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?
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 $S$.