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Let $\psi:M\longrightarrow N$ be $C^\infty$, bijective immersion, the $\psi$ is a diffeomorphism. I am having trouble in proving this statement.

What I have done so far is this : By inverse function theorem, it is enough to prove $d\psi:M_m\longrightarrow N_{\psi(m)}$ is an isomorphism $\forall\ m\in M$. We already have that $d\psi$ is injective for all $m$. So it is enough to prove that $d\psi$ is surjective $\forall\ m\in M$. Suppose there is a point in $M$ where $d\psi$ is not surjective, then this means that $dim\ M=p<d=dim\ N$. Let $(U,\phi)$ be a coordinate system on $N$ such that $\phi(U)= \mathbb{R}^d$. Since $\psi$ maps $M$ onto $N$, $\phi\circ\psi(M)=\mathbb{R}^d$. Now I am supposed to get a contradiction from this by proving range $\phi\circ\psi$ has measure zero in $\mathbb{R}^d$, or by any other means (for which the second countability of $M$ is crucial). But I am not able to get this.

Any help will be appreciated!

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  • $\begingroup$ Can we assume that $M$ is connected? $\endgroup$ Commented Mar 16, 2014 at 19:34
  • $\begingroup$ I suppose so, though it is not given. But if for $M$ connected there is an argument, maybe I could generalize. $\endgroup$ Commented Mar 16, 2014 at 19:52
  • $\begingroup$ As long as the dimension of $M$ is constant (and you have second countability), it doesn't really matter. Connectedness guarantees that the dimension is constant even if that is not demanded in the definition of a manifold. If the constantness of dimension isn't required [I'm not aware of anybody doing that on purpose, but I have witnessed people giving definitions of manifolds that allow different dimensions for different components], we can have bijective immersions that aren't homeomorphisms ($\mathbb{R}^n \cup \mathbb{R}^{n-1}\cup\mathbb{R}^n \to \mathbb{R}^n$ could be one). $\endgroup$ Commented Mar 16, 2014 at 20:01
  • $\begingroup$ A diffeomorphiss is in particular an homeomorphism. Under the standard definitions, a bijective immersion need not be an homeomorphism, though (irrational geodesics in a torus or the figure $8$ in the plane appropriately paramtrized are standard examples) You need some extra hypothesis of a global nature (that the map be an homeo works, of course, but I think that it be proper should be enough) $\endgroup$ Commented Mar 16, 2014 at 20:28
  • $\begingroup$ @MarianoSuárez-Alvarez I understand bijective as meaning that $\psi(M) = N$ (and $\psi$ being injective, of course), not "bijective to $\psi(M)$". Then $M$ being paracompact and of constant dimension implies that it is a homeomorphism. $\endgroup$ Commented Mar 16, 2014 at 20:59

1 Answer 1

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Suppose we had $\dim M < \dim N$. Then for every $m \in M$ there are coordinate neighbourhoods $U_m$ of $m$ and $V_m$ of $\psi(m)$ such that in those charts, $\psi$ is the embedding $x \mapsto (x,0)$ of $\mathbb{R}^{\dim M}$ into $\mathbb{R}^{\dim N} \cong \mathbb{R}^{\dim M} \times \mathbb{R}^{\dim N - \dim M}$. Let $K_m$ be the subset of $U_m$ corresponding to the closed unit ball of $\mathbb{R}^{\dim M}$. Then $\psi(K_m)$ is a compact subset of $N$ with empty interior. Since $M$ is second countable, it has a countable basis $\mathcal{B} = \{ B_n : n\in \mathbb{N}\}$ of open sets each of which is contained in some such $K_m$. But then

$$\psi(M) = \psi \left(\bigcup_{n=0}^\infty \overline{B_n}\right) = \bigcup_{n=0}^\infty \psi(\overline{B_n})$$

is a countable union of compact sets with empty interior. Since $N$ is locally compact, hence a Baire space, it follows that $\psi(M)$ has empty interior, and in particular, $\psi$ is not surjective.

Thus if we have a surjective immersion $\psi\colon M \to N$, we must have $\dim M = \dim N$, and such an immersion is a local diffeomorphism. If it is bijective, it is hence a diffeomorphism.

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    $\begingroup$ This is great! Just what I needed! Thank you! $\endgroup$ Commented Mar 16, 2014 at 20:59
  • $\begingroup$ Please verify. Surjective immersions are local diffeomorphisms? $\endgroup$
    – user636532
    Commented Jul 23, 2019 at 13:42
  • $\begingroup$ For the record, $\psi(M)$ has empty interior because meagre sets in Baire spaces have empty interior. This is because (i) subsets of meagre sets are again meagre and (ii) a space $X$ is Baire if and only if all the non-empty open subsets of $X$ are of the 2nd category in $X$. $\endgroup$ Commented Jan 6, 2023 at 13:40
  • $\begingroup$ Another way (perhaps overkill) to see that if $\psi:M\to N$ is a smooth bijective map between smooth manifolds then $\dim M=\dim N$ is Netto's theorem. $\endgroup$ Commented Jan 6, 2023 at 16:19

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