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Is a triangle (its sides and the region enclosed by its sides) in a 2D Euclidean space $\mathbb{E}^2$ a manifold? I was thinking to use the identity mapping as its charts, but for each point on the sides of the triangle, there is no neighborhood of it can be mapped to an open subset in $\mathbb{E}^2$.

Thanks!

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    Just to add the more general result that every topological space X that is homeomorphic to a manifold M, can be given manifold charts by pulling back the charts of M using the homeomorphism; in this case, the triangle is homeomorphic to a circle by smoothing out the corners, so the triangle can be given manifold charts by pulling back those of $S^1$. Still, as Ryan said, with the subspace topology, it is not a manifold2011-05-15

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Here's a variant of my comment:

A triangle is a topological manifold with boundary (and so can be made abstractly into a smooth manifold with boundary). But as a subspace of Euclidean space it is not a smooth manifold with boundary. On the other hand, a triangle is a "smooth manifold with corners".

There are a variety of stratified enhancements on the manifold concept. Manifold with no boundary is the "base" manifold concept. You can add boundary or various other stratifications and at some point you can let your space degenerate to the point that anything is more or less a "manifold with enough degeneracies..." Common terms for highly-degenerate manifold types are things like "stratified spaces" and "manifolds with singularities" or "pseudomanifolds". Orbifolds are another variation on this thread of ideas.