Hatcher Exercise 2.1.7 We wish to obtain S^3 as the quotient space of \Delta^3, the tetrahedron, by identifying faces of its boundary. The construction is as follows: Label the vertices of \Delta^3 as \set{0,1,2,3} (these do not refer to equivalence classes, this is simply a way to refer to the faces of \Delta^3): We will use simple concatenation to refer to a face, i.e. [012]. We start with four unique vertices, six unique edges, four unique faces, and one unique three-simplex. As a first step, identify [013] \sim [213] (note the orientation of the identification!) – call this class R. This induces the edge identifications [03] \sim [23] (call this class a), [01] \sim [21] (call this class b), and vertex identifications [0] \sim [2] (call this class v). The result now has three unique vertices, four unique edges, three unique faces, and one unique three simplex. Now, identify the other two faces [021] \sim [023] (call this class S). This induces edge identifications [21] \sim [23] and [01] \sim [03] (so all edges in b get identified to ones in a, so just call this whole class a, which now has four members), and the vertex identifications [1] \sim [3] (call this class w). We now have two unique vertices, three unique edges (put the two remaining unidentified edges in classes b and c), two unique faces, and one unique three-simplex. So, we finally have: % For a geometric argument as to why this is homeomorphic to S^3, 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 [1] and [3] as the antipodal tips. The cones themselves are [021] and [023]. Then, this is homeomorphic to D^3, and the faces represent the upper and lower hemisphere. When, we identify them directly, the quotient space becomes S^3 (this works the same as taking D^2 and identifying the upper semi-circle D^1 with the lower semi-circle D^1, we seal these ‘lines’ together and close off the space yielding S^2 – this generalizes to our higher dimensional case). Finally, we compute the homology. From the equivalence classes above, we have \Delta_0 = \set{v,w}, \Delta_1 = \set{a,b,c}, \Delta_2 = \set{R,S}, and \Delta_3 = \set{\mathcal{T}}. For \delta_0, we have \im \delta_0 = \set{0} and \ker \delta_0 = \Delta_0. For \delta_1, we have % So \im \delta_1 = \la w-v \ra. On the other hand, \ker \delta_1 = \la b,c \ra. For \delta_2, we have % So, \im \delta_2 = \la b,c \ra and \ker \delta_2 = \set{0}. For \delta_3, we have % So \im \delta_3 = \set{0} and \ker \delta_3 = \Delta_3. \delta_4 and higher are the zero map. Then we have: %