Show there exists a unique map $g$ such that $g \circ f_{2} = h$

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I was wondering if somebody could give me some help on this question. Any hints etc. would be greatly appreciated.

Let $0 < b < a$. Define a smooth map $h: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ by $h(\theta,\phi)=((a+b\cos \phi)\cos \theta , (a+b\cos \phi)\sin \theta, b\sin \phi)$ . Show that there exists a unique map $g: \mathbb{T}^2 \rightarrow \mathbb{R}^3$ such that $g \circ f_{2} =h$, where $f_{2}: \mathbb{R}^2 \rightarrow \mathbb{T}^2 = \mathbb{S}^1 \times \mathbb{S}^1 \subset \mathbb{R}^2 \times \mathbb{R}^2$, where $\mathbb{S}^1$ is the unit circle and $f_{2}(t_{1}, t_{2})=(f(t_{1}),f(t_{2}))$ with $f:\mathbb{R} \rightarrow \mathbb{S}^1$ the exponential map given by $f(t)=e^{2 \pi it}$. Further deduce that $g$ is a smooth map.

So far I have managed to show that the rank of the linear map $Dh(\theta, \phi):\mathbb{R}^2 \rightarrow \mathbb{R}^3$ is 2 for all $(\theta, \phi) \in \mathbb{R}^2$. However I am unsure on how to proceed from here.

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Write $f_2\colon \mathbb R^2\rightarrow \mathbb S^1\times \mathbb S^1, (\theta, \phi) \mapsto \bigl((\cos\theta, \sin\theta), (\cos\phi, \sin\phi)\bigr)$ and define $$ g((c_\theta,s_\theta), (c_\phi, s_\phi)) := \bigl((a+bc_\phi)c_\theta, (a+bc_\phi)s_\theta, bs_\phi\bigr) $$ for $(c_\theta,s_\theta), (c_\phi,s_\phi)\in \mathbb S^1$. Then you have $g\circ f_2 = h$ and $g$ is clearly unique with this property, since $f_2$ is surjective. Hence, it remains to show that $g$ is smooth.
For this, note that $f_2$ defines locally inverses of the charts in an atlas for $\mathbb S^1\times \mathbb S^1$. Taking the identity as the only chart in an atlas for $\mathbb R^3$, it follows from $\operatorname{id}\circ g\circ f_2 = g\circ f_2= h$ that $g$ is smooth.