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Is the unit square $\partial I^2$ (i.e. the square with vertices $(0,0), (0,1), (1,0), (1,1) \in \mathbb R^2$) a smooth manifold?

I guess it shouldn't be smooth because it has "corners", but i have trouble actually finding an explicit atlas which "makes sense" and which contains two coordinate charts which are not compatible.

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    It's actually a tricky business to find manifolds which do not have any smooth structure. You have to $g$o $t$o dimension 4 $t$o find the first example.2012-02-07

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I initially thought this question was about $I^2$, but I can give a definite answer for $\partial I^2$, which is that this question doesn't make sense. Note that as a topological space, $\partial I^2$ is homeomorphic to the unit circle $S^1$ (in particular, it is a topological manifold!), which can be equipped with a smooth structure in a fairly straightforward way (e.g. using the exponential map $e^{ix} : \mathbb{R} \to S^1$). So it's not clear what we would mean by the statement that $\partial I^2$ isn't smooth.

One way to make this intuition precise is to think of $\partial I^2$ as the image of $S^1$ under a continuous map $S^1 \to \mathbb{R}^2$. Then the statement you want is this: no such map can be an injective immersion. (Edit, 12/10/15: An earlier version of this answer claimed that no such map can be smooth. In fact this is false; a counterexample can be constructed by slowing down as you hit each corner using a bump function.)

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    @user: no, we can't. Arbitrary subsets of a smooth manifold inherit at best a topology; they don't inherit a smooth structure in any reasonable sense (indeed they need not even be topological manifolds) in general. We can ask, for example, whether they're immersed submanifolds, but this is extra structure (namely the data of the immersion).2014-02-06