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I'm trying to find the homology groups of this space X below.

The easiest part is $H_0(X)=\mathbb Z$, because $X$ is connected. In order to find $H_1(X)$ I think the easiest method is finding its fundamental group, but I don't know the fundamental group of this space, it's my first problem.

My second problem is to find $H_2(X)$, I don't know how to triangulate this picture and I don't know how to use Mayer-Vietoris in this case.

enter image description here

I really need help here, any help or comment would be appreciated.

Thanks a lot

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$H_0(X) = \mathbb{Z}$ since it's connected.

$H_2(X) = \mathbb{Z}^2$ since there are two non-contractable spheres

$H_1(X) = \mathbb{Z}$ since the torus has two non-contractible curves, but now one can be contracted along the sphere.


Let $a$ be the torus, and $b\cup b' = S^2$ be the upper and lower hemispheres.

$\partial a = a_1 + a_2 - a_1 - a_2 = 0$

Let $a_1 = \mathbb{T}^2 \cap S^2$ be the meridian of the sphere (as well as the torus)

$\partial b = a_1 =\partial b'$

Let $a_{12}$ be a point of $a_1 =\mathrm{torus}\cap \mathrm{sphere}$:

$\partial a_1 = \partial a_2 = \partial b_1 =a_{12} - a_{12}=0$


$H_0 = \ker\partial / \mathrm{im} \,\partial = \mathrm{span}\langle a_{12}\rangle = \mathbb{Z}$

$H_1 = \ker\partial / \mathrm{im} \,\partial = \mathrm{span}\langle a_1, a_2\rangle / \mathrm{span}\langle a_1\rangle = \mathbb{Z}$

$H_2 = \ker\partial = \mathrm{span}\langle a,b-b'\rangle = \mathbb{Z}^2$



You may also observe the meridian circle $\mathrm{sphere} \cap \mathrm{torus}$ is contractible to a point.

In Hatcher Chapter 0, example 0.8 (page 12) shows us the sphere with two points identified is the homotopy equivalent to the wedge of a sphere and a circle.

$$ S^2 \simeq S^2 \vee S^1$$

For your example, $ \mathbb{T}^2 \cup_{S^1} S^2 \simeq S^2 \vee S^2 \vee S^1$.

Their homology groups will also be the same. $H_0 = \mathbb{Z}, H_1 = \mathbb{Z}, H_2 = \mathbb{Z}^2$.

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  • $\begingroup$ we don't have to triangulate this space before? thank you for your answer :-) $\endgroup$ – user42912 Jan 24 '13 at 22:46
  • $\begingroup$ Here's a nice book on homology of surfaces and graph theory maths.ed.ac.uk/~aar/papers/giblin.pdf , except your example is two surfaces attached along a circle. $\endgroup$ – cactus314 Jan 25 '13 at 0:12
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    $\begingroup$ I'm using a CW complex instead of a triangulation to make the calculation easier. Were any steps unclear? $\endgroup$ – cactus314 Jan 25 '13 at 0:13

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