Let $X$ be the subspace $[0,1] \cup [2,3]$ of $\mathbb{R}$, and let $Y$ be the subspace $[0,2]$ of $\mathbb{R}$. The map $p \colon X \to Y$ defined by $$ p(x) \colon= \begin{cases} x \ &\mbox{ for } \ x \in [0,1], \\ x-1 &\mbox{ for } \ x \in [2,3] \end{cases} $$ is surjective and continuous. This much is clear to me.

But Munkres asserts that this map is also a closed (and hence a quotient) map. How to verify this?

Now if $A$ is the subspace $[0,1) \cup [2,3]$ of $X$, then the map $q \colon A \to Y$ obtained by restricting $p$ is continuous and surjective. Alright!

Now what stumps me is the following assertion by Munkres:

The set $[2,3]$ is saturated with respect to $q$. How to show this?


You can avoid compactness by taking advantage of the very simple nature of $p$. Let $p_0=p\upharpoonright[0,1]$ and $p_1=p\upharpoonright[2,3]$; then $p_0$ and $p_1$ are homeomorphisms of $[0,1]$ onto itself and of $[2,3]$ onto $[1,2]$, respectively. In particular, $p_0$ and $p_1$ are closed maps from $[0,1]$ to itself and from $[2,3]$ to $[1,2]$

Let $F$ be a closed subset of $X$. Let $F_0=F\cap[0,1]$ and $F_1=F\cap[2,3]$; then

$$p[F]=p[F_0]\cup p[F_1]=p_0[F_0]\cup p_1[F_1]\;.$$

$F_0$ is closed in $[0,1]$, so $p[F_0]$ is closed in $[0,1]$ and therefore in $\Bbb R$. Similarly, $F_1$ is closed in $[2,3]$, so $p_1[F_1]$ is closed in $[1,2]$ and hence in $\Bbb R$. Thus, $p[F]=p_0[F_0]\cup p_1[F_1]$ is closed in $\Bbb R$ and hence in $Y$.

For the other question, recall that $S\subseteq A$ is saturated with respect to $q$ iff $q^{-1}\big[q[S]\big]=S$. Here $q\big[[2,3]\big]=[1,2]$, and $q^{-1}\big[[1,2]\big]=[2,3]$, so $[2,3]$ is $q$-saturated.

  • $\begingroup$ the definition I know of a saturated subset is the following: Let $X$ and $Y$ be topological spaces, let $p \colon X \to Y$ be a surjective map, and let $C$ be a subset of $X$. Then $C$ is saturated with respect to $p$ if, for each $y \in Y$, the set $C$ contains every set $p^{-1}(\{ y \})$ that it intersects. Now how to show that $S$ is saturated if and only if $q^{-1}\left[ q[S] \right] = S$? $\endgroup$ – Saaqib Mahmood Apr 27 '15 at 4:21
  • $\begingroup$ @Saaqib: The two are equivalent. In fact, it would be a good (and quite easy) exercise to prove it, since it may give you a clearer picture of what’s going on. $\endgroup$ – Brian M. Scott Apr 27 '15 at 4:23
  • $\begingroup$ yes, @Brian M. Scott, I've managed to figure it out. $\endgroup$ – Saaqib Mahmood Apr 27 '15 at 4:43

Since $X$ is compact any continuous map from $X$ to a Hausdorff space is closed.

About the second question $q^{-1}([1,2])=[2,3]$, so $[2,3]$ is saturated with respect to $q$

  • $\begingroup$ I would like to avoid the use of compactness at this stage as it is to come later in Munkres; so please give an argument without the notion of compactness or connectedness. $\endgroup$ – Saaqib Mahmood Apr 26 '15 at 18:03

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