By Sard’s theorem, there is no smooth surjective map from a second-countable manifold onto a manifold with higher dimension. However, without second-countability, the identity map from ($ \mathbb{R} $, discrete) onto ($ \mathbb{R} $, usual) is a counterexample.

Question. What if we require connectedness instead of second-countability? i.e., is there a smooth surjective map from a connected manifold onto a manifold with higher dimension?

Since a connected paracompact manifold is second-countable, a counterexample must be non-paracompact (if exsits). I have no idea how to construct such a manifold, or how to prove non-existence.


1 Answer 1


Here is an example of a connected surface with a smooth map onto the threefold $\mathbb R^2\times S^1$.

Consider the Prüfer manifold $X=\left(\coprod_{a\in\mathbb R}\mathbb R^2_a \right)/{\sim}$, where each $\mathbb R^2_a$ is a copy of the real plane, the elements of which I want to denote by $(x,y;a)$ with $x,y\in \mathbb R$, and where the equivalence relation $\sim$ identifies $(x,y;a)\in\mathbb R^2_a$ with $(x',y';b)\in\mathbb R^2_b$ if and only if $y=y'> 0$ and $a+xy=b+x'y'$. Let me denote the class of $(x,y;a)\in \mathbb R^2_a$ in $X$ by $[x,y;a]$.

The obvious map $\pi\colon\coprod_{a\in\mathbb R}\mathbb R^2_a \to X$ defines the smooth structure on $X$, i.e., a map $f\colon X\to Y$ is smooth if and only if the composite $f\circ\pi$ is smooth. In other words, $f$ is smooth if and only if its restriction to each $\mathbb R^2_a$ is smooth. In particular, $[x,y;a]\mapsto y$ defines a smooth function $X\to \mathbb R$, but the maps defined by $x$ and $a$ are well-defined and smooth only on the locus where no glueing is happening, i.e., where $y\leq 0$.

Let $h \colon\mathbb R\to\mathbb R_{\geq 0}$ be a surjective smooth function satisfying $h(x)=0$ for all $x\geq 0$ and $h(x)>0$ for all $x<0$, e.g., $$h(x)=\begin{cases}e^{\tfrac{1}{x} - x} & x<0\\0& x\geq 0.\end{cases}$$

There with, we define the map $f\colon X\to\mathbb R^2\times S^1$, $[x,y;a]\mapsto (a+xy,ah(y),e^{iy})$. The first and third components are clearly well-defined and smooth. The second component is well-defined even though $a$ does not define a global function, because on the locus where $a$ is not well-defined, $y>0$ and so $ah(y)=0$. Therefore, $f$ is smooth.

To see that $f$ is surjective, let $(u,v,e^{iy})\in \mathbb R^2\times S^1$ be an arbitrary point. Shifting $y$ by $-2\pi$ as often as necessary, we may assume that $y<0$. Define $a=\frac{v}{h(y)}$ and $x=\frac{uh(y)-v}{yh(y)}$ and observe that $f([x,y;a])=(u,v,e^{iy})$. Thus, $f$ is surjective, as claimed.

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    $\begingroup$ By the way, this was fun to think about. I first tried to come up with a curve mapping onto a surface, but didn't manage and then I seemed to remember that there was some reason why there is nothing like the Prüfer manifold in dimension one. (Right?) After all, the idea is to take uncountably many "usual" manifolds, glue them somewhere to make the thing connected and then cover a one dimension-higher-manifold using the uncountably many layers, just like the OP did with discrete $\mathbb R$ mapping surjectively onto $\mathbb R$. I still wonder whether there is a curve mapping onto a surface... $\endgroup$
    – Ben
    Sep 24, 2020 at 10:30
  • $\begingroup$ Thanks for an excellent counterexample! I also wonder whether there is a smooth surjective map from a connected one-dimensional manifold onto a surface. By classification of one-dimensional manifolds, the problem is reduced to whether there is a smooth surjective map from the long line onto $ \mathbb{R}^2 $ though I don’t know how to tackle with it. $\endgroup$
    – o-ccah
    Sep 24, 2020 at 14:56
  • $\begingroup$ There is not; at least not into $\mathbb R^2$, since every continuous function $L\to\mathbb R$ is eventually constant. But I was thinking, perhaps onto a compact $2$-fold. I don’t know how to do it either, though. $\endgroup$
    – Ben
    Sep 24, 2020 at 15:09
  • $\begingroup$ Sorry, I was misunderstanding. $\endgroup$
    – o-ccah
    Sep 24, 2020 at 15:13
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    $\begingroup$ Actually, it's not so hard to see that there is also no smooth surjective map from the long line onto a compact manifold of higher dimension, for such a map had to be eventually constant and so a subspace diffeomorphic to a usual interval would be enough to cover the higher-dimensional manifold, which is clearly impossible. $\endgroup$
    – Ben
    Sep 28, 2020 at 14:56

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