How to prove that a mapping is continuous? I have the following bijection:
$$f: V = [0, 1) \to \{(x, y) \in \mathbb{R}^2 | x^2 + y^2 = 1\} =U$$
$$f(t) = (\cos(2\pi t), \sin(2\pi t))$$
And I want to prove that it's continuous (for every open set of $U$, the inverse image $f^{-1}(U)\subset V$ is the open subset of V). How can I do this ?
 A: You need a topology on each of $V$ and $U$.  Usually you inherit the topology from the ambient space.  I will assume the usual subspace topology:  


*

*A subset $B$ of $V$ is open if it is a countable union of sets of the form $[0,1) \cap (x,y)$, $x,y\in\mathbb{R}$  

*A subset $A$ of $U$ is open if it is a countable union of sets of the form $ U \cap B(x,r)$ where $B(x,r)$ is the open ball centered at $x\in\mathbb{R}^2$ of radius $r$
Now we can say that $f$ is continuous if its inverse maps open sets of $U$ to open sets of $V$.  So, assume $A \subseteq U$ is open. The key insight is given an open ball $B(x,r)$, its intersection with $U$ (if non-empty) is an arc of the unit circle (draw a picture to see it). This means that $A$, if non-empty, is the countable union of arcs $C_n$ of the unit circle, say
$$ A = \bigcup_{n\in\mathbb{N}} C_n $$
So 
$$ f^{-1} (A) = \bigcup_{n\in\mathbb{N}} f^{-1} (C_n ) $$
So all that's left to show is that each $f^{-1} (C_n )$ is open in $V$. But notice that we can identify any point on $U$ with the usual "polar coordinates"
$$ x = \cos \theta,\,\,y=\sin\theta,\,\,\theta\in [0,2\pi)$$
which is equivalent to 
$$ x = \cos 2\pi t,\,\,y=\sin 2\pi t,\,\,t\in [0,1)$$
and that the mapping preserves the order (ie. if $\theta_2 \geq \theta_1$, then $t_2 \geq t_1$).  Hence arcs of the circle can be identified with (potentially left closed) intervals $(\theta_1,\theta_2)$,which can be identified with (potentially left closed) intervals $(t_1, t_2)$, which are open in $V$.  Hence $f^{-1} (A)$ is the countable union of open intervals
A: Hint: It is enough to consider the topology of real plane in which the open sets are given by the epsilon-delta definition, same as for continuity.
