Show that $(X \times Y)^{\ast}$ with the resulting quotient topology is homeomorphic to $Y$

Question

Let $X \times Y$ be partitionde into subsets of the form $X\times \lbrace y \rbrace $ for all $y\in Y$.If we let $ (X \times Y)^{\ast}$ denote the collection of sets in the partition, Show that $(X \times Y)^{\ast}$ with the resulting quotient topology is homeomorphic to $Y$

My attempt

Consider the map $h: (X \times Y)^{\ast} \longrightarrow Y $ given by $h(X \times \lbrace y \rbrace)=y$, notice that $g:Y \longrightarrow (X \times Y)^{\ast}$ given by $g(y)=X \times \lbrace y \rbrace$ is the inverse of $h$ since

$(h \circ g)(y)=h(g(y))=h(X \times \lbrace y \rbrace)=y=id_Y$

$(g \circ h)(X \times \lbrace y \rbrace)=g(h(X \times \lbrace y \rbrace))=g(y)=X \times \lbrace y \rbrace=Id_{(X \times Y)^{\ast}}$ therefore the function is bijective.

$h$ is continuous function since is a projection $g$ is continuous since $C=(X \times \lbrace y \rbrace)$ closed in $(X \times Y)^{\ast}$ implies $g^{-1}(C)=y$ closed.

Answer 1

This is close, but there are a couple of problems. The first is that $h$ isn’t actually the projection map $\pi_Y:X\times Y\to Y$. The domain of $h$ is $(X\times Y)^*$, which is $\big\{X\times\{y\}:y\in Y\big\}$: the elements of this set are the sets $X\times\{y\}$ with $y\in Y$, while the elements of $X\times Y$, the domain of $\pi_Y$, are ordered pairs $\langle x,y\rangle$ with $x\in X$ and $y\in Y$. Thus, we can’t simply say that $h$ is continuous because $\pi_Y$ is continuous.

Let $U\subseteq Y$ be open;

$$h^{-1}[U]=\big\{X\times\{y\}:y\in U\big\}\,,$$

and we want to show that this is open in $(X\times Y)^*$. If

$$q:X\times Y\to(X\times Y)^*:\langle x,y\rangle\mapsto X\times\{y\}$$

is the quotient map, then $h^{-1}[U]$ is open in $(X\times Y)^*$ if and only if $q^{-1}\big[h^{-1}[U]\big]$ is open in $X\times Y$. And clearly

$$q^{-1}\big[h^{-1}[U]\big]=q^{-1}\left[\big\{X\times\{y\}:y\in U\big\}\right]=X\times U$$

is open in $X\times Y$, so $h$ is indeed continuous.

As a bit of an aside, it is certainly true that $h$ and $\pi_Y$ are closely related. Let $U$ be any open set in $Y$. Then

$$\pi_Y^{-1}[U]=X\times U=\bigcup\big\{X\times\{y\}:y\in U\big\}=\bigcup h^{-1}[U]\,,$$

and it’s easy to check that $h\circ q=\pi_Y$. Thus, $$q^{-1}\big[h^{-1}[U]\big]=(h\circ q)^{-1}[U]=\pi_Y^{-1}[U]\,,$$ which of course is open because $\pi_Y$ is continuous.

The proof that $g$ is continuous also has a problem. If you’re going to do this by showing that $g^{-1}[C]$ is closed in $Y$ for each closed $C\subseteq(X\times Y)^*$, you really do need to consider all closed subsets of $(X\times Y)^*$. In fact you’ve not considered even one: $X\times\{y\}$ is a single point of $(X\times Y)^*$.

I would show instead that $g^{-1}[U]$ is open in $Y$ for each open $U\subseteq(X\times Y)^*$, because you know exactly which subsets of $(X\times Y)^*$ are open: $U\subseteq(X\times Y)^*$ is open if and only if $q^{-1}[U]$ is open in $X\times Y$. Check that $q^{-1}[U]=X\times h[U]$; this implies that $q^{-1}[U]$ is open in $X\times Y$ if and only if $h[U]$ is open in $Y$. Finally, $g^{-1}=h$, so $U\subseteq(X\times Y)^*$ is open if and only if $q^{-1}[U]$ is open in $X\times Y$, which is the case if and only if $g^{-1}[U]$ is open in $Y$.

Notice that this is actually more than we needed: we were trying to show that if $U$ is open in $(X\times Y)^*$, then $g^{-1}[U]$ is open in $Y$, and we actually showed that the implication goes both ways and hence that both $g$ and $g^{-1}(=h)$ are continuous and therefore homeomorphisms. In other words, the work before the aside was unnecessary: our proof that $g$ is continuous also shows that $h$ is continuous. I kept the first part for two reasons. First, of course, I wanted to address your attempt. But there’s also a useful lesson here: once you’ve proved something, it’s worth going back over your argument to see whether it can be tightened up or presented more clearly. Here we could actually omit half of it.