Is this an identification map or not? The definition of identification map is the following:
A surjective map $f: X \to Y$ is an identification map iff $U$ in $Y$ is open if and only if $f^{-1}(U)$ is open in $X$.
I have an idea of what identification spaces should look like. Like for example you have a space and you identify all the points in one subset. That should give a new space. The problem is that this seems to violate the definition above. For consider this example:
Let $X$ be the unit disk $D^2$ centered at the origin in $\mathbb R^2$ with the subspace topology and let $S \subset D^2$ be an open ball around the origin of radius $1/2$. Define $Y$ to be the quotient space $X/\sim$ where $x \sim x'$ iff $x,x' \in S$. Let $f: X \to X/\sim$ be the map $x \mapsto x$ if $x \notin S$ and $x \mapsto (0,0)$ otherwise. Then $f^{-1}((0,0)) = O$ and $f$ is not an identification even though it should. What is going on here?  
 A: You do have a perfectly good identification space; it’s just that when you identify a non-closed set to a point, you get results that may at first seem a little odd.
The equivalence relation $\sim$ needs to be defined on all of $X$: $x\sim y$ iff $x=y$ or $\{x,y\}\subseteq S$. Then the points of $X/\sim$ are the sets $\{x\}$ for $x\in X\setminus S$, and the set $S$. Let $Y=\{\langle 0,0\rangle\}\cup(X\setminus S)$; $Y$ is a closed annulus together with one extra point, the origin. Let $\varphi:Y\to X/\sim$ be defined by $\varphi(x)=\{x\}$ if $x\in X\setminus S$, and $\varphi(\langle 0,0\rangle)=S$; then $\varphi$ is a bijection. Let $$g=\varphi^{-1}\circ f:X\to Y\;;$$ instead of talking about $X/\sim$ and $f$, I’ll talk about $Y$ and $g$, pretending that $Y$ is the quotient and $g$ the quotient (or identification) map. (You did this implicitly; I’m being a little more careful, at least mentioning that $Y$ isn’t actually the same as $X/\sim$.)
By definition a set $U\subseteq Y$ is open in $Y$ iff $g^{-1}[U]$ is open in $X$. If $U$ is disjoint from the closed disk of radius $\frac12$ centred at the origin, then $g^{-1}[U]=[U]$, so $U$ is open in $Y$ iff it’s open in $X$. Thus, if $\overline{S}$ is the Euclidean closure of $S$, the quotient (or identification) topology on $Y\setminus D$ is just Euclidean. What are open nbhds of $\langle 0,0\rangle$ in $Y$? If $\langle 0,0\rangle\in U\subseteq Y$, then $$g^{-1}[U]=S\cup g^{-1}[U\setminus\{\langle 0,0\rangle\}]=S\cup\big(U\setminus\{\langle 0,0\rangle\}\big)=S\cup U\;.$$ If $U=\{\langle 0,0\rangle\}$, this is just $S$, which is open in $X$, so $\langle 0,0\rangle$ is an isolated point in $Y$. I’ll leave it to you to check that if a point $p$ lies on the circle $C$ of radius $\frac12$ centred at the origin, and $p\in U\subseteq Y$, then $g^{-1}[U]$ is open in $X$ iff $\langle 0,0\rangle\in U$ and $U\setminus\{\langle 0,0\rangle\}$ is open in the relative Euclidean topology on $X\setminus S$. Thus, every point of $C$ is an accumulation point of $\{\langle 0,0\rangle\}$, and $Y$ is not a $T_1$ space, let alone Hausdorff: $\operatorname{cl}_Y\{\langle 0,0\rangle\}=\{\langle 0,0\rangle\}\cup C$.
This really isn’t surprising when you think it through: every point of $C$ is in the closure of $S$ in $X$, and the identification doesn’t do anything to $X\setminus S$, so we should expect these points still to be in the closure of the point-formerly-known-as-$S$.
