I have a smooth submersion $p:\Bbb T^2\to\Bbb S^1$. I want to prove the following assertion, with possibly all the details.
There is a diffeomorphism $\phi :\mathbb S^1\times \mathbb S^1\to \Bbb T^2$ and an integer $d\geq 1$ such that $$p\circ \phi(z_1,z_2)=z_1^d.$$
This means that $p\circ \phi$ is like projecting onto the first coordinate and applying a covering space of finite degree $d$.
What I did so far: Since $p$ is a submersion (hence an open map) and $\Bbb T^2$ is compact, $p(\Bbb T^2)$ is both an open subset of $\Bbb S^1$ and compact, so $p$ is onto. Also since $\Bbb T^2$ is compact, $p$ is proper. I think that Ehresmann's lemma says something like "a proper surjective submersion is a fiber bundle", so using this fact I know that $p$ is a fiber bundle. The fiber $F$ of this fiber bundle has to be a compact $1$-dimensional manifold, so there is $d\geq 1$ such that $$F\simeq \coprod_{i\in\{1,\dots,d\}}\Bbb S^1.$$ Here I am stuck because the rest of the proof should use techniques of fiber bundle which I don't really master. I think I can fixe a covering space $q:\Bbb R\to \Bbb S^1$ and I can pull back $p$ as follows: $$\require{AMScd} \begin{CD} X @>{f}>> \Bbb T^2\\ @V{h}VV @VV{p}V\\ \Bbb R @>q>> \Bbb S^1 \end{CD}$$ Then $h:X\to \Bbb R$ has to be a trivial $F$ bundle so there is a diffeomorphism $\Phi:\Bbb R\times F\to X$ such that $h\circ \Psi=\pi_1$. Also I think I am supposed to use the fact that $f$ is a covering space but I'm not even sure this is true.
It would be great if somebody coud explain me the details and tell me if I maid a mistake somewhere. A related question I asked before.
You're well on your way there. There may be some way to finish using the space $X$ constructed above, but I propose a slightly more direct route once we have $F \simeq \sqcup_d \, \mathbb{S}^1$.
Choose a point $x \in \mathbb{S}^1$, let $F_x$ be the fiber over that point, and let $W = \mathbb{S}^1 \setminus\{x\}$. Then $p: \mathbb{T}^2 \to \mathbb{S}^1$ restricts to a fiber bundle $\tilde{p}: (\mathbb{T}^2 \setminus F_x) \to W \simeq (0, 1)$. Since $(0, 1)$ has only trivial fiber bundles, we see $(\mathbb{T}^2 \setminus F_x) \simeq (W \times F)$, with $\tilde{p}$ corresponding to $(t, y) \mapsto t$. Now we see that $(W \times F) \simeq \sqcup_{i=0}^d\, ((0, 1) \times \mathbb{S}^1)_i \simeq (\sqcup_i (0, 1)_i) \times \mathbb{S}^1$. The picture we have so far is that $p^{-1}(W)$ looks like $\mathbb{T}^2$ with $d$ "vertical" circles removed.
Edit: I previously wrote to use the homotopy lifting property here in order to fill in the vertical circles, but it seems very messy. I think one should now repeat the construction again starting with $y \in \mathbb{S}^1$, $y \ne x$, $W' = \mathbb{S}^1 \setminus \{x\}$. I think one should be able to carefully construct the identification of $p^{-1}(W')$ with $\sqcup_d W' \times \mathbb{S}^1$ in such a way that this identification agrees with the first map we constructed (for $W$) on $p^{-1}(W \cap W')$. Here is where we really track the construction of the fiber bundle, so I'll leave it to you to verify that this compatibility is possible. This gives a submersion $\iota: \mathbb{S}^1 \times \mathbb{S}^1 \to \mathbb{T^2}$ which is not necessarily given in the form of $\phi$ above. Now you might appeal to: every submersion $\mathbb{S}^1 \to \mathbb{S}^1$ is smoothly homotopic to $(z \mapsto z^d)$ for some $d$.