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I need to find homology groups for the following simplicial complex:

$RP^2$ # $RP^2$ # $\Delta$ # $\Delta$

How to do it? If I am not mistaken, $RP^2$ is not orientable - so we cannot just sum the groups.

How to solve the problem?

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What is $\Delta$? – Gerben May 10 '11 at 19:26
You can use the Mayer Vietoris sequence. Not sure what you mean by "not orientable --> we cannot sum the groups". Summing the groups generally works only if the intersection has trivial homology. – Alon Amit May 10 '11 at 19:35
$\Delta$ is a triangle/disk. 2-dimensional simplex. – user10732 May 10 '11 at 19:38

Taking the connected sum of a surface and the 2-disk is the same as puncturing the surface. So your surface is the Klein bottle twice punctured. Its fundamental group, by van Kampen, is free on three generators, so its first homology is $\mathbb{Z}^3$. It has non-empty boundary, so its second homology is trivial.

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Indeed, the surface is homotopy equivalent to a wedge of three circles! – Grumpy Parsnip May 10 '11 at 20:46

Edit: Now that I see $\Delta$ is a disk, the surface in question has a nonempty boundary, and my answer below assumed the boundary was empty.

I'm not sure what $\Delta$ is, but if you can write your surface as an identification space of a polygon, it's easy to construct a cell complex, and therefore a chain complex that will calculate the homology. For example, $RP^2\# RP^2\# T$, where $T$ is a torus, is the quotient of a octagon where the boundary edges are glued by the pattern $aabbcdc^{-1}d^{-1}$. The chain complex has one generator in degree $0$, $4$ in degree 1 and $1$ in degree $2$. The boundary operator is zero on edges and is equal to $2a+2b$ on the unique generator in degree $2$. So $\partial_2$ is injective, implying $H_2=0$. Also $H_1=\mathbb Z^4/im(\partial_2)$, which you can verify is $\mathbb Z_2\oplus \mathbb Z\oplus \mathbb Z\oplus \mathbb Z$.

Alternatively you can appeal to the classification of surfaces, and figure out whether your surface is a connected sum of some number of projective planes (in the nonorientable case) or tori (in the orientable case), and just look up or figure out the answer for these examples.

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Could you please explain to me how one could identify a surface to an identification of space of polygons in general? – El Moro May 10 '11 at 20:35
@El Moro: There's probably a good place on the web to find this. I know that Massey's book on algebraic topology has a nice treatment. Basically, you cut along simple closed curves to get a punctured disk, and then cut along some properly embedded arcs to get a polygon. Gluing the cuts back together is the identification space topology. – Grumpy Parsnip May 10 '11 at 20:43

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