Quotients of Subspaces and Subspaces of Quotients

I'm currently at Section 9 in Willard, on quotient spaces and the strong topology.

I've still got an open question on stackexchange for 9A.2, which seems to require using compactness to prove the second part. (The text hasn't covered compactness yet! So that's a pain.)

Exercise 9D took me a while, and it feels messy because of notation. So I'll write it up here, which should make it neater and force me to explain it better.

Requisite background: 

Now, in this text quotient spaces are understood in the context of decompositions. I greatly dislike this word, and would much rather use the word partition. So if you're like me and have only ever heard the word "partition" before, and not "decomposition," then just read the word "partition" in your head every time they word "decomposition" shows up.

A decomposition $\mathscr{D}$ of a topological space $X$ is a collection of disjoint subsets of $X$ whose union is $X$.

In other words. It's a partition. A subset of the decomposition is a set of equivalence classes, effectively.

We assign a topology to the decomposition by determining that a set $\mathscr{E}$ of equivalence classes (a subset of $\mathscr{D}$) is open in $\mathscr{D}$ iff the union of those equivalence classes is an open set in $X$. That is, $\bigcup \lbrace E | E\in \mathscr{E} \rbrace$ is open in $X$.

We also note that this endows $\mathscr{D}$ with the quotient topology induced by the map P(x), which maps x to the equivalence class containing x. (I call this the Partition map in my head, though it appears nowhere in the text.)

If we have endowed the decomposition with such a topology, we naturally call it the decomposition space.

Okay, on to the problem...

When we're talking quotient spaces, there's two ways you could naturally think of to "restrict" the decomposition to some subset A of the original space X.

The first way:
Let's denote as $\mathscr{D}_A$ the decomposition of A defined by $F\in \mathscr{D} \Rightarrow F\cap A \in \mathscr{D}_A$. The topology on $\mathscr{D}_A$ thus follows from the subspace topology on A.
That is, $\mathscr{U}\subset \mathscr{D}_A$ is open iff $\bigcup \lbrace F | F\in \mathscr{U}\rbrace$ is open in A, as a subspace of X.

The second way:Let's denote our second alternative, $\mathscr{D}|A = \lbrace y \in \mathscr{D} : A\cap y \neq \emptyset \in X \rbrace$ . This is a subset of our original space $\mathscr{D}$ endowed with the subspace topology from $\mathscr{D}$, composed of all equivalence classes which "touch" our set A. 

Evidently, sometimes these two topologies are not identical. The text gives an example which makes this obvious: Consider the set $\lbrace (x, 0) \in \mathbb{R}^2| x < 0\rbrace \cup \lbrace (0,1)\rbrace $; for which $\mathscr{D}_A$ has an isolated point, and $\mathscr{D}|A$ does not.

However, if A in X is a union of elements of our original $\mathscr{D}$, then $\mathscr{D}_A$ is homeomorphic to $\mathscr{D}|A$, which I will prove here. 

For this homeomorphism, I'm going to explicitly define a map. It may help to remind yourself what $P(x): X \rightarrow \mathscr{D}$ means. Also remember common restriction notation; I will abuse it a little and use $(P|A)(x): A\rightarrow \mathscr{D}_A$. That is, $(P|A)(x)$ maps x onto the equivalence class considered as a subset of A, not necessarily the full equivalence class in $\mathscr{D}$.

Now for our homeomorphism! Define $\phi: \mathscr{D}_A \rightarrow \mathscr{D}|A$ by:
$$ \phi (x) = P( (P|A)^{-1}(G)) \text{ for } G\in \mathscr{D}_A$$
Let's take a moment to make sure the map is well-defined, one-to-one, and onto.

Well-defined: If $G\in \mathscr{D}_A$ then $(P|A)^{-1}(G)$ is actually a big ol' subset of X (well, the subset G!), and I'd want to make sure that every element in this subset gets sent to the same element of $\mathscr{D}$, and more specifically, an element of $\mathscr{D}_A$. Since G in $\mathscr{D}_A$ is defined to be equal to $F\cap A$ for some set $F \in \mathscr{D}$ then every element in G is also in F. Every element of F gets sent to $F\in \mathscr{D}$ by P() so regardless of the element we consider in G, it will get sent to this equivalence class F. Additionally, since $G=F\cap A$ is nonempty, F must be in $\mathscr{D}|A$ by definition.

Onto: If $F\in \mathscr{D}|A$, this means $F\in \mathscr{D}$ and $F\cap A$ is nonempty. But then $F\cap A$ is an element of $\mathscr{D}_A$ which maps to F, so F is in the image of $\phi$

One-to-One: Suppose $\phi(G_1) = \phi(G_2) = F \in \mathscr{D}|A$ for $G_1, G_2 \in \mathscr{D}_A $ Then $(P|A)^-1(G_1)$ and $(P|A)^-1(G_2) \subset F$ in X.

We didn't need that A is a union of elements of D in order to prove that the map is well-defined, one-to-one, and onto! We only need it to prove that it is a homeomorphism.

Quick lemma I am going to call the lemma of inclusion: With the assumed information in this problem, $F\in \mathscr{D}$ and $F \cap A$ is nonempty, then $F \subset A$. Since A is a union of elements of $\mathscr{D}$ and the elements of $\mathscr{D}$ are disjoint, any element $F\in \mathscr{D}$ is either entirely contained in A, or has no intersection with A. This follows immediately.

So, now that we have that squared away, to ensure it's a homeomorphism, I will prove that a set is open in $\mathscr{D}_A$ iff its image is open in $\mathscr{D}|A$

($\Rightarrow$) Let $\mathscr{U}\in \mathscr{D}_A$ be an open set, and consider $\phi(\mathscr{U})$. 

$$\begin{align}
\phi(\mathscr{U}) &= P( (P|A)^{-1} (\mathscr{U}))\\
&= P(V\cap A)
\begin{split} &\quad \text{for some V} \subset X \text{ that is open, since}\\
&\quad \mathscr{U}\text{ being open implies its preimage in A is open in A by definition}\\
\end{split}\\[1em]
&= P(V) \cap P(A)

\begin{split}&\quad P(V\cap A) \subset P(V \cap A) \text{is well known; and, in this case,}\\
&\quad F\in P(V)\cap P(A) \Rightarrow F\subset A \text{ (by inclusion lemma) and }\\
&\quad  \exists x \in F \text{such that} x\in V \text{and} x \in A \text{ so }P(F) \in P(V\cap A)\\
\end{split}
\end{align}$$

Since $P(V)$ is open in $\mathscr{D}$, and $P(A) = \mathscr{D}|A$, we see that $P(V) \cap \mathscr{D}|A$ is therefore open in the subspace topology inherited from $\mathscr{D}$.


($\Leftarrow$) Let $\mathscr{V} \subset \mathscr{D}|A$ be open in $\mathscr{D}|A.$ Then there exists some $\mathscr{W} \in \mathscr{D}$. Consider $\phi^{-1} (\mathscr{V})$

$$\begin{align}
\phi^{-1}(\mathscr{V}) &= \phi^{-1}(\mathscr{W}\cap \mathscr{D}|A) \\
&= P|A( P^{-1} (\mathscr{W}\cap \mathscr{D}|A)\\
&=P|A(P^{-1}(\mathscr{W})\cap P^{-1}(D|A)) \quad\\
&= P|A(P^{-1}(\mathscr{W}) \cap A) \quad  \text{using the inclusion lemma}\\
\end{align}$$

Now since $\mathscr{W}$ is open in $\mathscr{D}$ its preimage in X is open, which implies  $P^{-1}(\mathscr{W}) \cap A$ is open as a subset of A under the subspace topology!

Therefore $\phi^{-1}(\mathscr{V})$ is open in $\mathscr{D}_A$

We conclude that $\phi$ is a homeomorphism.

PHEW! Now that we are done, I bet you can see why the notation really messed with me in this proof. This was hell to type up, but it definitely helped me make sure I really knew why what I was doing worked.