Homeomorphism of Quotient Space I am relatively new to topology and thought one should be able to prove the following:
Let $X = A \cup B$ be a topological space and $A,B \subset X$ (Maybe we need them to be closed subspaces?). Then $X/A \cong B/(A \cap B)$.
This at least seems pretty intuitively, but i do not really know how to prove that. A hint/solution would be amazing!
I tried something along the lines of:
Since $X = A \cup B$, the identity on X yields a homeomorphism $X \cong A \cup B$. This induces a homeomorphism $X/A \cong (A \cup B)/A$. Now I would like the union to respect quotients, such that $(A \cup B)/A \cong A/A \cup B/(A \cap B)$. Now the left space is a singleton and i kind of get what I want.
 A: Let us first go back to the definition. If $A$ is a subset of $X$, let $\sim_A$ be the equivalence relation on $X$ defined by
$$
  x \sim_A y \iff \text{$x =y$ or ($x\in A$  and $y \in A$)}
$$
This is well-defined, even if $A$ is the empty set. Now, the quotient space $X/A$ is by definition $X/{\sim_A}$. Let $0$ denote the the equivalence class $[A]$ of $\sim_A$. Then, as a set,
$$
  X/{\sim_A} = \begin{cases}
  X & \text{if $A$ is empty}\\
  (X - A) \cup \{0\} & \text{if $A$ is nonempty}
\end{cases}
$$
The open subsets of $X/{\sim_A}$ are the subsets of $X-A$ open in $X$ and the subsets $U \cup \{0\}$ such that $U \cup A$ is open in $X$.
Back to your question. Since $X = A \cup B$, one gets $X - A = B - (A \cap B)$. Thus, as sets, $X/A$ and $B/(A \cap B)$ are both equal to $B$ if $A \cap B$ is empty and to $(B - (A \cap B)) \cup \{0\}$ otherwise. They are also isomorphic as topological spaces: the open sets are in both cases the sets $U$ or $U \cup \{0\}$ such that $U \cup (A\cap B)$ is open in $B$.
