Unsure how to prove that this function is homeomorphism: Consider the sets: $S^{1} = \{(x,y) \in \mathbb{R^2}:x^2+y^2=1\}$ and $Q = \{(x,y)\in\mathbb{R}^2:|x|+|y|=1\}$ and the function: $f:Q\to S^1$ given by:
$f(x,y) = (\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})$. I want to show that this function is a homeomorphism. Given the definition of a homeomorphism is a continuous, bijective function with a continuous inverse. To prove it is bijective I have said: Let $(a,b),(c,d) \in Q$ so that: $|a|+|b|=1$ and $|c|+|d|=1$. Suppose that $f(a,b) = f(c,d)$. From this I have gathered that $\frac{a}{b} = \frac{c}{d}$, but I don't know how to deduce that $a = c$ and $b = d$. Furthermore it is very difficult to find an inverse.
 A: You were given 
$$
f : Q \rightarrow S^1 ~;~ (x,y) \mapsto \left( \frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2} } \right)
$$
(Result from analysis:) $f$ is continuous on $\mathbb{R}^2 \setminus \{(0,0)\}$ as a composition of continuous maps on this domain. Thus it is also continuous on the subset $Q \subset \mathbb{R}^2 \setminus \{(0,0)\}$ with the subset-topology.
Now let $g$ be
$$
g : S^1 \rightarrow Q ~;~ (x,y) \mapsto \left( \frac{x}{ \vert x \vert + \vert y \vert }, \frac{y}{ \vert x \vert + \vert y \vert } \right)
$$
By the same argument as above, we can also conclude that $g$ is continuous. We will see now, that $g$ is also the inverse of $f$. For $(x,y) \in Q$ we have
\begin{align}
g(f(x,y)) 
&= 
g \Big( \underbrace{\frac{x}{\sqrt{x^2 + y^2}}}_{=:~a}, \underbrace{\frac{y}{\sqrt{x^2 + y^2}}}_{=:~b} \Big)
=
g(a,b)
\\&= \left( \frac{a}{ \vert a \vert + \vert b \vert }, \frac{b}{ \vert a \vert + \vert b \vert } \right)
\end{align}
Where we have
$$
\vert a \vert + \vert b \vert
=
\frac{\vert x \vert}{\sqrt{x^2 + y^2}} + \frac{\vert y \vert}{\sqrt{x^2 + y^2}}
= \frac{1}{\sqrt{x^2 + y^2}}
$$
since $\vert x \vert + \vert y \vert = 1$. So it follows
\begin{align}
g(f(x,y)) 
&= 
\left( a \sqrt{x^2 + y^2} ,~ b \sqrt{x^2 + y^2} \right)
=
(x,y)
\end{align}
Similarly, you can check that $f(g(x,y)) = (x,y)$. So we found a continuous inverse of $f$, which shows that $f$ is a homeomorphism.
A: Hint:  How many times does a ray with vertex at the origin intersect $S^1$ and how many times does it intersect $Q$?  Does this ray only intersect at $(x,y)$ and $f(x,y)$?  Does this make $f$ a bijection?  Does this actually make a bijection between open sectors of the plane and open intervals in each of $S^1$ and $Q$?
A: The easy form: after proving that $f$ is bijective (for example, like in the answer of Eric Towers) use that Continuous Bijection from ($T_1$) Compact to Hausdorff is Homeomorphism. The essential idea: $f^{-1}$ is continuous because $f$ is closed because $f$(compact) $=$ compact.
