) is path-connected if and only if \(Y\) is path-connected as required.  □

  • 1.11 Proposition.   The subset \(\R \setminus \{0\}\) is not path-connected and so \(\R \setminus \{0\}\not \cong S^1\).

  • Proof. This is true because there is no path in \(\R \setminus \{0\}\) from \(-1\) to 1. This may be proved by contradiction. Suppose. for contradiction, that \(\sigma \colon [0,1]\to \R \setminus \{0\}\) is a path from \(-1\) to 1 so that \(\sigma (0)=-1\) and \(\sigma (1)=1\). Then \(i\circ \sigma \colon [0,1] \to \R \setminus \{0\} \to \R \) is a continuous function with values \(-1\) and \(1\) for which \(0\) is not a value. This contradicts the intermediate value property of the function \(\sigma \) (Theorem 0.23(b) in the Background Material) since \(-1<0<1\) and so gives the necessary contradiction. Hence \(\sigma \) cannot exist, as required and so \(\R \setminus \{0\}\) is not path-connected. It follows that \(\R \setminus \{0\}\not \cong S^1\) since \(S^1\) is path-connected and path-connectedness is a topological property.  □

  • 1.12 Problem.   Are \(S^1\) and \([0,1)\) homeomorphic? There is a continuous bijection \(f\colon [0,1)\to S^1\) defined by \(f(x) = (\cos 2\pi x,\sin 2\pi x)\).

    More generally, is \(S^1\) homeomorphic to any subset of \(\R \)?

Path-components
  • 1.13 Definition.   Suppose that \(X\) is a subset of a Euclidean space.

    • (a) Given \(\x \in X\), we may define a path \(\e _{\x }\colon [0,1]\to X\) by

      \[\e _{\x }(s) = \x \quad \mbox {for $0\leq s\leq 1$}.\]

      This is called the constant path at \(\x \).

    • (b) Given a path \(\sigma \colon [0,1]\to X\) in \(X\) we may define a path

      \[\overline {\sigma }(s) = \sigma (1-s)\quad \mbox {for $0\leq s\leq 1$}.\]

      This is called the reverse path of \(\sigma \) and is a path from \(\sigma (1)\) to \(\sigma (0)\).

    • (c) Given paths \(\sigma _1\colon [0,1]\to X\) and \(\sigma _2\colon [0,1]\to X\) in \(X\) such that \(\sigma _1(1) = \sigma _2(0)\) we may define a path \(\sigma _1\ast \sigma _2\colon [0,1]\to X\) by

      \[\sigma _1\ast \sigma _2(s) = \left \{\begin {array}{l}\sigma _1(2s)\quad \mbox {for $0\leq s\leq 1/2$,}\\\sigma _2(2s-1)\quad \mbox {for $1/2\leq s\leq 1$.}\end {array}\right .\]

      This is called the product of the paths \(\sigma _1\) and \(\sigma _2\) and is a path from \(\sigma _1(0)\) to \(\sigma _2(1)\).

    [Note that \(\sigma _1\ast \sigma _2\) is well-defined and continuous at \(t=1/2\) by the conditions on \(\sigma _1\) and \(\sigma _2\).]

  • 1.14 Proposition.   Given \(X\), a subset of a Euclidean space, we may define an equivalence relation on \(X\) as follows: for \(\x \), \(\x '\in X\), \(\x \sim \x '\) if and only if there is a path in \(X\) from \(\x \) to \(\x '\).

  • Proof. We check the conditions for an equivalence relation (Definition 0.15).

    The reflexive property. For each point \(\x \in X\), \(\x \sim \x \) using the constant path \(\e _{\x }\).

    The symmetric property. Suppose that \(\x \) and \(x'\in X\) such that \(\x \sim \x '\). Then there is a path \(\sigma \) in \(X\) from \(\x \) to \(\x '\). The reverse path \(\overline {\sigma }\) is then a path in \(X\) from \(\x '\) to \(\x \) and so \(\x '\sim \x \) as requirec.

    The transitive property. Suppose that \(\x \), \(\x '\) and \(\x \in X\) such that \(\x \sim \x '\) and \(\x '\sim \x  \). This means that there is a paths \(\sigma _1\) in \(X\) from \(\x \) to \(\x '\) and a path \(\sigma _2\) in \(X\) from \(\x '\) to \(\x  \). Then the product path \(\sigma _1\ast \sigma _2\) is a path in \(X\) from \(\x \) to \(\x  \) and so \(\x \sim \x  \) as required.  □

  • 1.15 Definition.   Given \(X\), a subset of a Euclidean space, the equivalence classes of the equivalence relation in Proposition 1.14 are called the path-components of \(X\). We write \(\pi _0(X)\) for the set of path-components of \(X\) and \([\x ]\) for the path-component of a point \(\x \in X\).

  • 1.16 Example.   \(\pi _0(\R \setminus \{0\})=\{(-\infty ,0),(0,\infty )\}\).

  • 1.17 Proposition.   Homeomorphic sets have the same number of path-components.

  • Proof. Suppose that \(X\) and \(Y\) are homeomorphic subsets of Euclidean spaces. Then there is a homeomorphism \(f\colon X\to Y\). It it will be shown that this continuous function induces a bijection \(f_*\colon \pi _0(X) \to \pi _0(Y)\) by \(f_*([\x ]) = [f(\x )]\). This implies that \(\pi _0(X)\) and \(\pi _0(Y)\) have the same cardinality which is what we have prove.

    The function \(f_*\) is well-defined because, if \([\x ] = [\x ']\) then \(\x \sim \x '\) and so there is a path \(\sigma \colon [0,1]\to X\) in \(X\) from \(\x \) to \(\x '\). It follows that \(f\circ \sigma \colon [0,1]\to X\to Y\) is a path in \(Y\) from \(f(\x )\) to \(f(\x ')\) and so \(f(\x )\sim f(\x ')\), i.e. \([f(\x )]=[f(\x ')]\).

    The function \(f_*\) is a bijection since it is easily checked that \((f^{-1})_*\colon \pi _0(Y)\to \pi _0(X)\), the function induced by the inverse \(f^{-1}\colon Y\to X\), is an inverse for \(f_*\) (Exercise). \(\Box \)  □

Cut-points in subsets of Euclidean space
  • 1.18 Definition.   Suppose that \(X\) is a subset of some Euclidean space. Then a point \(p\in X\) is called a cut-point of type \(n\) of \(X\) or an \(n\)-point of \(X\) if its complement \(X \setminus \{p\}\) has \(n\) path-components.

  • 1.19 Example.  

    • (a) In \([0,1)\) each \(x\in (0,1)\) is a 2-point and 0 is a 1-point.

    • (b) In the subset of \(\R ^2\) given by the coordinate axes, \(\{\,(x_1,x_2)\in \R ^2\mid x_1 = 0 \mbox { or }x_2=0\,\}\), \((0,0)\) is a 4-point whereas all other points are 2-points.

    • (c) In \(S^1\) every point is a 1-point.

  • 1.20 Theorem.   Homeomorphic sets have the same number of cut-points of each type.

  • Proof. Let \(X\) and \(Y\) be homeomorphic subsets of Euclidean spaces. Then there is a homeomophism \(f\colon X\to Y\). Suppose that \(\p \in X\) is an \(n\)-point of \(X\). Then \(f\) induces a homeomorphism \(X\setminus \{\p \}\to Y\setminus \{f(\p )\}\) and so these subsets have the same number of path-components by Proposition 1.17. Hence \(f(\p )\) is an \(n\)-point of \(Y\).

    This shows that \(f\) induces a bijection between the \(n\)-points of \(X\) and the \(n\)-points of \(Y\) and so they must have the same number of \(n\)-points.  □

  • 1.21 Example.   \([0,1)\) and \(S^1\) are not homeomorphic since \([0,1)\) has some 2-points (all of its points apart from \(0\)) whereas \(S^1\) has none.

Exercises
  • 7.  A subset \(A \subset \R ^n\) of Euclidean space is said to be convex if for all pairs of points \(\x \), \(\x '\in A\), the straight line segment \(\{ (1-t)\x + t\x '\mid 0\leq t \leq 1\}\) between \(\x \) and \(\x '\) is a subset of \(A\). Prove that a convex subset is path-connected. Is the converse true?

  • 8.  Prove that if \(f\colon X\to Y\) is a continuous surjection between subsets of Euclidean spaces and \(X\) is path-connected then \(Y\) is path-connected. [Theorem 1.9]

  • 9

    • (a) Use Theorem 1.9 to prove that \(S^1\) is connected.

    • (b) Prove that \(S^2\) is path-connected.

      [Hint: Use spherical polar coordinates to define a continuous surjection \(\R ^2 \rightarrow S^2\).]

  • 10

    • (a) Suppose that \(f\colon X\to Y\) and \(g\colon Y\to Z\) are continuous maps between subsets of Euclidean spaces. Prove that \((g\circ f)_* = g_*\circ f_*\colon \pi _0(X) \to \pi _0(Z)\) (using the notation in the proof of Proposition 1.17).

    • (b) Prove that the identity map on \(X\) induces the identity map on \(\pi _0(X)\), i.e. \((I_X)_* = I_{\pi _0(X)}\colon \pi _0(X)\to \pi _0(X)\).

    • (c) Hence complete the proof of Proposition 1.17 by proving that if \(g\colon Y\to X\) is the inverse of a homeomorphism \(f\colon X\to Y\) then \(g_*\colon \pi _0(Y)\to \pi _0(X)\) is the inverse of \(f_*\colon \pi _0(X) \to \pi _0(Y)\).

  • 11.  Use a cut-point argument to show that the three intervals \([0,1]\), \((0,1)\) and \([0,1)\) are topologically distinct (i.e. no two are homeomorphic).

  • 12.  Suppose that \(p\) and \(q\) are distinct points in \(X\), a path-connected subset of a Euclidean space. We call \(\{p,q\}\) an \(n\)-pair or a cut-pair of type \(n\) if the set complement \(X\setminus \{p,q\}\) has \(n\) path-components. Prove that homeomorphic sets have the same number of cut-pairs of each type.

Other applications of path-connectness
  • 1.22 Theorem (The Brouwer Fixed Point Theorem in dimension 1).   Suppose that \(f\colon [-1,1]\to [-1,1]\) is a continuous map. Then \(f\) has a fixed point, i.e. there exists a point \(t\in [-1,1]\) such that \(f(t) = t\).

  • Proof. Suppose for contradiction that \(f\) does not have a fixed point. Then \(f(t)\not = t\) for all \(t\in [-1,1]\). Thus we may define a function \(g\colon [-1,1] \to \{-1,1\}\) by \(g(t) = (f(t)-t)/|f(t)-t|\). This is a continuous function from basic real analysis. However, since \(f(-1)>-1\) and \(f(1)<1\) it follows that \(g(-1)=1\) and \(g(1)=-1\). Hence \(g\) is a surjection. Hence, by Proposition 1.9, \(\{-1,1\}\) path-connected which contradicts the Intermediate Value Theorem (as in the proof of Proposition 1.11). Hence \(f\) has a fixed point.  □

  • 1.23 Theorem (The Borsuk-Ulam Theorem in dimension 1).   Suppose that \(f\colon S^1 \to \R \) is a continuous function. Then there is a point \(\x \in S^1\) such that \(f(\x ) = f(-\x )\).

  • Proof. Exercise. Try a similar proof to that of Theorem 1.22.  □

1.24 Definition. A subset \(A\subset \R ^n\) is bounded if there is a real number \(R\) such that \(\x \in A\implies |\x |\leq R\).

  • 1.24 Theorem (The Pancake Theorem).   Let \(A\) and \(B\) be bounded subsets of \(\R ^2\). Then there is a (straight) line in \(\R ^2\) which divides each of \(A\) and \(B\) in half by area.

Remark. The statement of this result assumes that \(A\) and \(B\) each have a well-defined area. In this course we ignore the technical difficulties associated with defining the area of a subset of \(\R ^2\) (the subject of integration and measure theory).

  • Outline Proof. Since \(A\) and \(B\) are bounded there is a real number \(R\) such that \(\a \in A\Rightarrow |\a |\leq R\) and \(\x \in B\Rightarrow |\x |\leq R\).

    Suppose that \(\x \in S^1\). For \(t\in [-R,R]\) let \(L_{\x ,t}\) denote the straight line through \(t\x \) perpendicular to \(\x \). Let \(v(t)\in [0,1]\) be the proportion of the area of \(A\) on the same side of \(L_{\x ,t}\) as \(R\x \). Then \(v\colon [-R,R]\to [0,1]\) is a continuous decreasing function with \(v(-R) = 1\) and \(v(R)=0\). By the Intermediate Value Theorem there exists \(t\in [-R,R]\) such that \(v(t) = 1/2\). This \(t\) may not be unique but it is not difficult to show that \(v^{-1}(1/2) = \{\,t\mid v(t)=1/2\,\} = [\alpha , \beta ]\), a closed interval. Let \(f_A(\x ) = (\alpha +\beta )/2\). Then the line \(L_{\x ,f_A(\x )}\) bisects \(A\).

    The function \(f_A\colon S^1\to \R \) can be shown to be continuous. Furthermore \(f_A(-\x ) = -f_A(\x )\) (since \(L_{\x ,f_A(\x )}\) and \(L_{\x ,f_A(-\x )}\) are the same line so that \(f_A(\x )\x = f_A(-\x )(-\x )\)).

    Similarly, using the region \(B\), we may define a continuous function \(f_B\colon S^1 \to \R \) such that \(f_B(-\x ). = -f_B(\x )\) and \(L_{\x ,f_B(\x )}\) bisects \(B\).

    Let the continuous function \(f\colon S^1\to \R \) be given by \(f(\x ) = f_A(\x ) - f_B(\x )\).

    By the Borsuk-Ulam Theorem, there exists \(\x _0\in S^1\) such that \(f(\x _0)=f(-\x _0)\). But \(f(-\x _0) = f_A(-\x _0) - f_B(-\x _0) = -f_A(\x _0) + f_B(\x _0) = -f(\x _0)\). Hence \(f(\x _0) = -f(\x _0)\) so that \(f(\x _0)=0\). This means that \(f_A(\x _0) - f_B(\x _0) = 0\) so that \(f_A(\x _0)=f_B(\x _0)\).

    From the definition of \(f_A\) and \(f_B\) it follows that the line \(L_{\x _0,f_A(\x _0)} = L_{\x _0,f_B(\x _0)}\) bisects both of \(A\) and \(B\) and so is the line whose existence is the claim of the theorem.  □