Homeomorphisms between spaces $a).$ $\mathbb R$  and  the  parabola $y=x^2$ .
$b).$  circle  and  ellipse  in  $\mathbb R^2$
$c).$ Any  two  open  balls in  $\mathbb R^n$  are   homeomorphic.
Intuitively , I  can  see  the  pairs  being  transformed  into  one  another .. Obviously   that  is  not  much  useful.  How  do  I  prove  them  with  analytically  with   suitable  bi-continuous  functions  $?$
Sorry  for  there  is  no  work/effort  in  this  post , I  did  not  know  what  to  do .
 A: a) $\Phi:\mathbb R\longrightarrow \{(x,x^2)\mid x\in\mathbb R\}$ define by $$\Phi(x)= (x,x^2).$$
b) $\Psi : \{(x,y)\mid x^2+y^2=1\}\longrightarrow \{(x,y)\mid \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\}$ define by $$\Psi(x,y)=\left(\frac{x}{a},\frac{y}{b}\right)$$
A: This will be a bit hand-waving (as is apparently the custom of topologists :) ), but:
a) Send $x\in \mathbb{R}$ to $(x,x^2)\in \mathbb{R}^2$. See the following for a more complete answer: 
Prove that real line and parabola are topologically same but different geometrically?
b) Translate the circle to have its center at the origin (this is clearly a homeomorphism), then scale in the $x$-direction, then scale in the $y$-direction (these are also homeomorphisms, unless you scale by zero of course, but that will never happen). Now you should have your ellipse, just rotate it and translate to its designated position. Since compositions of homeomorphisms are homeomorphisms, the above yields a homeomorphism.
c) Same as in (b): Translate to origin, scale (but now in all directions), and translate to the correct position.
I leave it to you to verify that translations and nonzero scalings are homeomorphisms.
