Homology groups of the space obtained from $\mathbb{D}^2$ by first deleting the interiors of two disjoint subdisks in the interior of $\mathbb{D}^2$

Question

Homology groups of the space obtained from $\mathbb{D}^2$ by first deleting the interiors of two disjoint subdisks in the interior of $\mathbb{D}^2$ and then identifying all three resulting boundary circles together via homeomorphisms preserving clockwise orientations of these circles.

This problem has already been posted here Computing the homology groups., but all approaches to solve this problem is using CW-complexes. I would like to know if it is possible to calculate the homology groups of this space using the succession of mayer-vietoris, for this I have tried to take the open U as the space by removing a hole and V a small disk covering the hole, these spaces serve me ? Which would be more convenient?

I would also like to know what this space looks like, that is, what known space this space is homotopically equivalent to. Thank you.

Answer 1

Regarding your first question, I do think that the calculation of the homology of this space can be done with a Mayer-Vietoris sequence, however I do not understand the particular decomposition that you propose so I don't have anything to say about that.

Regarding your second question, I see no reason to expect that this space, which I shall denote $X$, is homotopy equivalent to anything simpler than itself.

There are a lot of things that one can say positively about $X$, using tools of algebraic topology and of combinatorial and geometric group theory. Just to list a few of these things:

• $X$ does not deformation retract to any proper subset of itself;
• $X$ has a negatively curved metric and is therefore an Eilenberg-MacClane space.. A good reference for this material is the book "Metric Spaces of Non-Positive Curvature" by Bridson and Haefliger. In brief outline, one puts a hyperbolic metric on the two holed disc with totally geodesic boundary such that the boundary components are all of the same length, and one glues the boundary components together by isometries. The result is a "locally $\text{CAT}(0)$" metric on $X$, which lifts to a globally $\text{CAT}(0)$ metric on its universal covering space, hence the universal cover is contractible.
• Its fundamental group $\pi_1(X)$ is Gromov hyperbolic (as a consequence of the second point).
• $\pi_1(X)$ has cohomological dimension 2 (also as a consequence of the second point, plus the fact that $X$ is 2-dimensional);
• $\pi_1(X)$ is a one-ended group;
• No finite index subgroup of $\pi_1(X)$ is isomorphic to the fundamental group of a closed surface.

These are not hard to verify if you know these tools, although learning the tools takes some time.

One can conclude from these things that the homotopy type of $X$ is rather complicated relative to the simplicity of its topological description, and for that reason I cannot think that there is any simpler space in the homotopy type of $X$.

Answer 2

I think you can do a Mayer-Vietoris argument by viewing your space as the union of the following two pieces. $A$ is the original disk minus the closures of the two disks that got removed and minus the boundary of the original disk. In other words, it's the space you want minus the circle that resulted from identifying the three boundary circles. $B$ is a little neighborhood of that circle; it's the image in your space of three disjoint annuli in the original disk, one along each boundary circle. So $A\cup B$ is the space you want. You can compute the homology of $A$ because it's just an open disk minus two smaller closed disks in its interior; it's a deformation retract of a doubly punctured disk. You can compute the homology of $B$ because it has a circle as a deformation retract (just let the annuli shrink to the boundary circle(s)). And you can compute the homology of $A\cap B$ because it's just three disjoint open annuli. Finally, you need to figure out the homomorphisms induced in homology by the inclusion maps between these pieces; that's where you'll use the details of how the three boundary circles were identified with each other.