Hatcher Exercise 2.1.7


We wish to obtain as the quotient space of , the tetrahedron, by identifying faces of its boundary. The construction is as follows:

Label the vertices of as (these do not refer to equivalence classes, this is simply a way to refer to the faces of ):

We will use simple concatenation to refer to a face, i.e. . We start with four unique vertices, six unique edges, four unique faces, and one unique three-simplex. As a first step, identify (note the orientation of the identification!) – call this class . This induces the edge identifications (call this class ), (call this class ), and vertex identifications (call this class ). The result now has three unique vertices, four unique edges, three unique faces, and one unique three simplex. Now, identify the other two faces (call this class ). This induces edge identifications and (so all edges in get identified to ones in , so just call this whole class , which now has four members), and the vertex identifications (call this class ). We now have two unique vertices, three unique edges (put the two remaining unidentified edges in classes and ), two unique faces, and one unique three-simplex. So, we finally have:

For a geometric argument as to why this is homeomorphic to , consider the following: Imagine performing the first identification by taking the two faces and pulling them around toward each other, using the edge incident to both of them as a pivot. After this, we end up with a double cone, with vertices and as the antipodal tips. The cones themselves are and . Then, this is homeomorphic to , and the faces represent the upper and lower hemisphere. When, we identify them directly, the quotient space becomes (this works the same as taking and identifying the upper semi-circle with the lower semi-circle , we seal these ‘lines’ together and close off the space yielding – this generalizes to our higher dimensional case).

Finally, we compute the homology. From the equivalence classes above, we have , , , and .

For , we have and .

For , we have

So . On the other hand, .

For , we have

So, and .

For , we have

So and . and higher are the zero map. Then we have: