The tetrahedron is a topological manifold I have been thinking that simplicial complexes can not be given topological manifold structure since a simplicial complex is a union of simplices of different dimensions, hence there may be are points with neighborhoods homeomorphic to different dimensional euclidian spaces. Then I've been told that the  tetrahedron which is a simplicial complex is also a $2$-dimensional topological manifold. what about vertices of the tetrahedron and the points on the edges, do they have neighborhoods homeomorphic to $\mathbb R^2$ ?
 A: The usual hierarchy of manifolds 'with structure' is that smooth manifolds and piecewise linear manifolds are contained in the class of piecewise differentiable manifolds (although there is an equivalence between piecewise linear and piecewise differentiable because of Whitead's triangulation theorem) which are contained in the class of topological manifolds. There are a few other kinds of manifolds which fit somewhere into this hierarchy, but these are the main ones. Piecewise linear manifolds are probably exactly what you think they are, they are roughly the manifolds you can construct by gluing together disks such that they overlap in a piecewise linear way. They can be defined in terms of an atlas whose transition functions are all piecewise linear.
You should hopefully be able to see that the tetrahedron is a manifold whose faces can be glued together in such a way that the open 'flaps' on the faces overlap in a piecewise linear way, and so the tetrahedron is a PL manifold. As the class of PL manifolds is contained in the class of topological manifolds, this means that the tetrahedron is also a topological manifold, and in fact this is easiest to see by noting that the tetrahedron is homeomorphic to a sphere, and so in particular every point in the tetrahedron has a neighbourhood which is homeomorphic to $\mathbb{R}^2$ - the definition of a topological manifold.
