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Suppose $f: U\to V$ is a $C^1$ surjection from a $k$ dimensional manifold $U$ onto a $k$ dimensional simply connected manifold $V$ with everywhere nonzero Jacobian or nonsingular differential. Is $f$ diffeomorphic?

Hans
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  • If $U$ is compact, then Stack of Records Theorem gives that $f$ is a covering map. So... maybe? – J. Moeller Feb 25 '19 at 16:41
  • @Prototank: The stack of records theorem implies local not global diffeomorphism. The map bijective map $[0,1)\to S^1$ satisfies the stack of records theorem, yet it is not a global diffeomorphism. – Hans Feb 25 '19 at 18:22
  • According to https://math.stackexchange.com/questions/860351/a-proper-local-diffeomorphism-between-manifolds-is-a-covering-map, if $f$ is proper, it's a covering map. So, in this case $f$ is a diffeomorphism (assuming $U$ is connected). In particular, this applies if $U$ is a closed manifold. Thus, a counterexample must have $U$ non-compact. – Jason DeVito - on hiatus Feb 25 '19 at 20:32
  • @JasonDeVito: I do not see how the covering map leads to diffeomorphism for connected $U$. How do you explain the map $x\in [0,1]\to (\cos(2\pi x),\sin(2\pi x))$ being not diffeomorphic? – Hans Feb 25 '19 at 20:47
  • Well, first, I was assuming $U$ and $V$ have no boundary. Second, if you extend the domain to be $(-\epsilon, 1+\epsilon)$ for any finite $\epsilon > 0$, then this map is not proper (and it is not a covering no matter what you do with the domain, unless you make the domain all of $\mathbb{R}$) – Jason DeVito - on hiatus Feb 25 '19 at 21:13
  • @JasonDeVito: The smooth mapping from $S^1$ to a figure $8$ loop has both $U$ and $V$ without boundary. It satisfies the theorem you cited, with the map being proper and covering. Yet it is not a diffeomorphism. Aside from this example, I just do not see how that theorem, particularly covering map, helps with the problem. Second, why do I need to extend the domain for my original example $U=[0,1]$? That example satisfies the theorem you cited. – Hans Feb 25 '19 at 22:29
  • @Hans: The figure 8 loop isn't a manifold and it's not simply connected. What does it mean to "satisfy a theorem"? – Jason DeVito - on hiatus Feb 25 '19 at 23:43
  • @JasonDeVito: You are right that figure $8$ is not a manifold. My bad. I see your point to extend the interval so as to include the boundary. Can you explicate why you say the map for the extended $(-\epsilon,1+\epsilon)$ is not proper? It seems the preimage of a compact subset of the circle is compact in the interval, and thus the map is proper. By satisfying "a theorem" I mean it satisfies the premise of the theorem and therefore its conclusion. I am looking into Moishe Cohen's answer. – Hans Feb 26 '19 at 01:02
  • The map is not proper because $K = S^1$ is a compact subset of $S^1$, but $f^{-1}(K) = (-\epsilon, 1 + \epsilon)$ is not compact. – Jason DeVito - on hiatus Feb 26 '19 at 05:09
  • @JasonDeVito: Oh darn. I was focusing on the compact proper subsets of $S^1$. Thanks. So then, how does covering map lead to injectiveness and thus diffeomorphism? – Hans Feb 26 '19 at 08:07
  • This comes from general results about covering spaces. Namely, given a (nice enough) connected space $X$ with basepoint $x_0 \in X$, to each subgroup $H\subseteq \pi_1(X,x_0)$ there is a unique connected covering space $X_H\xrightarrow{\pi_H} X$ with basepoint $x_h$ (with $\pi(x_h) = x_0)$ for which $\pi_\ast(\pi_1(X_H, x_0')) = H$. Now, if $X$ is simply connected, i.e., $\pi_1(X,x_0) = {1}$, then the only possibility for $H$ is $H = {1}$. So, there is a unique connected covering of $X$. Since the identity map $X\rightarrow X$ is a covering, this must be the only one.... – Jason DeVito - on hiatus Feb 27 '19 at 14:39
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    In summary, if $V$ is simply connected, the only connected covering it has is itself. So, if $U\rightarrow V$ is a covering, it must be the case that $U\cong V$. – Jason DeVito - on hiatus Feb 27 '19 at 14:40

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