Im not sure if I have read a real definition of a parametrization before, so here is what I thought. A subspace $A$ of a topological space $(X, \mathcal{O})$ is called a parameterized, if there exists a continuous function $f$ satisfies $\text{im}(f)=A$. The gap I am having here is what the domain of $f$ is supposed to be.
If this is somewhat correct, then a parametrization I came up with for the torus is given by
$$\varphi:[0,2\pi) \times [0,2\pi) \to \mathbb{R}^3, (\vartheta,\phi) \mapsto ((R+r\cos(\vartheta))\cos(\phi),(R+r\cos(\vartheta))\sin(\phi),r\sin(\vartheta)),$$ where $0<r<R$. That is, one can define the torus $T$ to be
$$T:=\{(x,y,z) \in \mathbb{R}^3 \ | \ \exists \vartheta,\phi \in [0,2\pi) : x=(R+r\cos(\vartheta))\cos(\phi), y=(R+r\cos(\vartheta))\sin(\phi),z=\sin(\vartheta)\}.$$
With this definition how can one show that the torus is homeomorphic to $S^1\times S^1$? The above function $\varphi$ can't help solving this since $[0,2\pi) \times [0,2 \pi)$ is not homeomorphic to $S^1 \times S^1$ by compactness arguments.
Trying to solve this I found this question which doesn't make any sense to me. In the question it says that $g$ is a bijection, which can't be true because $(2\pi,2 \pi)$ is mapped to the same point that $0$ is mapped to. Furthermore the answer just switches things around in a way that makes no sense to me. We still start with a generalization of my $\varphi$, that is called $f$ and suddenly claim that $f$ is a function that has $S^1 \times S^1$ as its domain which is compact.
So my $2$ questions are: $(1)$ What would be a correct definition of parametrization?
$(2)$ How can I construct such a homeomorphism?
EDIT.
I was completely lost but I think that I have a solution thanks to the comments. We regard the same $\varphi$ as above but modified to have $[0,2\pi] \times [0,2\pi]$ as its domain. We know that $[0,1] \times [0,1] \cong [0,2\pi]\times[0,2\pi]$ via $(x,y) \mapsto (2\pi x,2\pi y)$. We can now define equivalence relations as follows. Define $\sim$ on $[0,1] \times [0,1]$ via $$(a,b) \sim (c,d) \iff (a,b) = (c,d) \lor a=0,c=1 \lor b=0,d=1 \lor a=1,c=0 \lor b=1,d=0.$$
Analogously define $\sim'$ on $[0,2\pi] \times [0,2\pi]$. Then we have that this induces a homeomorphism $$f:[0,1]\times [0,1]/\sim \to [0,2\pi] \times [0,2\pi]/\sim'.$$ By the universal property of quotient spaces we get a continuous $$\overline\varphi:[0,2\pi] \times [0,2\pi]/\sim' \to T,[(x,y)] \to \varphi(x,y),$$ where $T$ is defined as the imagine of $\varphi$. Then $\overline\varphi$ is surjective which is inherited by $\varphi$ and is injective by definition of the equivalence relation. Since the domain is compact and the codomain is hausdorff, we get that its a homeomorphism. Using that $[0,1] \times [0,1]/\sim \cong S^1 \times S^1$ this proves the claim.
Best Answer
What you have done is correct. There is only one little gap: The injectivity of $\bar \varphi$ relies on the injectivity of $\varphi : [0,2\pi) \times [0,2\pi) \to \mathbb R^3$. This fact may seem obvious, but nevertheless requires a formal proof.
However, let me suggest an alternative approach without using any parameterization. In the definition of $\varphi$ the angles $\vartheta$ and $\phi$ do not play a role on their own, but are only used to describe points $(\cos \vartheta, \sin \vartheta) \in S^1$ and $(\cos \phi, \sin \phi) \in S^1$. Thus we can directly define
$$h : S^1 \times S^1 \to \mathbb R^3,h((x_1,y_1),(x_2,y_2)) = ((R+rx_1)x_2,(R+rx_1)y_2,ry_1) .$$
$\operatorname{im}(h)$ is the torus $T$ defined in your question. Clearly $h$ is continuous (it is the restriction of a continuous map on $\mathbb R^4$). Let us show that $h$ is injective (this proves that $h : S^1 \times S^1 \to T$ is a continuous bijection and therefore a homeomorphism).
So let $h((x_1,y_1),(x_2,y_2)) = h((x'_1,y'_1),(x'_2,y'_2))$. Thus
$(R+rx_1)x_2 = (R+rx'_1)x'_2$
$(R+rx_1)y_2 = (R+rx'_1)y'_2$
$ry_1 = ry'_1$.
From 3. we get $y_1 = y'_1$. Summing the squares of 1. and 2. and noting that $x_1^2 + y_1^2 =1$, $(x'_1)^2 + (y'_1)^2 =1$ gives
But $R > r > 0$ and $x_1, x'_1 \ge -1$, thus $R+rx_1, R+rx'_1 > 0$ so that we must have $R+rx_1 = R+rx'_1$, i.e. $x_1 = x'_1$. But now 1. and 2. imply $x_2 = x'_2, y_2 = y'_2$.