Let $A = [0,1] \times [0,1]$.
Let $B = A \setminus_{\sim}$, where $\sim$ is generated by $(t,0) \sim (t,1) \ \forall t \text{ and } (0,s) \sim (1,s) \ \forall s$.
Consider further $S^1 \times S^1 = \{ (x,y,z,u) \in \mathbb{R}^2 \times \mathbb{R}^2 : x^2 + y^2 = 1, z^2 + u^2 = 1 \}$.
How can I show then that the function
$f \colon B \to S^1 \times S^1, (x,y) \mapsto (\exp(2 \pi ix), \exp(2 \pi iy))$
is continuous and bijective.
It basically follows from the fact that $e: \Bbb R \to \Bbb C\simeq \Bbb R^2; e(x)= \exp(2\pi ix)$ obeys $e(x)=e(y)$ iff $x - y = k 2 \pi$ for some $k \in \Bbb Z$, e.g. from properties of the sine/cosine, etc.
Then we can conclude that $f(x,y)=f(x',y')$ iff $e(x) = e(x')$ and $e(y)=e(y')$ iff $x -x' \in \Bbb Z$ and $y-y' \in \Bbb Z$ and the latter can only happen (as $x,y \in [0,1]$) when $x,x' \in \{0,1\}$ and $y,y' \in \{0,1\}$ or $(x,y) \sim (x',y')$.
The implication $(x,y) \sim (x',y') \to f(x,y)=f(x',y')$ (the same class implies the same value),for $f$ considered to be defined on $A=[0,1]^2$, gives us that $f$ is well-defined on $B=A{/}\sim$, because the choice of representative of a class does not affect the function value on that class. The universal property of quotients than gives that $f$ as defined on $B$ is continuous iff $f$ defined on $A$ is, which it is, as all component functions are (even) differentiable.
The implication $f(x,y)=f(x',y') \to (x,y) \sim (x',y')$ gives us the injectivity when defined on $B$: the same value implies the same class.
Compactness of $A$ and thus $B$ plus Hausdorffness of $S^1 \times S^1$ implies that we have a homeomorphism between $B$ and $S^1 \times S^1$.