Quotient map for usual topology Let $X=[0,1]\bigcup(2,3]$ , let $Y=[0,2]$ and suppose $X$ and $Y$ have the usual topologies.  Define $f:X\mapsto Y$ by $f(x)=x$ if $x\in[0,1]$ and $f(x)=x-1$ if $x\in(2,3]$. Is $f$ a quotient map?
Well I need to show that $f$ is surjective because by definition of $f$. The topology $\mathscr U$ on Y then $$\mathscr U = [{U\in\mathscr U : f^{-1}(U)\in\mathscr T}].$$  So f continuous.  My concern is at $1^{+}$ because from $[0,1]$ $f$ is surjective, but  is $f$ at the domain $(2,3]$. Is it enough to say that the image of $f((2,3])=(1,2]$ and $f([0,1])=[0,1]$ so $$f([0,1])\bigcup f((2,3])=[0,1]\bigcup(1,2]=[0,2]=Y.$$ 
 A: Small notational point: you mean that $f:X\to Y$. The arrow $\mapsto$ is used to show what $f$ does to an individual element of $X$:
$$f:X\to Y:x\mapsto\begin{cases}x,&\text{if }x\in[0,1]\\
x-1,&\text{if }x\in(2,3]\;.\end{cases}$$
Saying that $f$ is surjective ‘by definition of $f$’ is rather vague. Specifically, let $y\in Y=[0,2]$. If $y\in[0,1]$, then $f(y)=y$; if not, then $y\in(1,2]$, and $f(y+1)=y$. In either case $y$ is in the range of $f$, so $f$ is surjective.
The topology on $Y$ is the topology that it inherits from the real line, so that’s the topology that you use in checking continuity of $f$. You do not at this point know whether that is also the quotient topology on $Y$. I’ll leave it to you to check that $f$ is continuous at every point of $X$ except possibly $1$; this follows easily from the fact that both the identity map $\Bbb R\to\Bbb R:x\mapsto x$ and the translation $\Bbb R\to\Bbb R:x\mapsto x+1$ are continuous. Now $f(1)=1$, so we have to check that for any $\epsilon>0$, there is a $\delta>0$ such that $f\big[(1-\delta,1+\delta)\cap X\big]\subseteq(1-\epsilon,1+\epsilon)\cap Y$. And this is true: for $\epsilon<1$ we can take $\delta=\epsilon$, since $(1-\epsilon,1+\epsilon)\cap X=(1-\epsilon,1]$, and $f\big[(1-\epsilon,1]\big]=(1-\epsilon,1]\subseteq(1-\epsilon,1+\epsilon)$. (Why don’t I need to worry about $\epsilon\ge 1$?)
Now you can worry about whether $f$ is actually a quotient map, i.e., whether it’s true that a set $U\subseteq Y$ is open in $Y$ if and only if $f^{-1}[U]$ is open in $X$. HINT: Consider the set $[0,1]$.
A: A quotient map has to satisfy: $f^{-1}U$ is open in $X$ iff $U$ is open in $Y$. 
Hint: What can you say about the set $f^{-1}[0,1]$?
A: What is a "$1^+$"? The domain consists of numbers, not numbers with decorations. $f$ is indeed a surjective map.
However, your use of "$1^+$", however, conveys an important idea -- the trick is to figure out what you really mean by the idea and relate it to the problem.
In the codomain, every open neighborhood of $1$ contains real numbers greater than $1$ -- your use of $1^+$ is, presumably, some expression of this fact, and the fact you bring it up is that you recognize there is something suspicious about how $f$ relates to these things.
So, what can you say about the relationship of $f$ to open neighborhoods $1$, the numbers greater than $1$ contained in them, and what it means for $f$ to be a quotient map?
