Simplicial Complexes, Triangulation general question. I am taking a first course in topology, and I am struggling with simplicial complexes. Specifically the triangulation of subspaces of $ \mathbb{R}^n $ confuses me. If you could help me on the following points I would be very grateful.


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*In general how do you construct a triangulation of a subspace? I have been given a very basic example, where the 2-sphere is triangulated, but how do we go about doing this for more complicated subspaces, such as the "Dunce hat" space? Additionally how do we prove or disprove the existence of such a triangulation?

*Do you have any book recommendations which would help with triangulation specifically and with simplicial complexes in general?
EDIT: drew a picture for barycentric subdivison, but drawing it took ages. will find an easier way for the next one.

Clearly this does not give a simplicial complex. I can see how the next subdivison does.
 A: There is no general recipe for this, but here is a construction which might help you. Suppose that $X$ is obtained by gluing certain simplices along faces (via affine maps). The result (which is a cell complex) may or may not be a simplicial complex, as you can see by looking at the dunce hat. You can, however, imitate the 2nd barycentric subdivision for $X$ as follows. Subdivide each face of $X$ barycentrically; you obtain a new complex $X'$, which still can fail to be simplicial (this happens in the dunce hat case). However, if you subdivide $X'$ barycentrically again, you obtain a cell complex $X''$ which carries a natural structure of a simplicial complex as intersection of any two faces is either empty or a face. Try it with the dunce hat and see what you get. You can also use this if faces of $X$ are not simplices, but, say, polygons: You first triangulate each polygon and then repeat the 2nd barycentric subdivision construction above. (There are more efficient way, but this one will work.) Try it with the torus obtained by gluing opposite sides of the square in the standard fashion. 
