A typical example of Homeomorphism The set $\mathbb{R}^2-\{(0,0)\}$ with the usual topology is:

(A) Homeomorphic to the open unit disc in $\mathbb{R}^2$
(B) the cylinder $\{(x,y,z)\in \mathbb{R}^3/ x^2+y^2=1 \}$
(C) the paraboloid $\{(x,y,z)\in \mathbb{R}^3/ z=x^2+y^2 \} $
(D) the paraboloid $\{(x,y,z)\in \mathbb{R}^3/ z=xy \} $

I had tried which topological property is not hold under homeomorphism. I'm little bit confused in graph of (b),(c) and (d) because I have not any software related to 3d graphs. Please give me suggestion. Is this true $\mathbb{R}^2-\{(0,0)\}$ is homeomorphic to
$S^2 = \{x\in \mathbb{R}^3/ \|x\|=1\}$ ? if it is not true then give some hint for option (a).Thanks in Advance.
 A: It's going to be hard to find an invariant that distinguishes these spaces from $ \Bbb R^2 \setminus \{(0,0)\} $, if you only know compactness, connectedness, and path-connectedness. If you knew what the fundamental group of a space was, this would be rather straightforward. As is, your best bet would be to show that these spaces are homeomorphic to a space that you know is not homeomorphic to $ \Bbb R^2 \setminus \{(0,0)\} $.
In particular, I would try that for (A), (C), and (D).
A hint for the cylinder: do you see that $ \Bbb R^2 \setminus \{(0,0)\} $ is homeomorphic to an open disc with a closed disc removed from the interior?

The fundamental group of a space is a group whose elements are closed curves of the space.
Well, that's not entirely true: we say that two closed curves are equivalent if you can continuously vary one to the other and we talk about the two interchangeably.
To make this set into a group we need a group operation. Here, we choose beforehand a basepoint in the space where our closed curves will begin and end. To make a new closed curve from two others, we will concatenate them, that is, we travel along the first one and then along the second one. It's not terribly hard to show that this turns our set into a group.
For example, it can be shown that the fundamental group of $ \Bbb R^2 \setminus \{(0,0)\} $ is isomorphic to the additive group of integers.
To see the intuition behind this, take a look at a closed curve of $ \Bbb R^2 \setminus \{(0,0)\} $, if it does not go around $(0,0)$, then we could shrink it down so that the entire curve is just a constant point. However, if the curve does go around $(0,0)$, then the number of times (and in what direction) the curve goes around the origin determines what curves it is equivalent with.
A: The space $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homeomorphic to $(a)$, $(c)$, or $(d)$, but it is homeomorphic to $(b)$.
To prove that $\mathbb{R}^2 \setminus \{(0,0)\}$ is homeomorphic to the cylinder $(b)$, define the following map. For the circle of radius $r$ around the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$, map this circle homeomorphically onto the circle of the cylinder at $z = \tan\left(\frac{\pi r}{r+1} - \frac{\pi}{2} \right)$. This covers all possible values of $z$, hence the whole cylinder.
Next, observe that $\mathbb{R}^2$ is homeomorphic to the topological spaces $(a)$, $(c)$, and $(d)$ via the maps:
$(a)$ Let $D$ denote the open unit disk. Then the map $D \to \mathbb{R}^2$ given by $x \mapsto x/(1-\|x\|)$ is a homeomorphism.
$(c)$ Let $P_1$ denote the paraboloid. Then the map $\mathbb{R}^2 \to P_1$ given by $(x,y) \mapsto (x,y,x^2+y^2)$ is a homeomorphism.
$(d)$ Let $P_2$ denote the paraboloid. Then the map $\mathbb{R}^2 \to P_1$ given by $(x,y) \mapsto (x,y,xy)$ is a homeomorphism.
Thus, to prove $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homeomorphic to $(a)$, $(b)$, or $(c)$, it suffices to prove that $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homeomorphic to $\mathbb{R}^2$.
To do this, we use the notion of homotopic curves. In a topological space $X$, two closed curves (continuous maps $\gamma : [0,1] \to X$ with $\gamma(0) = \gamma(1)$) are homotopic if they can be deformed continuously from one to the other. For instance, in $\mathbb{R}^2$, the unit circle and the circle of radius 2 (centered at the origin) are homotopic, by translating each point on the unit circle to its corresponding point on the circle of radius 2, along the line connecting them. Likewise, the unit circle is homotopic to the constant curve at the origin.
More opaquely, a homotopy between closed curves $\gamma_1$ and $\gamma_2$ is a continuous map $H:[0,1]\times[0,1] \to X$ such that $H(0,t) = \gamma_1(t)$ and $H(1,t) = \gamma_2(t)$ for all $t\in [0,1]$.
Since homotopies are continuous deformations, if $f: X \to Y$ is a homeomorphism and $\gamma_1$, $\gamma_2$ are homotopic in $X$, then the closed curves $f\circ\gamma_1$ and $f\circ\gamma_2$ are homotopic in $Y$.
Let us use this fact. The intuition is that any closed curve in $\mathbb{R}^2$ is homotopic to a point, whereas this is not true for closed curves enclosing the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$. We will prove this latter fact now. 
Lemma: A closed curve enclosing the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homotopic to a point.
Proof:
Suppose $\gamma_1$ is a closed curve in $\mathbb{R}^2 \setminus \{(0,0)\}$ enclosing the origin, and $\gamma_2$ is the constant curve at some point $c\in\mathbb{R}^2$. For the sake of contradiction, assume $H(s,t)$ is a homotopy between $\gamma_1$ and $\gamma_2$. Let $I$ be the subset of $[0,1]$ such that for every $x\in I$, the closed curve $H(x,t)$ encloses the origin. Since $[0,1]$ is compact, we see that the distance $|H(x,t) - H(x+\delta, t)|$ reaches a maximum as $t$ ranges over $[0,1]$, and this maximum can be made as small as desired by changing $\delta$. Hence, we can choose $\delta$ so that if $x\in I$ then $x+\delta \in I$. It follows that $I$ is an open subset of $[0,1]$. Now, take a convergent sequence $\{x_n\} \to x$ in $I$. If the distance $|H(x_n, t)|$ is bounded below as $t$ ranges over $[0,1]$, then $H(x,t)$ must also enclose the origin. If this distance gets arbitrarily small: let $t_n \in [0,1]$ be such that $|H(x_n, t_n)|$ is smallest. Since $[0,1]$ is compact, there is a convergent subsequence $t_{n_k} \to t$. Then $|H(x,t)| = 0$, which is a contradiction. Thus, $I$ is a closed subset of $[0,1]$. As $[0,1]$ is connected, $I$ must be all of $[0,1]$, and thus the homotopy cannot exist.
At last, we can prove $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^2 \setminus \{(0,0)\}$. Suppose $f: \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}^2$ is a homeomorphism. Choose a closed curve $\gamma$ enclosing the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$. Let $H$ be a homotopy between $f\circ \gamma$ and a point in $\mathbb{R}^2$. Then $f^{-1} \circ H$ is a homotopy between $\gamma$ and a point in $\mathbb{R}^2 \setminus \{(0,0)\}$. By the lemma, this is impossible.
The ideas here regarding homotopy of curves form the basis for algebraic topology, and the theory of the fundamental group. The fundamental group of a topological space $X$ is the set of equivalence classes of closed curves, under the equivalence relation of homotopy. What we proved here is essentially that the fundamental groups of $\mathbb{R}^2$ and $\mathbb{R}^2 \setminus \{(0,0)\}$ are nonisomorphic.
As a closing remark, the fundamental group of $\mathbb{R}^2$ is 0 because every closed curve is homotopic to a point. What is the fundamental group of $\mathbb{R}^2 \setminus \{(0,0)\}$? Can you prove your guess?
