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I'm looking for conditions that ensure the closure of an embedded manifold is a manifold with boundary.

The specific case I'm interested is as follows. Consider an oriented $C^1$ surface ${\cal S}$ in $\Bbb R^3$ such that its closure is known to be a $C^0$ surface with boundary. Moreover, suppose we know that the normal to ${\cal S}$ has a continuous extension to the closure. Does this imply that the closure is a $C^1$ manifold with boundary?

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    $\begingroup$ But for instance, what about a plane domain in {z=0}, interior to a continuous nondifferentiable curve? $\endgroup$
    – Pietro Majer
    Commented Feb 17 at 8:36
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    $\begingroup$ Ah, good one. Thanks for the counterexample! Going back to the larger question of what conditions are necessary to ensure the closure is a $C^1$ surface, what if we also assume that the boundary is a $C^1$ curve. Is that enough to ensure the closure is a $C^1$ manifold? $\endgroup$
    – Brian Seguin
    Commented Feb 18 at 0:42
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    $\begingroup$ Take a $C^1$ function on $(0,1]$ which has a continuous but not differentiable extension to $[0,1]$. Now let the manifold be the open “rounded” square (the corners rounded smoothly), and move its $z$ coordinate by that function w.r.t. the $x$-coordinate: The normal has a continuous extension to the boundary, the boundary is a $C^1$-curve, but the manifold is not smoother than continuous at the boundary $\endgroup$
    – Martin Väth
    Commented Feb 18 at 11:00
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    $\begingroup$ @MartinVäth Could you give more details? it is nor clear to me... thx $\endgroup$
    – Pietro Majer
    Commented Feb 18 at 11:07
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    $\begingroup$ @PietroMajer This is my intuition as well, but it is unclear to me exactly how to use the local inverse theorem here since $\bar{\cal S}$ is closed and this theorem deals with open sets. If you know how the details would go, I'd be grateful if you could writeup an answer. If you do, it might be better to do so under this other post as it asks specifically this question: mathoverflow.net/questions/464387/… $\endgroup$
    – Brian Seguin
    Commented Feb 18 at 23:56

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