Exercise 22.1

Question

Check the details of Example 3.

Dan Whitman · Accepted Answer

\def\a{\alpha}
\def\b{\beta}
\def\d{\delta}
\def\D{\Delta}
\def\e{\epsilon}
\def{ho}

Recall that Example~22.3 includes a function $p: {\mathbb{R}} 	o A$ where $A = \left\{a,b,cight\}$ is a three-point set.
  The function is defined by
  \begin{gather*}
    p(x) = \begin{cases}
      a & x > 0 \
      b & x < 0 \
      c & x = 0 \,.
    \end{cases}
  \end{gather*}
  We are then asked to verify that the quotient topology on $A$ induced by $p$ is that indicated by the following diagram:
  \begin{center}
    \begin{tikzpicture}
      % Points
      \fill (-2,0) circle (0.06);
      \fill (0,0) circle (0.06);
      \fill (2,0) circle (0.06);

% Point labels
      
ode [above] at (-2,0) {$a$};
      
ode [above] at (0,0) {$b$};
      
ode [above] at (2,0) {$c$};

% Single point subsets
      \draw (-2,0) circle (0.5);
      \draw (0,0) circle (0.5);

% Two-point subsets and full set
      \draw (-1, 0) circle (2 and 1);
      \draw (0, 0) circle (3.3 and 1.5);
    \end{tikzpicture}
  \end{center}
  \begin{proof}
    Clearly the diagram illustrates the topology $\mathcal{T} = \left\{\varnothing, \left\{aight\}, \left\{bight\}, \left\{a,bight\}, Aight\}$ on $A$.
    We also note that $p$ is surjective and that $\mathcal{T}$ is the unique topology such that $p$ is a quotient map.
    First, obviously, $\varnothing$ and $A$ must be open in the quotient topology $\mathcal{T}$ since it is a topology.
    Then, clearly the following sets
    \begin{align*}
      {p}^{-1}\left(\left\{aight\}ight) &= (0, \infty) \
      {p}^{-1}\left(\left\{bight\}ight) &= (-\infty, 0) \
      {p}^{-1}\left(\left\{a,bight\}ight) &= (0, \infty) \cup (-\infty, 0)
    \end{align*}
    are all open in ${\mathbb{R}}$ so that $\left\{aight\}$, $\left\{bight\}$, and $\left\{a,bight\}$ should be open in the quotient topology since $p$ is a quotient map.
    On the contrary, the sets
    \begin{align*}
      {p}^{-1}\left(\left\{cight\}ight) &= \left\{0ight\} \
      {p}^{-1}\left(\left\{a,cight\}ight) &= [0, \infty) \
      {p}^{-1}\left(\left\{b,cight\}ight) &= (-\infty, 0]
    \end{align*}
    are all clearly not open in ${\mathbb{R}}$ (in fact they are all closed) so that $\left\{cight\}$, $\left\{a,cight\}$ and $\left\{b,cight\}$ should not be open in the quotient topology.
    As we have considered all eight of the possible subsets of $A$, this shows the desired result.
  \end{proof}

Exercise 22.1

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Exercise 22.1

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