Simplicial homology of a tetrahedron

I'm studying the basics of homology on Nakahara, Geometry, Topology and Physics and I'm trying to work out the (simplicial) homology group of the tetrahedron described by the complex $K=\{0,1,2,3,(01),(02),(03),(12),(13),(23),(012),(013),(023),(123)\}$.

I have already calculated $H_2(K)=\mathbb Z$, but I'm stuck on $H_1(K)$.
I know $H_0(K)=\mathbb Z$, because the tetrahedron is connected, but I don't want to use this result.
I found

$$Z_0(K)=C_0(K)=\{i0+j1+k2+l3\}\\ Z_1(K)=\{i[(01)+(12)+(20)]+j[(02)+(21)+(13)+(30)]+k[(02)+(23)+(30)]\}\\ Z_2(K)=\{i[(012)-(013)+(023)-(123)]\} \\ B_0(K)=\{(-a-b-c)0+(a-d-e)1+(b+d-f)2+(c+e+f)3\} \\ B_1(K)= \{(i+j)(01)+(k-i)(02)+(-j-k)(03)+(i+l)(12)+(j-l)(13)+(k+l)(23)\}\\ B_2(K)= 0$$

Typically, in the previous examples, Nakahara tries to construct a surjective homomorphism from the cycles group to what will be discovered to be the homology group.
This works for $H_0(K)$: define $$f\colon Z_0(K) \to \mathbb Z \qquad\text{such that}\qquad i0+j1+k2+l3 \mapsto i+j+k+l$$ Then, the kernel of $f$ is the set of $0$-chains whose coefficients sum to $0$: this is precisely $B_0(K)$. By the first isomorphism theorem, $Z_0(K)/\ker f\simeq \mathbb Z$.

Now, my question: is it possible to construct a similar argument to conclude that $H_1(K)=0$?
I would say no. Also, I feel that this technique is somewhat tricky... maybe there is a more standard and reliable procedure, like the one described here?

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Better to pick an orientation for each, so you don't include $(20)$ and $(02)$, but write $-(02)$ for $(20)$ in $C_1$. But perhaps that is just shorthand here? – Thomas Andrews Jul 25 '13 at 17:39
Also, is the bondary of $(01)=(1)-(0)$ or $(0)-(1)$? Seems like you are using $(1)-(0)$, but wanted to check. – Thomas Andrews Jul 25 '13 at 17:49
@ThomasAndrews: The boundary of (01) is (1)-(0), right. Also, I wrote (20) only for evidentiating the structure of the generators of $Z_1$. In my calculations, I always used the same orientation ((01), (023) and so on). =) – Andrea Orta Jul 25 '13 at 17:56

It is much easier to just show that $B_1=Z_1$ directly, I think. When $H_i(X)\not\cong 0$ you often have to come up with some clever argument and isomorphisms to figure it out what form it takes, but when $H_i(X)\cong 0$, you just need to show that every element in $Z_i$ is in $B_i$.
You just need to show that each of your generators for $Z_1$ is in $B_1$, which is easy - the only remotely hard one is \begin{align}(02)-(12)+(13)-(03)&=\left(-(01)+(02)-(12)\right) + \left((01)+(13)-(03)\right)\\&=-\partial(012)+\partial(013)\end{align} or maybe some sign adjustment is required there, depending on the definition. (edit: fixed)