Computing simplicial homology of the 2-sphere I've been essentially taking a crash course in simplicial homology over the past 2 days. I would like to compute the homology groups $H_{k}$ of the 2-sphere. I have a somewhat rigorous argument given below and I'd love if someone could tell me if this argument is sound. 
We can triangulate the 2-sphere using the shell of a $3$-simplex, and let's call this $T$ and its vertices $a$, $b$, $c$, and $d$. We know that $H_{k}(S^{2}) \cong H_{k}(T)$. We take a look at the chain complex
$$ \cdots \xrightarrow{\partial_{4}}  0 \xrightarrow{\partial_{3}} C_{2}(T) \xrightarrow{\partial_{2}} C_{1}(T) \xrightarrow{\partial_{1}} C_{0}(T) \xrightarrow{\partial_{1}} 0. $$
We know that $T$ is path connected and so $H_{0}(T) \cong \mathbb{Z}$. Obviously, $\text{im} (\partial_{3}) = 0$ and the only generator for $\ker (\partial_{2})$ is $-bcd+acd-abd+abc$. So, we get $H_{2}(T) = \ker (\partial_{2})/\text{im} (\partial_{3}) \cong \mathbb{Z}$. Now, I would like to get that $H_{1}(T) = 0$. I checked where the basis elements of $C_{2}(T)$ and $C_{1}(T)$ are sent through $\partial_{2}$ and $\partial_{1}$ respectively and found that $\text{rank} \ \text{im}(\partial_{2}) = \text{rank} \ker (\partial_{1}) = 3$, and using the fact that $\text{im} (\partial_{2}) \subset \ker(\partial_{1})$ we get that $\text{rank} \ H_{1}(T) = 0$ and we're done.
Something that bothers me about the way I did this is that I resorted to using a fair amount of linear algebra to show that $H_{1}(T) = 0$. I imagine that chasing elements would be pretty tedious if this proof was done for the $n$-sphere, which is what I'll do next. Is there a nice result or slick way to go about showing $H_{1}(T) = 0$ (less looking at individual elements)? Thanks!
 A: Yes. The truth is, it is very rare to compute the homology of a space "explicitly" like this. This may work for small, finite examples, but as soon as you have a big space this becomes awful.
Instead, we have many tools (theorems) at our disposal to compute the homology of a space. One of them is the Mayer–Vietoris theorem. It roughly says that if your space $X$ can be decomposed as the union of two open subsets $U$, $V$, and such that you know the homologies $H_*(U)$, $H_*(V)$, and $H_*(U \cap V)$, and if you know how they are related to one another under inclusion, then you can know the homology of $X$.
For the sphere, the standard way to do this is to view $S^n$ as the union of the two hemispheres $U$ and $V$. Then $U$ and $V$ are both contractible (hence $H_*(U)$ and $H_*(V)$ vanish in positive degree), and $U \cap V = S^{n-1}$. So using induction, you find the homology of $S^n$, just knowing the homology of $S^0$, which is very easy.
But it's a good exercise to try and compute the homology "explicitly" as you did. This way, when you learn about the powerful theorems, you can appreciate how much effort they save you.

However, don't be surprised that you had to resort to a lot of linear algebra. As abstract as things become, in the end, we're just computing kernels and images... We can't escape it.
A: Because your triangulation consists explicitly of the shell of a 3-simplex, you ought to be able to argue directly that each element of a basis 2-cycle in that shell is also a 2-face, and obtain the rank image equality immediately. This argument should generalize to show that $H_{n-2}(T)=0$ in n dimensions.
