# is triangle a manifold?

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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It's a topological manifold with boundary. But as a subspace of Euclidean space it is not a smooth manifold with boundary. On the other hand, it is a "smooth manifold with corners". There are a variety of stratified enhancements on the manifold concept. Manifold with no boundary is the "base" concept, then 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..." –  Ryan Budney Oct 3 '10 at 3:01
@Ryan Budney: Very good comment, but I think it should be given as a regular answer instead of just as a comment (however trivial the fact may appear to you). Otherwise, the question will remain on the list of unanswered questions even though a perfectly good answer has been obtained. –  Hans Lundmark Oct 3 '10 at 13:07
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 manifold –  gary May 15 '11 at 18:25