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.

  • $\begingroup$ You mean $p\circ\phi$ is like projection onto the first coordinate and applying a $d$-fold covering. $\endgroup$
    – user403337
    Jul 20, 2019 at 14:22
  • $\begingroup$ @ChrisCuster $p\circ \phi$ is the projection onto the first coordinate composed with a $d$-fold covering. I wrote "$p$ is like projecting onto the first coordinate and applying a covering space of finite degree $d$" because "$p$ is like $p\circ \phi$". I'll change it anyway. $\endgroup$ Jul 20, 2019 at 14:39

1 Answer 1


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 $d$.

  • $\begingroup$ Thanks for your answer. I don't get one part of your argument, why does the homotopy lifting property allow us to extend $\tilde{i}$? $\endgroup$ Jul 20, 2019 at 22:34
  • $\begingroup$ It's basically what Nicolas Hemelsoet said in their answer to your other question you linked to. I'll expand my answer. $\endgroup$ Jul 21, 2019 at 1:29
  • $\begingroup$ I changed my approach while editing. I also believe that the statement you're seeking might not literally be true, for the simple reason that every submersion from the circle to itself doesn't have to be of the form $(z \mapsto z^d)$ but is only smoothly homotopic to a map of this form. $\endgroup$ Jul 21, 2019 at 21:17
  • $\begingroup$ Thank you for spending some time on this problem. I am not sure that the statement is true as it is right now, but I strongly believe it is. What should be the correct statement in your opinion? Also I don't understand completely your answer sorry I'll try to think about it in details later. $\endgroup$ Jul 22, 2019 at 22:16

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