Question on showing homeomorphism between spaces in Topology The following question comes from a section in a topology text having to do with contracting subsets.  
Prove that the following spaces are homeomorphic:
(a) $\mathbb{R}^2$
(b) $\mathbb{R}^2/I$
(d) $\mathbb{R}^2/I^2$
I am uncertain if I am asked to show homeomorphsim between the plane, an unit interval and the unit square.  I did not even know it is possible that homeomorphism between any two of these three spaces.
Can someone show me how such is possible?  Thank you in advance.
[Edit:  I have edited the post to include the page where the question comes from.  I am not certain if the question preceding the one I am asking in the included page will be of relevance to the one I am asking. My apologies for the confusion]
 A: For a point $(x,y)$ in the plane let $|(x,y)|=\sqrt{x^2+y^2}$ denote the usual Euclidean norm. 
$\def\R{\Bbb R^2}$
Part (c) of the question (as shown on the included picture by OP) asking for a homeomorphism between $\R$ and $\R/D^2$ is easiest (where $D^2$, I assume is the unit disc, $D^2=\{(x,y):|(x,y)|\le1\}$). 
In this case let $h(x,y)=(0,0)$ if $|(x,y)|\le1$, and 
let $h(x,y)=(x,y)\cdot(|(x,y)|-1)$ if $|(x,y)|\ge1$. 
Note that here $(0,0)$ as an element of $\R/D^2$ is used to represent the point to which all points of $D^2$ were sent (or smashed to). So, $h:\R\to\R/D^2$, and $h$ is easily shown to be a homeomorphism. Indeed $h$ is constant on $D^2$ so (its restriction) is continuous. Also, clearly $h$ is continuous on the set $E=\{(x,y):|(x,y)|\ge1\}$. The intersection of $D^2$ and $E$ is the unit circle and the two definitions of $h$ agree on the unit circle so $h$ is continuous by the Pasting Lemma. One has to show also that $h$ is a bijection and $h^{-1}$ is continuous ... left to the reader. 
The notation $I=[0,1]$ is standard in most topology books. But, for the present purposes (for part (d) only) I prefer to use $I=[-1,1]$, (it is not really much different, but is technically easier to work with). 
In case (d) we need to show that $\R$ and $\R/I^2$ are homeomorphic. It is perhaps a more common exercise to show that $D^2$ and $I^2$ are homeomorphic, 
so part (d) in this way becomes closely related to part (c). At any rate, one could exhibit a homeomorphism, using a different norm on $\R$ (the maximum norm), that is, let $||(x,y||=\max\{|x|,|y|\}$. Note that $I^2$ is the same as the (closed) "unit ball" with respect to this norm, i.e. $I^2=\{(x,y):||(x,y)||\le1\}$. Define a homeomorphism $g:\R\to\R/I^2$ by 
$g(x,y)=(0,0)$ if $(x,y)\in I^2$, and $g(x,y)=(x,y)\cdot(||(x,y)||-1)$ if $||(x,y)||\ge1$. 
It could be shown in a similar manner that if $C$ is a compact and convex subset of $\R$, and if $C$ has a non-empty interior then $\R$ and $\R/C$ are homeomorphic. Indeed one could use $C$ to define a norm on $\R$ (but I won't go into details) and then use this norm to define a homeomorphism. 
Finally consider case (b), $\R/I$. 
This time I find it more convenient to work with 
$I=[0,1]$ but since it is understood that $I\subset\R$ we must formally use $I=[0,1]\times\{0\}=$ $\{(x,0):0\le x\le1\}=$ $\{(x,y):0\le x\le1,y=0\}$. 
Note that $I$ is compact and convex, but has empty interior. 
Nevertheless one could explicitly describe a homeomorphism $f:\R\to\R/I$. (I do not know of a "standard" way to do this, but here is one description that works.) 
If $|y|\ge1$ let $f(x,y)=(x,y)$.
Also, if $x\le0$ then let $f(x,y)=(x,y)$.
If $|y|\le1$ and $x\ge1$ let $f(x,y)=(x-(1-|y|),y)$.
If $|y|\le1$ and $0\le x\le1$ let $f(x,y)=(x\cdot|y|,y)$.
Again, the pasting lemma could be used to show that $f$ is continuous (and with some more work, that it is a homeomorphism $f:\R\to\R/I$.) Several cases need to be considered, but I would only verify one: when $|y|\le1=x$. Then $f(x,y)=(1-(1-|y|),y)=(|y|,y)$ and also $f(x,y)=(1\cdot|y|,y)=(|y|,y)$, so these definitions agree. (Note also that $f(1,1)=(1,1)$ whether you use the definition $f(x,y)=(x,y)$, or $f(x,y)=(x-(1-|y|),y)$, or $f(x,y)=(x\cdot|y|,y)$. ) 
(The following is an abandoned attempt to describe a homeomorphism $f:\R\to\R/I$, where $I=[-1,1]\times\{0\}=$ $\{(x,0):-1\le x\le1\}=$ $\{(x,y):-1\le x\le1,y=0\}$, but I changed my mind and described the above instead. If $|x|\le1$ let $f(x,0)=(0,0)$. If $|x|\ge1$ let $f(x,0)=(x\cdot(|x|-1),0)$. There are different way to deal with $|y|\neq0$. Let $f(x,y)=(x,y)$ if $|y|\ge1$. Finally, if $0\le|y|\le1$ consider two cases depending on whether $|x|\le|y|$ or $|x|\ge|y|$. If $0\le|x|\le|y|\le1$ then let ...) 
