In an exercise I have given a topological space $A\cup_{A\cap B}B$. The only preconditions I have are that $X$ is a topological space and $A,B$ are subsets of $X$ such that $X=A\cup B$.
In the lecture we defined a construction like that but did this via maps: Let $f : C \to Y$ and $g : C \to Z$ be two continuous maps between topological spaces. We refer to the topological space $Y\cup_C Z := (Y \sqcup Z)/\sim $ where $f(c) \sim g(c)$ for all $c \in C$ as the pushout of $f$ and $g$ .
In the considered case I guess we assume $f$ and $g$ to be the inclusion maps from $A\cap B$ into $A$ or $B$. Isn't $A\cup_{A\cap B}B$ then just the same as $A\cup B$?
Here's an example to show that the (bijective, continuous) map $A \cup_{A\cap B} B\to X, [x] \mapsto X$ is not a homeomorphism in general. Let $A$ and $B$ such that $A \cap B =\varnothing$. Then $A \cup_{A\cap B} B$ is simply $A \sqcup B$ and in this case, the map is a homeomorphism if and only if both $A$ and $B$ are open in $X$.