Consider $\mathbb{R}$ with the Euclidean topology. Suppose we have an equivalence relation on $\mathbb{R}$ such that the equivalence classes are $\mathbb{Z}$ and single non-integer points. Let $q: \mathbb{R}\to Y$ be the corresponding quotient map, so $Y$ is given by identifying all the integers to a single point, say $\alpha$, and is given the quotient topology.
I am interested in proving that $q$ is a closed map, but I am pretty much stumped. So any hint/help will be useful.
Thanks.
Let $C \subseteq \mathbb R$ be a closed set, we need to prove that $q^{-1}(q(C))$ is closed and then from properties of quotient maps we can deduce that $q(C)$ is close, and indeed:
if $C \cap \mathbb Z = \emptyset$ than from the definition of q we have that $q^{-1}(q(C)) = C$ wich is closed by the assumption
if $C \cap \mathbb Z \neq \emptyset$ than from the definition of q we have that $q^{-1}(q(C)) = C \cup \mathbb Z$ wich is closde as a finite union of closed sets.
and thus $q(C)$ is close and thus $q$ is a closed map