I've attached a model for the Möbius band. The vertices are $a$ and $b$. The ones that eventually get glued together are given the same letter. The edges are labeled $A$, $B$ and $C$. There are two edges labelled $A$, and they have arrows. The must be glued so that the arrows agree, i.e. you need to give a half twist before you glue. The arrows on $B$ and $C$ are only there because we must orient simplices. Finally, $\alpha$ is the one face. I don't know how to put a clockwise arrow around it, so please add one.
To find the homology groups, we must look at the images and the kernels of the boundary maps. Consider the series of maps $0 \to F \to E \to V \to 0$, where $F$ stands for faces, $E$ for edges and $V$ for vertices. In between each is a boundary map.
- Consider $\partial : 0 \to F$. The image and the kernel are both $0$.
- Consider $\partial : F \to E$. We have $\partial \alpha = 2A+B+C$ and so the image is non-empty. There was only one face, so the image is isomorphic to $\mathbb{Z}$. The only face had a non-zero image so the kernel is $0$.
- Consider $\partial : E \to V$. We have $\partial A = b-a$, $\partial B = a-b$ and $\partial C = a-b$. Up to an integer, the images are all $a-b$ and so the image is one dimensional: $\mathbb{Z}$. There were three edges, and the image was one dimensional, so the kernel must be two dimensional: $\mathbb{Z}^2$.
- Consider $\partial : V \to 0$. We have $\partial a = \partial b = 0$ and so the image is $0$. There are two vertices, and so the kernel must be $\mathbb{Z}^2.$
We can put all of this together. The group $H_2(M,\mathbb{Z})$ is given by the quotient of the kernel of $F \to E$ by the image of $0 \to F$, i.e. $0/0 \cong 0$. The group $H_1(M,\mathbb{Z})$ is given by the quotient of the kernel of $E \to V$ by the image of $F \to E$, i.e. $\mathbb{Z}^2/\mathbb{Z} \cong \mathbb{Z}$. The group $H_0(M,\mathbb{Z})$ is given by the quotient of the kernel of $V \to 0$ by the image of $E \to V$, i.e. $\mathbb{Z}^2/\mathbb{Z} \cong \mathbb{Z}$. Hence:
\begin{array}{ccc}
H_2(M,\mathbb{Z}) &\cong& \{0\} \\
H_1(M,\mathbb{Z}) &\cong& \mathbb{Z} \\
H_0(M,\mathbb{Z}) &\cong& \mathbb{Z}
\end{array}
Sure. Those orientations are fine.
Using a single vertex is a bit shaky...more on this later.
Your matrix looks OK at first glance. rows 3-8 and columns 1-5 give a submatrix whose determinant is obviously nonzero, so its rank is at least 5, and as you observe, the last column is a linear combination of earlier ones, so the rank is exactly 5.
Your computation of $H_1$ is OK, but it's not really a great thing to look at, is it? I mean, is there a $Z/2Z$ factor in there? It's hard to tell.
It turns out that $H_1$ is actually $\Bbb Z \oplus \Bbb Z \oplus \Bbb Z \oplus \Bbb Z$, so let's see how to get there.
From the last item in the quotient (the generator $i - h - a$) we can say that in our group, $i$ is the same as $h+a$, so let's just get rid of it:
\begin{align}
H_1(X, \mathbb{Z})
&= \langle a, b, c, d, e, f, g, h, i \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b, i-h-a \rangle \\
&= \langle a, b, c, d, e, f, g, h \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b \rangle \\
\end{align}
and after that, you can do the same with h, and then $g$, then $f$, then $e$:
\begin{align}
H_1(X, \mathbb{Z})
&= \langle a, b, c, d, e, f, g, h \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b \rangle \\
&= \langle a, b, c, d, e, f, g \rangle / \langle e-d-c, d-e+f, c-f+g \rangle \\
&= \langle a, b, c, d, e, f \rangle / \langle e-d-c, d-e+f \rangle \\
&= \langle a, b, c, d, e \rangle / \langle e-d-c\rangle \\
&= \langle a, b, c, d \rangle \\
\end{align}
at which point the group is evidently the free abelian group on four generators. You can probably, at this point, see how to do all those operations by messing with integer row operations on matrices, but I figured I'd do it out without that.
Back to item 2: what you've got here is not actually a simplicial complex, because each 1-simplex should have as boundary a pair of 0-simplices, but your 1-simplexes all have $v - v$ as their boundaries, and that's not allowed in the definitions.
On the other hand, it all worked out OK, right? How can that be? Well, you've kind of computed the cellular homology of the 2-hold torus, and there's a great theorem that says that this gives the same result as the simplicial homology. But do to it right, you really should turn your octagon into a 16-gon, then put a concentric octagon inside, and a vertex at the very center, and then confirm that every triangle, for instance, has three distinct vertices. Your matrix will be much larger...but the operations on it will go nice and fast and very soon you'll get rid of most of the rows and have something no more complicated than the one you have above.
Best Answer
The matrix of $d^1\colon \mathbb Z\{a,b,c,d\}\to\mathbb Z\{v,w\}$ is obtained from your triangulation as $$ \begin{pmatrix} 1 & -1 & 1 & 0 \\ -1 & 1 & -1 & 0\end{pmatrix}, $$ which has rank $1$ and hence the kernel is free of rank $3$. It is generated by $d$, $a+b$ and $a-c$. Hence, we get $$ H_1 = \frac{\langle d,a+b,a-c\rangle}{\langle a+b-d,a-c+d\rangle} = \frac{\langle d,a+b-d,a-c+d\rangle}{\langle a+b-d,a-c+d\rangle} \cong\langle d\rangle\cong\mathbb Z. $$