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I want to try to find out a solution of this problem using Mayer-Vietoris sequence. I want to compute the homology groups of space X obtained from $D^2$ by first deleting the interior of two disjoint subdisks in the interior of $D^2$ and then identifying all three resuting boundaries together via homeo that preserve the clockwise orientation of these circles (basically 2.2.9 in Hatcher).

So I know a lot of solutions use cellular homology but this is not covered in my exam so I might not be able to use it. Here is my approach using the idea of one of the answers that I found here https://math.stackexchange.com/a/3304215/752801.

A deformation retraction onto the wedge of 2 circles so we can find the homology group of it. B deformation retracts onto the common boundary so it has the homology group of a circle. Now, as the answer mentioned, $A \cap B$ is 3 open disjoint annulus. What is the homology groups of $A \cap B$ ? Most probably that $H_2(A \cap B)=0.$

Here's the reduced Mayer-Vietoris sequence that we are going to get: $$ 0 \mapsto H_2(X) \mapsto H_1(A\cap B) \mapsto (\mathbb{Z} \times \mathbb{Z}) \times \mathbb{Z}) \mapsto H_1(X) \mapsto 0.$$

My question is how can we find $H_1(A \cap B)$ and once we found it, how can we find the map $H_1(A \cap B) \mapsto (\mathbb{Z} \times \mathbb{Z}) \times \mathbb{Z})$.

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$A \cap B$ is three open disjoint annuli. Each annulus is homotopy-equivalent to a circle. So $H_1(A \cap B) = \mathbb Z^3$. Each $\mathbb Z$ in this direct sum is generated by a 1-cycle that wraps one of the three annuli. Let's give these generators names:

  • Let's say that $\alpha = (1, 0, 0) \in \mathbb Z^3$ wraps the big annulus near the outer boundary of the main disk in a clockwise direction.
  • Let's say that $\beta = (0, 1, 0) \in \mathbb Z^3$ wraps the little annulus encircling the left-hand cut-out disk in a clockwise direction.
  • Let's say that $\gamma = (0, 0, 1) \in \mathbb Z^3$ wraps the little annulus encircling the right-hand cut-out disk in a clockwise direction.

$A$ is the original disk minus the closures of the two circles cut out minus the boundary of the original disk. It does indeed look like a figure-of-eight, which means that $H_1(A) = \mathbb Z^2$.

  • Let's say that $\eta = (1, 0) \in \mathbb Z^2$ is the generator that encircles the left-hand cut-out disk in a clockwise direction
  • Let's say that $\zeta = (0, 1) \in \mathbb Z^2$ is the generator that encircles the right-hand cut-out disk in a clockwise direction.
  • Thus a loop that runs encircles both cut-outs in a clockwise direction is represented by the class $\eta + \zeta = (1, 1) \in \mathbb Z^2$.

$B$ is a little neighbourhood around the circle formed by identifying the three boundary circles. So $H_1(B) = \mathbb Z$.

  • Let's say that $\epsilon = 1 \in \mathbb Z$ is the generator that wraps this circle in a clockwise direction.

In the Mayer-Vietoris sequence $$ 0 \to H_2 (X) \to H_1(A \cap B) \to H_1(A) \oplus H_1 (B) \to H_1(X) \to 0, $$ the map $H_1(A \cap B) \to H_1(A) \oplus H_1 (B)$ is induced by the inclusion maps $i : A \cap B \hookrightarrow A$ and $j : A \cap B \hookrightarrow B$.

We can visualise how these inclusion maps act on the generators of the various first homology groups.

The map $i_\star : H_1(A \cap B) \to H_1 (A)$ sends $$\alpha \mapsto \eta + \zeta, \ \ \beta \mapsto \eta, \ \ \gamma \mapsto \zeta.$$

The map $j_\star : H_1(A \cap B) \to H_1 (B)$ sends $$ \alpha \mapsto \epsilon, \ \ \beta \mapsto \epsilon, \ \ \gamma \mapsto \epsilon.$$

I encourage you to draw the pictures of these cycles and convince yourself of these inclusions!

This should be enough detail for you to figure out the kernel and cokernel of the map $H_1(A \cap B) \to H_1(A) \oplus H_1 (B)$!

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  • $\begingroup$ Thank you so much! Now, as I can see this map is injective so H_2(X)={0} and H_1(X)=<\eta, \xi, \epsilon>/<\eta + \xi + \epsilon, \eta+ \epsilon, \xi + \epsilon>. How can we see that this is going to be Z^2? $\endgroup$
    – user752801
    Commented Jul 29, 2020 at 15:23
  • $\begingroup$ @m96 I don't think $H_1(X)$ is $\mathbb Z^2$. I think $H_1(X) = 0$. Consider the subgroup $G = \langle \eta + \xi + \epsilon, \eta + \epsilon, \xi +\epsilon \rangle$. We can see that $\xi = (\eta + \xi + \epsilon) - (\eta + \epsilon) \in G$, and $\eta = (\eta + \xi + \epsilon) - (\xi + \epsilon) \in G$, and $\epsilon = (\eta + \xi + \epsilon) - \eta - \xi \in G$. So $G = (\eta, \xi, \epsilon) = \mathbb Z^2$. Hence $H_1(X) = \mathbb Z^2 / G = 0$. $\endgroup$
    – Kenny Wong
    Commented Jul 29, 2020 at 18:47

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