A subset $S$ of a smooth manifold $M$ is called a weakly embedded submanifold (at least in Lee) if it admits a smooth structure making the inclusion an immersion, and such that for any other smooth manifold $N$, a map $N \to S$ is smooth iff its composition with $S \hookrightarrow M$ is smooth. Such a smooth structure on $S$ is clearly unique. To get a better feel for this property I am asking for the following:

Is there an example of a subset $S$ of smooth manifold $M$ which admits a unique smooth structure making the inclusion $S \hookrightarrow M$ an immersion, but which is not a weak embedding?


1 Answer 1


Nice question. Here's an example.

Let $S\subset\mathbb R^2$ be the union of the $x$-axis with the positive half of the $y$-axis, with the smooth structure induced by the immersion $F\colon \mathbb R\times \{0,1\}\to \mathbb R^2$ given by \begin{align*} F(x,0) &= (x,0),\\ F(x,1) &= (0,e^x). \end{align*} It's a fairly straightforward exercise to prove that this is the only topology and smooth structure making $S$ into an immersed submanifold of $\mathbb R^2$.

Now consider the smooth map $\phi\colon \mathbb R\to \mathbb R^2$ given by \begin{equation*} \phi(t) = \begin{cases} (e^{-1/t^2},0), & t>0,\\ (0,0), & t=0,\\ (0,e^ {-1/t^2}) & t<0. \end{cases} \end{equation*} Note that the $x$-axis is an open subset of $S$ in its submanifold topology. But the preimage of the $x$-axis under $\phi$ is $[0,\infty)$, which is not open in $\mathbb R$, so $\phi$ is not continuous as a map into $S$.

  • $\begingroup$ It occurred to me that my question is not that interesting and your example shows why: Whether or not your example fits the criteria of the example I was looking for hinges on whether you define manifolds to second countable or paracompact. I definitely prefer assuming paracompactness because I want to be able to view any set as a discrete manifold. In this case $S$ with the discrete topology would give a second smooth structure. If you in assume manifolds to second countable, then your smooth structure is indeed the only one. $\endgroup$ Feb 21, 2016 at 23:34
  • 1
    $\begingroup$ @AdrianClough: Yes, this is one of the reasons I always assume manifolds are second-countable. Also that they have a fixed dimension -- if you allow different components of $S$ to have different dimensions, then you get another second-countable smooth structure -- with three $1$-dimensional components and one $0$-dimensional one. $\endgroup$
    – Jack Lee
    Feb 21, 2016 at 23:44

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