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In the Euclidean space $\mathbb R^2$, the function of absolute values $y=|x|$ determines a $1$-dimentional topological manifold $M$. Define the natural projection map $$p:\mathbb R^2\to \mathbb R,$$ $$(x,y)\mapsto x,$$ let $f:=p|_M$ be the restriction of $p$ on $M$, then $M$ can be covered by two charts $U,V$, where $U=f^{-1}(-1,+\infty)$ and $V=f^{-1}(-\infty,1)$, the transition map between $U$ and $V$ is the identity map, which is clearly a smooth function, so according to the definition of differential (or smooth) manifolds (see for example Wiki page ), can we get the conclusion: the function image of $y=|x|$ is a differential manifold?

Tom
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    You can use a single chart, for that matter. So, yes, you have a smooth structure on $M$, but with that structure it is not an embedded (smooth) submanifold of $\Bbb R^2$. – Ted Shifrin May 21 '23 at 15:51
  • @TedShifrin, so, can we say $M$ as a subest of $\mathbb R^2$ is not a differential manifold, but $M$ can be given a smooth structure? – Tom May 21 '23 at 16:05
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    @Tom: You can say that it is not a differentiable submanifold of $R^2$ but, as a topological space, admits a smooth structure, i.e. is homeomorphic to a smooth manifold. – Moishe Kohan May 21 '23 at 16:26
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    If you're comfortable with groups, this is kind of like noticing that ${0,1}$ can be made into a group, i.e. $\mathbb Z_2$, but it is not a subgroup of $\mathbb Z$, even though it is a subset. – JonathanZ May 21 '23 at 17:02
  • Thanks@MoisheKohan@JonathanZ supports MonicaC, your comments are really helpful. – Tom May 22 '23 at 09:33

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