If $p:X\to Y$ is quotient map and $Z$ is a locally compact Hausdorff space, then $p\times I_z$ is a quotient map

Question

The problem from my topology book is as follows.

If $p:X\to Y$ is quotient map and $Z$ is a locally compact Hausdorff space, then $p\times I_z$ : $X\times Z \to Y\times Z$ is a quotient map where $I_z$ is an identity map on $Z$.

I'm stuck in proving that if $ (p\times I_z)^{-1}[U]$ is open in $X\times Z$, then $U$ is open in $Y\times Z$. How are locally compactness and Hausdorffness applied to this problem? Somebody help me please...

Thanks.

Answer 1

Assuming you know the following fact (also known as Kuratowski's theorem):

1. Suppose $X$ is compact and $Y$ is any space, then the projection map $p: X \times Y \rightarrow Y$ defined by $p(x,y) = y$ is a closed map.

Then we can prove: (I've changed notation because I adapted my write-up on another forum)

2. Let $X$ be a locally compact space (where locally compact means that every $x$ has a base of compact neighbourhoods), and let $Y, Z$ be topological spaces, and $g: Y \rightarrow Z$ is a quotient map. Define $f: X \times Y \rightarrow X \times Z$ to be $\operatorname{id}_X \times g$, i.e. $f(x,y) = (x, g(y))$. Then $f$ is a quotient map as well. (All quotient maps are assumed to be onto).

First a simple fact we will use a few times:

3. If $A \subset X, y \in Y, B \subseteq X \times Z$, then $$ A \times \{y\} \subseteq f^{-1}[B] \text{ iff } A \times g^{-1}[\{ g(y) \}] \subseteq f^{-1}[B]$$

Right to left is clear, as $\{y\} \subseteq g^{-1}[\{ g(y) \}]$, and left to right: let $y'$ be in $g^{-1}[\{ g(y) \}]$, and $x \in A$, then $f(x,y') = (x, g(y')) = (x, g(y)) = f(x,y)$, and as $(x,y)$ is in $A \times \{y\}$, we know that $f(x,y)$ is in $B$, and so $f(x,y')$ is in $B$ as well, and so $(x, y') \in f^{-1}[B]$ for all $x \in A$, $y' \in g^{-1}[\{ g(y)\} ]$.

Now for the proof of 2: Now, let $W \subseteq X \times Z$ be such that $f^{-1}[W]$ is open. We want to show that $W$ is open, and then $f$ is a quotient map.

So let $(x_0, z_0)$ be in $W$. Let $y_0$ be such that $g(y_0) = z_0$ (as $g$ is onto). Let $C$ be a compact neighbourhood of $x_0$, such that $C \times \{y_0\} \subseteq f^{-1}[W]$ which is possible, as the latter set is open, and $X$ is locally compact.

By 3. we see that in fact $C \times g^{-1}[\{ z_0 \}] \subseteq f^{-1}[W]$.

Define $$V = \{z \in Z : C \times g^{-1}[\{ y \}] \subseteq f^{-1}[W] \}$$

By the previous, $z_0$ is in $Z$. Clearly, $C \times V \subseteq W$ (for, if $x \in C, z \in W$, then let $y$ be such that $g(y) = z$. Then by 3. again, $A \times \{y\} \subseteq f^{-1}[W]$ and so $f(x,y) = (x,z)$ is in $W$ ).

So if $V$ would be open, $C \times V$ is the required neighbourhood of $(x_0 , z_0)$ that sits inside $W$, showing that $(x_0, z_0)$ is an interior point of $W$, and $W$ is open. To check that $V$ is indeed open, it suffices, as $g$ is a quotient map, to check that $g^{-1}[V]$ is open in $Y$.

Now

Now:

$$g^{-1}[V] = \{y \in Y: g(y) \in V \} = \left\{y \in Y: C \times g^{-1}[\{ g(y) \}] \subseteq f^{-1}[W] \right\} = \text{ (by 3.) } \left\{y \in Y : C \times \{y\} \subseteq f^{-1}[W] \right\} = Y \setminus p[ (C \times Y) \setminus f^{-1}[W] ]$$

where $p: C \times Y \rightarrow Y$ is a closed map by 1. and $(C \times Y) \setminus f^{-1}[W]$ is closed in $C \times Y$, so $g^{-1}[V]$ is indeed open, and so $V$ is open, and we are done.

Based on my write-up on ask a topologist which was in turn based on Engelking's proof in General Topology.

Answer 2

We have this map is continuous, and we only need prove that if $U$ a saturated open in $X \times Z$, then $ p\times i_z(U)$ is open in $Y\times Z$. For this we want to find a basis open in $Y \times Z$ which contains $y \times z \in p\times i_z(U)$.

Assume $x \times z \in (p \times i_z)^{-1}({y\times z})= p^{-1}(\{y\})\times z$ and we know that $x \times z\in U$. As a result, there are $U_x$ and $U_z$ opens, respectively, in $X$ and $Z$ that $U_x \times U_z \subseteq U$. As $Z$ is a hausdorff and loca1ly compact for each open $z \in U_z$, there is open $V$ contains $z$ and $z \in V\subseteq \bar V \subseteq U_z$, and this $\bar V$ is a compact set. As a result of above we have: $$ \{x\} \times \bar V \subseteq U, \; z \in V $$ We have $U$ is saturated so: $p^{-1}(\{y\}) \times \bar V \subseteq U$.

Now we define the $W$ as the following, and we will prove that this $W$ is open and saturated in $X$. $$ W=\{x | \{x\} \times \bar V \subseteq U \}$$ Lemma 1: $W$ is open in $X$.
Proof: Assume $x\in W$, so ${x}\times \bar V \subseteq U$. We know $\bar V$ is compact and with the use of Tube Lemma: $$\exists U_x, U_{\bar V} \;\text{opens respectively in}\; X, Z \;\text{which}\; x\in U_x, \bar V \subseteq U_{\bar V}$$ As a result, we can infer that $U_x \subseteq W$.

Lemma 2: $W$ is saturated in the map $p$.
Proof: It is obvous from the fact that $U$ is a saturated open set according to the map $p\times i_z.$ If $\{x\} \times \bar V \subseteq U$ then as $U$ is saturated: $$p^{-1}(\{p(x)\})\times \bar V \subseteq U \Rightarrow p^{-1}(\{p(x)\}) \subseteq W $$

We now have $W \times \bar V \subseteq U$, and $W$ is open saturated in $X$ that contains $p^{-1}(\{y\})$. Thus, $P(w) \times V $ is a open in $p\times i_z(U)$, which contains $y \times z$