Computing (Simplicial) Homology of 2 holed torus Here's the $\Delta$ complex. I would like to compute the simplicial homology for this complex ($\mathbb{Z}$-coefficients). There's several steps I'm not sure about. To start off with, I'm confused with orientations in $\Delta$-complexes; are the orientations marked in my picture (for the triangles) correct? Beyond that, here's what I have so far:
$C_0 = \langle v \rangle$ ($v$ is the vertex)
$C_1 = \langle a, b, c, d, e, f, g, h, i \rangle$
$C_2 = \langle A, B , C, D, E, F \rangle$
Our chain complex is:
$$...0 \xrightarrow{\partial_3} C_2 \xrightarrow{\partial_2} C_1 \xrightarrow{\partial_1} C_0 \xrightarrow{\partial_0} 0$$
$im(\partial_0) = 0, ker(\partial_0) = C_0$
$im(\partial_1) = 0$ since there's only one vertex (all generators begin and end with $v$) and thus $ker(\partial_1) = C_1$
$\partial_2$ is something I'm wary of. $\partial_2(A) = e-c-d$, etc, so I thought of expressing $\partial_2$ in terms of a $9\times6$ matrix:
$\begin{bmatrix} 0&0&0&0&-1&1 \\ 0&0&0&-1&0&1\\-1&0&1&0&0&0 \\-1&1&0&0&0&0\\1&-1&0&0&0&0\\0&1&-1&0&0&0\\0&0&1&-1&0&0\\0&0&0&1&-1&0\\0&0&0&0&1&-1 \end{bmatrix}$
and then simply putting it through an online calculator to get the null space and column space. Is this a valid technique (I know $\mathbb{Z}$ technically isn't a field/vector space but the methods should translate)? Is there a simpler way to do this? It feels rather strange to be using a calculator in algebraic topology.
Anyway, this yielded:
$ker(\partial_2) = \langle A+B+C+D+E+F \rangle$ (which I suppose was obvious)
and $im(\partial_2) = \langle e-d-c, d-e+f,c-f+g, h-g-b, i-h-a \rangle$ (basically the last column is the only one linearly dependent on the rest)
Finally $im(\partial_3) = 0$
So, finally, our homology groups are:
$H_0(X, \mathbb{Z}) = C_0 \cong \mathbb{Z}$
$H_1(X, \mathbb{Z}) = \langle a, b, c, d, e, f, g, h, i \rangle / <e-d-c, d-e+f, c-f+g, h-g-b, i-h-a \rangle$
$H_2(X, \mathbb{Z}) = \langle A+B+C+D+E+F \rangle \cong \mathbb{Z}$ again
This leaves me with two final questions: Does this seem right? And how do I simplify $H_1(X, \mathbb{Z})$!?
TLDR: My main concerns are about the following: Are the orientations of the triangles marked out in the picture correct/consistent? How does one simplify the $H_1$ I obtained (assuming it's correct)? And I would appreciate if any errors could be pointed out.
 A: *

*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.
