Week 7

Section 2.7 Week 7

Subsection Monday

I held a review session on Monday.

Subsection Wednesday

The take-home midterm was distributed on Wednesday. I did not hold class in order to give folks time to work on the exam.

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Subsection Friday

Subsubsection Connected components

The notion of connectedness breaks any topological space into its connected components, which we now define.

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Definition 2.7.1. Connected components.

Let \(X\) be a topological space. Declare \(x\sim_{\mathrm{conn}} y\) to mean there is a connected subspace \(A\subseteq X\) for which \(x,y\in A\text{.}\) The set of connected components is the set of equivalence classes:

\begin{equation*} [X] := X/{\sim_{\mathrm{conn}}}. \end{equation*}

This set is regarded as a topological space via the quotient topology from the surjection \(X\xrightarrow{x\mapsto [x]_{\mathrm{conn}}} [X]\text{.}\)

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Remark 2.7.2.

A connected component of a topological space \(X\) is an equivalence class \(C\subseteq X\) with respect to \(\sim_{\mathrm{conn}}\text{.}\) Unpacking the definition, two points lie in the same connected component if and only if they are both contained in some connected subspace of \(X\text{.}\)

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Definition 2.7.3. Totally disconnected spaces.

A topological space \(X\) is totally disconnected if the quotient map \(X\to [X],~x\mapsto [x]_{\mathrm{conn}}\) is a bijection ‚Äî equivalently, if the only connected subspaces of \(X\) are singletons.

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Example 2.7.4.

We have \([S^0] = \bigl\{\{-1\},\{+1\}\bigr\}\) and \([\mathbb{R}] = \{\mathbb{R}\}\text{.}\) The rationals \(\mathbb{Q}\) are totally disconnected.

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Lemma 2.7.5.

Let \(D\) be a discrete topological space. Every connected component of \(D\) is a singleton.

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Proof.

This is an exercise.

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Observation 2.7.6.

A topological space \(X\) is connected if and only if its set \([X]\) of connected components consists of a single element.

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Theorem 2.7.7. Universal property of connected components.

Let \(X\) be a topological space. For each continuous map \(f\colon X\to D\) to a discrete topological space, there is a unique continuous map \(\bar{f}\colon [X]\to D\) for which \(f(x) = \bar{f}([x]_{\mathrm{conn}})\text{:}\)

\begin{equation*} X \xrightarrow{f} D, \quad X \xrightarrow{x\mapsto [x]} [X] \xrightarrow{\exists!\,\bar{f}} D. \end{equation*}

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Proof.

By the universal property of the quotient topology, it suffices to verify that \(x\sim_{\mathrm{conn}} y\) implies \(f(x)=f(y)\text{.}\) Suppose \(x\sim_{\mathrm{conn}} y\text{.}\) Then there is a connected subspace \(A\subseteq X\) with \(x,y\in A\text{.}\) Since the continuous image of a connected space is connected, \(f(A)\subseteq D\) is connected. By Lemma 2.7.5, \(f(A)\) is a singleton, so \(f(x)=f(y)\text{.}\)

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Subsubsection Path-components and \(\pi_0\)

Definition 2.7.8. Path-components and \(\pi_0\).

Let \(X\) be a topological space.

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For \(x_0,x_1\in X\text{,}\) declare \(x_0\sim_{\mathrm{path}} x_1\) to mean there exists a continuous map \(\gamma\colon [0,1]\to X\) with \(\gamma(0)=x_0\) and \(\gamma(1)=x_1\text{.}\) (That \(\sim_{\mathrm{path}}\) is an equivalence relation is the content of Lemma 2.7.9.)
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A path-component of \(X\) is an equivalence class with respect to \(\sim_{\mathrm{path}}\text{.}\)
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The set of path-components of \(X\) is the quotient set

\begin{equation*} \pi_0(X) := X/{\sim_{\mathrm{path}}}. \end{equation*}

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We say \(X\) is path-connected if \(\pi_0(X)\) is a singleton.
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Lemma 2.7.9.

Let \(X\) be a topological space. The relation \(\sim_{\mathrm{path}}\) is an equivalence relation on \(X\text{.}\)

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Proof.

Reflexivity. For each \(x\in X\text{,}\) the constant map \(\mathrm{const}_x\colon [0,1]\to X\) satisfies \(\mathrm{const}_x(0)=x=\mathrm{const}_x(1)\text{,}\) so \(x\sim_{\mathrm{path}} x\text{.}\)

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Symmetry. Suppose \(x\sim_{\mathrm{path}} y\text{,}\) witnessed by \(\gamma\colon [0,1]\to X\text{.}\) The reverse path \(\bar\gamma\colon [0,1]\to X,~t\mapsto \gamma(1-t)\) is continuous and satisfies \(\bar\gamma(0)=\gamma(1)=y\) and \(\bar\gamma(1)=\gamma(0)=x\text{,}\) so \(y\sim_{\mathrm{path}} x\text{.}\)

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Transitivity. Suppose \(x\sim_{\mathrm{path}} y\) and \(y\sim_{\mathrm{path}} z\text{,}\) witnessed by \(\alpha,\beta\colon [0,1]\to X\) with \(\alpha(0)=x\text{,}\) \(\alpha(1)=y=\beta(0)\text{,}\) and \(\beta(1)=z\text{.}\) The concatenation

\begin{equation*} \alpha\ast\beta\colon [0,1]\to X,\quad t\mapsto \begin{cases} \alpha(2t) \amp \text{if } t\in [0,\tfrac{1}{2}], \\ \beta(2t-1) \amp \text{if } t\in [\tfrac{1}{2},1], \end{cases} \end{equation*}

is continuous by the gluing lemma, with \((\alpha\ast\beta)(0)=x\) and \((\alpha\ast\beta)(1)=z\text{,}\) so \(x\sim_{\mathrm{path}} z\text{.}\)

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Proposition 2.7.10. Functoriality of \(\pi_0\).

Let \(X\text{,}\) \(Y\text{,}\) \(Z\) be topological spaces.

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A continuous map \(f\colon X\to Y\) induces a map between sets

\begin{equation*} \pi_0(f)\colon \pi_0(X)\to\pi_0(Y),\quad [x]\mapsto [f(x)]. \end{equation*}

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\(\pi_0(\mathrm{id}_X) = \mathrm{id}_{\pi_0(X)}\text{.}\)
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For continuous maps \(f\colon X\to Y\) and \(g\colon Y\to Z\text{,}\)

\begin{equation*} \pi_0(g\circ f) = \pi_0(g)\circ\pi_0(f). \end{equation*}

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There is a canonical surjective map

\begin{equation*} X\to\pi_0(X),\quad x\mapsto [x], \end{equation*}

assigning to each point its path-component.

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For \(f\colon X\to Y\) continuous, the diagram of sets

\begin{equation*} \begin{array}{ccc} X \amp \xrightarrow{f} \amp Y \\ \downarrow \amp \amp \downarrow \\ \pi_0(X) \amp \xrightarrow{\pi_0(f)} \amp \pi_0(Y) \end{array} \end{equation*}

commutes.

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Remark 2.7.11.

Parts (1)‚Äì(3) of Proposition 2.7.10 together assert that \(\pi_0\colon\mathbf{Top}\to\mathbf{Set}\) is a functor.

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Proposition 2.7.12.

Let \(f\colon X\to Y\) be a continuous surjection between topological spaces. Then the induced map

\begin{equation*} \pi_0(f)\colon\pi_0(X)\to\pi_0(Y) \end{equation*}

is surjective. In particular, \(\operatorname{Card}(\pi_0(X))\geq\operatorname{Card}(\pi_0(Y))\text{,}\) and \(Y\) is path-connected whenever \(X\) is.

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Proof.

Consider the commutative diagram of Proposition 2.7.10(5). The vertical maps are surjective by part (4), and the top horizontal map is surjective by assumption. It follows that the bottom horizontal map is surjective.

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Proposition 2.7.13.

Let \(f\colon X\to Y\) be a homeomorphism between topological spaces. Then

\begin{equation*} \pi_0(f)\colon\pi_0(X)\xrightarrow{\;\cong\;}\pi_0(Y) \end{equation*}

is a bijection. In particular, if \(\operatorname{Card}(\pi_0(X))\ne\operatorname{Card}(\pi_0(Y))\text{,}\) then \(X\) and \(Y\) are not homeomorphic.

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Proof.

Since \(f\) is a homeomorphism, there is a continuous inverse \(f^{-1}\colon Y\to X\) with \(f\circ f^{-1}=\mathrm{id}_Y\) and \(f^{-1}\circ f=\mathrm{id}_X\text{.}\) Applying parts (2) and (3) of Proposition 2.7.10:

\begin{equation*} \pi_0(f)\circ\pi_0(f^{-1}) = \pi_0(f\circ f^{-1}) = \pi_0(\mathrm{id}_Y) = \mathrm{id}_{\pi_0(Y)}, \end{equation*}

and similarly \(\pi_0(f^{-1})\circ\pi_0(f)=\mathrm{id}_{\pi_0(X)}\text{.}\) Thus \(\pi_0(f^{-1})\) is an inverse to \(\pi_0(f)\text{.}\)

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Corollary 2.7.14.

Let \(X\) be a topological space and \(A\subseteq X\) a subspace. Suppose that for each \(x\in X\) there is a path \(\gamma\colon [0,1]\to X\) with \(\gamma(0)=x\) and \(\gamma(1)\in A\text{.}\) If \(A\) is path-connected, then \(X\) is path-connected.

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Proof.

This is an exercise.

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Proposition 2.7.15.

Let \(X\) and \(Y\) be topological spaces. The canonical map

\begin{equation*} \pi_0(X\times Y)\xrightarrow{\;\cong\;}\pi_0(X)\times\pi_0(Y),\quad [(x,y)]\mapsto ([x],[y]), \end{equation*}

is a bijection.

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Proof.

Surjectivity. For any \(([x],[y])\in\pi_0(X)\times\pi_0(Y)\text{,}\) the element \([(x,y)]\) maps to it, so the map is surjective.

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Injectivity. Suppose \([(x,y)]\) and \([(x',y')]\) map to the same element, so \([x]=[x']\) in \(\pi_0(X)\) and \([y]=[y']\) in \(\pi_0(Y)\text{.}\) Choose paths \(\alpha\colon [0,1]\to X\) and \(\beta\colon [0,1]\to Y\) with \(\alpha(0)=x\text{,}\) \(\alpha(1)=x'\text{,}\) \(\beta(0)=y\text{,}\) \(\beta(1)=y'\text{.}\) The path

\begin{equation*} (\alpha,\beta)\colon [0,1]\to X\times Y,\quad t\mapsto (\alpha(t),\beta(t)), \end{equation*}

is continuous (since each coordinate is) and connects \((x,y)\) to \((x',y')\text{.}\) Thus \([(x,y)]=[(x',y')]\text{.}\)

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