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:

$$\begin{array}{ll} \text{Vertices:} & v = \set{[0],[2]}, w = \set{[1],[3]} \\[0.1in] \text{Edges:} & a = \set{[01],[03],[21],[23]}, b = \set{[02]}, c = \set{[13]} \\[0.1in] \text{Faces:} & R = \set{[013],[213]}, S = \set{[021],[023]} \end{array}$$

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

$$\begin{align} & \delta_1(a) = \delta_1([01]) = [1] - [0] = (w) - (v) = w-v \\ & \delta_1(b) = \delta_1([02]) = [2] - [0] = (v) - (v) = 0 \\ & \delta_1(c) = \delta_1([13]) = [3] - [1] = (w) - (w) = 0 \end{align}$$

So $\im \delta_1 = \la w-v \ra$. On the other hand, $\ker \delta_1 = \la b,c \ra$.

For $\delta_2$, we have

$$\begin{align*} & \delta_2(R) = \delta_2([013]) = [13] - [03] + [01] = (c) - (a) + (a) = c \\ & \delta_2(S) = \delta_2([021]) = [21] - [01] + [02] = (a) - (a) + (b) = b \end{align*}$$

So, $\im \delta_2 = \la b,c \ra$ and $\ker \delta_2 = \set{0}$.

For $\delta_3$, we have

$$\begin{align*} & \delta_3(\mathcal{T}) = \delta_3([0123]) = [123] - [023] + [013] - [012] = (-R) - (S) + (R) - (-S) = 0 \end{align*}$$

So $\im \delta_3 = \set{0}$ and $\ker \delta_3 = \Delta_3$.

$$\delta_4$$ and higher are the zero map. Then we have:$$

\begin{align} & H_0 = \la v, w \ra / \la v-w \ra \cong \mathbb{Z} \ & H_1 = \la b, c \ra / \la b,c \ra \cong \set{0} \ & H_2 = \set{0} / \set{0} \cong \set{0} \ & H_3 = \la \mathcal{T} \ra / \set{0} \cong \mathbb{Z} \end{align}

$$