Homology groups of a tetrahedron I have been solving a lot of questions lately in preparation for an exam which is in about 3weeks. Well, I came across this question but I have no clue. I think I would need a lot of help.
I am asked to compute the homology groups of a tetrahedron (i.e a triangulation of the 2-sphere) indicating all the chain groups and boundary homomorphisms. I am familiar with computing the homology groups of some well-known surfaces like the Möbius strip, the Klein bottle, the torus and the projective plane, but I am not sure how to go about the tetrahedron. Any help would be greatly appreciated. 
 A: Let us finish the computation of $H_1(X)$ using Agustí's notation above: 
Now I have row reduced Agustí's matrix above and (if my calculations are correct) we have that the kernel of $\partial_1$ is isomorphic to the free abelian group on three generators:
$$\ker \partial_1 = \Bbb{Z}\bigg\langle \color{red}{[1,2] - [1,3] + [2,3]}, \color{green}{[1,2]-[1,4] + [2,4]}, \color{blue}{[1,3] - [1,4] + [3,4]}\bigg\rangle.$$
Now by direct calculation I find that:
$$\begin{eqnarray*} \partial_2([1,2,3]) &=& \color{red}{[2,3] - [1,3] + [1,2]} \\
\partial_2([1,2,4]) &=& \color{green}{[1,2] + [2,4] - [1,4]}\\
\partial_2([1,3,4]) &=& \color{blue}{[1,3] + [3,4] - [1,4]} \\
\partial_2([2,3,4]) &=& [2,3] + [3,4] - [2,4] .\end{eqnarray*}$$
Now when we quotient out by $\textrm{im} \partial_2$ in calculating $H_1(X)$, what you are basically doing is imposing relations on $\ker \partial_1$. What are those relations? Well, you are simply declaring now that $$\partial_2(\text{all 2-simplices}) = 0.$$
From this you immediately see that upon quotiening, you have sent all the generators to zero so that $H_1(X) = 0$. 
A: I don't know what notation you have been using, but the following could help -just in case you haven't arrived so far.
Name the vertexes of the tetrahedron $[1],[2],[3],[4]$, for instance. Then the set of edges can be denoted by $[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]$ and the one of faces as $[1,2,3],[1,2,4],[1,3,4],[2,3,4]$.
So the corresponding groups of $0,1$ and $2$ chains are the free abelian groups on these generators:
$$
\begin{eqnarray*}
C_0 &=& \mathbb{Z}\langle [1],[2],[3],[4]\rangle \\
C_1 &=& \mathbb{Z}\langle [1,2],[1,3],[1,4],[2,3],[2,4],[3,4]\rangle \\
C_2 &=& \mathbb{Z}\langle [1,2,3],[1,2,4],[1,3,4],[2,3,4]\rangle
\end{eqnarray*}
$$
and the boundary operators
$$
C_2 \stackrel{\partial_2}{\longrightarrow} C_1 \stackrel{\partial_1}{\longrightarrow} C_0
$$
can be computed as follows:
$$
\partial_1 [1,2] = [2]-[1] \ , \qquad \partial_2 [1,2,3] = [2,3]-[1,3]+[1,2] \ , \qquad \text{etc}\dots
$$
So you can represent these boundary operators by matrices. For instance, 
$$
\partial_1 =
\begin{pmatrix}
-1  & -1  & -1  &  0  &  0  &  0  \\
 1  &  0  &  0  & -1  & -1  &  0  \\
 0  &  1  &  0  &  1  &  0  & -1  \\
 0  &  0  &  1  &  0  &  1  &  1
\end{pmatrix}
$$
and pursue your computations from this point.
For instance, if we denote by $T$ the tetrahedron, then
$$
H_2 (T) = \mathrm{ker}\ \partial_2
$$
and some few elementary column transformations with the $\partial_2$ matrix gives us that
$$
H_2 (T) = \mathbb{Z} \langle [2,3,4] - [1,3,4] + [1,2,4] - [1,2,3] \rangle \cong \mathbb{Z}\ .
$$
That is, as we already knew, of course: $H_2(T)$ is $\mathbb{Z}$. But we've got more: an explicit generator for this second homology group. Namely, the 2-cycle $[2,3,4] - [1,3,4] + [1,2,4] - [1,2,3] $, which is, of course, the sum of the faces of the tetrahedron (with a sign). 
A: The simplicial complex for the tetrahedron $T$ is exactly the same as that of the filled tetrahedron $F$ (a.k.a. the $3$-simplex) except in degree $3$. As we know the homology of the $3$-simplex (hopefully!) we need only compute $H_2(T)$.
Now, the $2$-cycles in $T$ at the same as in $F$. Since the group of $2$-boundaries in $F$ is cyclic of rank $1$ (because the $3$-chains form such a group) and $H_2(F)=0$, we see at once that $H_2(T)=\mathbb Z$.
