Homology of a triangulation of the cube I am struggling with the following exercise from class:
Consider the cube $[0,1]^3$ and subdivide it in 48 tetrahedrons in the following way: We subdivide each face of the cube into eight triangles by adding a vertex in the center of the face and then adding an edge between this new vertex and the corners as well as the midpoints of the edges of the face. Then add a vertex to the center of the cube and finally take all the tetrahedrons having the vertex in the center as a vertex as well as three vertices of a triangle on one of the faces of the cube. We note this complex $C$.
(This is really complicated to explain and ever worse to draw on the computer but I think the idea is clear)
The aim is to calculate $H_1(C)$ with the help of the Euler-characteristic. In a first exercise, we have to calculate the Euler-characteristic as well as $H_0(C)$, which is easy since $C$ is connected.
In the second exercise, we have to show that both $H_2(C)$ and $H_3(C)$ are trivial and then deduce $H_1(C)$. At this point I have three questions:


*

*I tried to show that $H_3(C)$ is trivial using a "cancellation argument", i.e. I observed that each 2-simplex in the interior of the cube belongs to exactly two tetrahedrons. So one can show that each element in the kernel of $\partial_3$ must be of the form
$$
 n \sum_{T \in C^{3}} T.
$$
Then I tried to argue that the 2-simplices on the faces of the cube do not cancel out and thus $n = 0$. However, I could not come up with a rigorous argument for that. Is this a correct approach and what would be a rigorous argument for the last step?

*Is there an analogous argument for $H_2(C) = 0$? Or another simple argument?

*I suppose that eventually the aim is to conclude with the Euler-Poincaré Theorem. However, this theorem only gives me the free part of $H_1(C)$. How can I exclude the case that $H_1(C)$ has a torsion part? (Which is clear intuitively)
So far we have only introduced basic tools in class. So I cannot use too fancy results.
 A: For item 1, you say you already know that each 2-simplex in the interior of the cube belongs to exactly two tetrahedrons. So I am sure that you also know that each 2-simplex on the boundary of the cube belongs to exactly one tetrahedron. I'm sure you can apply that fact to finish the proof rigorously. I would add, thought, that until you specify orientations, your deduction that the kernel of $\partial_3$ has the form $n \sum_{T \in C^3} T$ is incomplete; the best you could say is that it has the form $\sum_{T \in C^3} n_T T$ such that the absolute value $|n_T|$ is constant. Still, though, that should be enough to make the proof go through.
For item 2, it's not going to be that simple. The kernel of $\partial_2$ is definitely nontrivial: for any tetrahedron $T$, the 2-chain $\partial_3 T$ is nontrivial and is in the kernel of $\partial_2$. What you have to prove is that the kernel of $\partial_2$ equals the image of $\partial_3$, in other words the kernel of $\partial_2$ is spanned by the 2-chains of the form $\partial_3 T$. One can do this, of course, by brute force linear algebra, and every beginner should be subjected to such torture at least once in their lives, although I'll admit that row reducing a $48 \times $(big number) matrix might be a bit much.
But there is a more conceptual way to do item 2 (which amounts to a more conceptual row reduction method). Think about what you know about the cube. It is contractible. But even more, you can construct a contraction in a very explicit manner using the given triangulation into 48 tetrahedra. You can actually enumerate those tetrahedra as $T_1,T_2,T_3,...,T_{48}$, and then prove for each $i=2,...,48$, that $T_i \cup \cdots \cup T_{i-1}$ intersected with $T_i$ is an explicitly contractible subcomplex of the boundary of $T_i$ that I will denote $\tau_i$: perhaps a single 2-simplex, or two 2-simplices, or three 2-simplices on the boundary of $T_i$, or perhaps a few 1-dimensional configurations. From this it follows that there is a deformation retraction from $T_1 \cup\cdots\cup T_{i-1} \cup T_i$ to $T_1 \cup\cdots\cup T_{i-1}$. If the idea of the triangulation itself is sufficiently clear to you --- you took 6 lines of text to explain it quite clearly, and that's not complicated at all --- then perhaps you can visualize the required enumeration $T_1,T_2,....,T_{48}$.
Then, given a 2-chain 
$$\sigma = \sum_{S \in C_2} n_S S
$$ 
contained in the kernel of $\partial_2$, can explicitly figure out the coefficients of a 3-chain $\sum_{i=1}^{48} n_i T_i$ whose $\partial_3$ image equals the given 2-chain. You'll figure out these coefficients in the order $n_{48}$, $n_{47}$,....,$n_1$. By induction, starting with $\sigma_{48}=\sigma$, you'll define $\sigma_{i-1}=\sigma_i - \partial_3(n_i T_i)$, and at the end you'll check that $\sigma_1$ is a multiply of $\partial T_1$ and so $\sigma_0=0$ and you'll be done.
Let's start with $n_{48}$. The tetrahedron $T_{48}$ will have exactly one triangular face $S$ on the boundary of the cube. Let $n_S$ be the coefficient of $\sigma$ in that triangular face. Let $n_{48}$ be either $n_S$ or $-n_S$, depending on orientations.
Then, by induction, assuming you've figured out $n_{48},...,n_{i+1}$, you let $n_i$ be plus or minus the coefficient in $\sigma_i$ of the 2-simplices in $\tau_i$.
