Let $0<r<a$ be fixed. For $(\alpha,\beta)\in[0,2\pi]^2$ we consider
$$ g(e^{i\alpha},e^{i\beta}) = ((a+r\cos\alpha)\cos\beta,(a+r\cos\alpha)\sin\beta,r\sin\alpha) $$
I would like to prove it is an homeomorphism from $S^1 \times S^1$ to $g(S^1 \times S^1)$.
The fact it is onto is clear. The fact the map is continuous is clear since each coordinate function is continuous. The conclusion should follow from the fact that a bijective and continuous map from a compact metric space to another is an homeomorphism.
However I have some difficulty to prove it is injective. I tried to usual machinery by considering $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$ such that $g(e^{i\alpha_1},e^{i\beta_1})=g(e^{i\alpha_2},e^{i\beta_2})$ but I get nothing interesting using the trigonometry formula I know and I wonder if it is the good way to prove the fact it is objective.
I would like to know if there is another way I should look for, more « economical », to prove that the map is injective please. Thank you a lot
Say we have some $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$ such that $g(e^{i\alpha_1},e^{i\beta_1})=g(e^{i\alpha_2},e^{i\beta_2})$, and let $g_1,g_2,g_3$ denote the component functions of $g$.
By $$g_3(e^{i\alpha_1},e^{i\beta_1})=g_3(e^{i\alpha_2},e^{i\beta_2})$$
we get $\sin(\alpha_1)=\sin(\alpha_2)$, and by
$$[g_1(e^{i\alpha_1},e^{i\beta_1})]^2+[g_2(e^{i\alpha_1},e^{i\beta_1})]^2=[g_1(e^{i\alpha_2},e^{i\beta_2})]^2+[g_2(e^{i\alpha_2},e^{i\beta_2})]^2$$
we obtain $(a+r\cos(\alpha_1))^2=(a+r\cos(\alpha_2))^2$. As $0<r<a$ we have $a+r\cos(\alpha_1)>0$ and $a+r\cos(\alpha_2)>0$, and so $$a+r\cos(\alpha_1)=a+r\cos(\alpha_2)\implies \cos(\alpha_1)=\cos(\alpha_2)$$
As $\alpha_1,\alpha_2\in [0,2\pi]$ and $\sin(\alpha_1)=\sin(\alpha_2), \space\cos(\alpha_1)=\cos(\alpha_2)$ we must have either $\alpha_1=\alpha_2$ or one of the angles be $0$ and the other $2\pi$. In any case, $e^{i\alpha_1}=e^{i\alpha_2}$.
Dividing $g_1$ and $g_2$ by $a+r\cos(\alpha_1)=a+r\cos(\alpha_2)$ yields $$\cos(\beta_1)=\cos(\beta_2) \space\text{ and } \space\sin(\beta_1)=\sin(\beta_2)$$
A similar argument as before ensures $e^{i\beta_1}=e^{i\beta_2}$, from where the map is injective.