Exercise 22.2

Question

\begin{enumerate}[label=(\alph*)]
\item Let $p\colon X 	o Y$ be a continuous map. If there is a continuous map
  $f\colon Y 	o X$ such that $p \circ f$ equals the identity map of $Y$, then
  $p$ is a quotient map.
\item If $A \subset X$, a \emph{	extbf{retraction}} of $X$ onto $A$ is a
  continuous map $r \colon X 	o A$ such that $r(a) = a$ for each $a \in A$.
  Show that a retraction is a quotient map.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
  If $V \subseteq Y$ with $U = p^{-1}(V) \subseteq X$ open, $f^{-1}(U) = f^{-1}(p^{-1}(V)) = (p \circ f)^{-1}(V) = V$ is open. Thus, $p$ is a quotient map.
\end{proof}
\begin{proof}[Proof of $(b)$]
  Let $\iota \colon A 	o X$ be the inclusion map; then, $r \circ \iota$ is the
  identity on $A$, hence $r$ is a quotient map by $(a)$.
\end{proof}

Dan Whitman · Answer

(a)
  \begin{proof}
    First, $f$ is a right inverse for $p$ by definition so that $p$ is surjective by Exercise~2.5 part (a).
    Suppose that $U$ is a subset of $Y$.
    If $U$ is open in $Y$ then ${p}^{-1}(U)$ is open in $X$ since $p$ is continuous.
    So suppose that $V = {p}^{-1}(U)$ is open in $X$.
    Since we have $p \circ f = i_Y$ is bijective and $i_Y = {i_Y}^{-1}$, it follows from Exercise~2.4 part~(a) that
    \begin{gather*}
      U = i_Y(U) = {i_Y}^{-1}(U) = {(p \circ f)}^{-1}(U) = {f}^{-1}({p}^{-1}(U)) = {f}^{-1}(V) \,.
    \end{gather*}
    Then, since $V$ is open in $X$, we have that ${f}^{-1}(V) = U$ is open in $Y$ since $f$ is continuous.
    This shows that $p$ is a quotient map by definition.
  \end{proof}

(b)
  \begin{proof}
    Suppose $X$ is a topological space, $A \subset X$, and $r: X 	o A$ is a retraction.
    Let $f: A 	o X$ be defined by $f(a) = a$ for all $a \in A$, i.e. $f$ is the identity function on $A$ with the range expanded to $X$.
    Now, $i_A$ is continuous (in fact it is a homeomorphism) by Exercise~18.3 so that $f$ is also continuous by Theorem~18.2 part~(e) since it is just $i_A$ with an expanded range.
    Then for any $a \in A$ we have that
    \begin{gather*}
      (p \circ f)(a) = p(f(a)) = p(a) = a
    \end{gather*}
    since $p$ is a retraction.
    Thus $p \circ f = i_A$, which shows that $p$ is a quotient map by what was shown in part~(a).
  \end{proof}

Exercise 22.2

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Exercise 22.2

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