continuitygeneral-topology
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 ?
Take $U=[0,\pi)$ for instance: it is open in $[0,2\pi)$ but its image by the inverse map is not open due to the image of $0$ having only half of a neighbourhood (see below).
What you might find confusing is that $U$ is indeed open in $[0,2\pi)$. But in $[0,2\pi)$, there is nothing to the left of $0$ that would be missing in $U$ preventing it from being open.
To show that $f$ is a homeomorphism, we can use the following. If $f:X\rightarrow Y$ is a continuous bijection where $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.
Then as @blamethelag suggested, you can prove continuity of $f$ by using results of product functions. (Here, I'll write $f(x)=(f_1(x),f_2(x))$.)
Let $U\times V$ be an open subset of $\mathbb{R}^2$. Then for every $x\in f^{-1}(U\times V)$, we have that $x\in f_1^{-1}(U)\cap f_2^{-1}(V)\subseteq f^{-1}(U\times V)$. Since $f_1$ and $f_2$ are continuous, $f_1^{-1}(U)\cap f_2^{-1}(V)$ is open in the domain of $f$, and this shows $f$ is continuous.
Best Answer
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:
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