Show that, the spiral and the half straight line in $\Bbb{R}^2$ are homeomorphic. We are given a spiral say the set $A=\{(e^x\cos x,e^x\sin x)|x\in\Bbb{R}\}$ and $B=\{(y,y)|y\ge0\}$. We have to find a homeomorphim $h:A\to B$.
Intuitively by looking at their graphs in $\Bbb{R}^2$, it seems to be homeomorphic as you can always transform a spiral into a half straight line by stretching from the end (think of a string of infinite length) and vice versa.
By I want to have a rigorous construction of $h$.
Let's write $h=(h_1,h_2)$, then we must have $h(e^x\cos x,e^x\sin x)=(y,y)$ for some $y\ge0$. 
Now, $h(e^x\cos x,e^x\sin x)=(h_1(e^x\cos x,e^x\sin x),h_2(e^x\cos x,e^x\sin x))=(y,y)$
So, we must have $h_1(e^x\cos x,e^x\sin x)=h_2(e^x\cos x,e^x\sin x)$
If I choose $h_1(x_1,x_2)=h_1(x_1,x_2)=x_1^2+x_2^2$ i.e. $h(x_1,x_2)=(x_1^2+x_2^2,x_1^2+x_2^2)$.
Then $h_1(e^x\cos x,e^x\sin x)=h_2(e^x\cos x,e^x\sin x)=e^{2x}$. Then choose $y=e^{2x}$.
Will that $h$ work? Can anyone verify my solution? Is there anything that I need to rectify? Thanks for help in advance.
 A: Your $h$ does not seem to work because it does not have $(0, 0)$ in its image.
In fact, I don't think your spaces are homeomorphic. If you remove the point $(0, 0)$ from the line, it remains connected but removing any point from the spiral would make it disconnected.
If you consider the open ray $\{(y, y) \mid y > 0\}$, then your solution would work with $$h(x_1, x_2) = (x_1^2 + x_2^2, x_1^2 + x_2^2).$$

EDIT: In fact, if we work with the open ray, it might be easier to construct the following homeomorphisms:


*

*$\Bbb R \sim A.$
Consider the map $f:\Bbb R \to A$ defined by $x \mapsto (e^x\cos x, e^x\sin x)$.
This is clearly a bijection. Its inverse is also easily calculatable as
$$(x, y) \mapsto \dfrac{1}{2}\ln(x^2 + y^2).$$
Note that $(0, 0) \notin A$ and so the above map is indeed well-defined. Now it is clear that both $f$ and its inverse are continuous and thus, we have the homeomorphism.

*$\Bbb R^+ \sim B$. (Where $\Bbb R^+ = (0, \infty)$.)
This is quite easy as we have the map $x\mapsto (x, x)$ in one direction and the map $(x, y)\mapsto x$ in the other.  

*$\Bbb R^+ \sim \Bbb R$.
This is again easy as we have the maps $e^x$ and $\ln x$.
