# Homology of Torus with two discs glued on the inside

I am trying to figure out the homology of the space $$X$$, where $$X$$ is a $$T^2$$ torus with two discs $$D_1,D_2$$ glued on the inside (see drawing).

What is the CW-complex structure of $$X$$?

I thought I would take one 0-cell $$p$$ and attach four 1-cells $$a,b,\alpha,\beta$$ where $$a,b$$ correspond to the 1-cells of $$T^2$$ in $$X$$. Then I would attach three 2-cells $$A,D_1,D_2$$, where $$A$$ is attached via the identification of the torus and $$D_1,D_2$$ are attached to $$\alpha,\beta$$ respectively. So we get the chain complex $$0\to\mathbb{Z}^3\to\mathbb{Z}^4\to\mathbb{Z}\to0.$$ Is that the right approach?

What are the boundry maps on the chain complex?

My try: The differentials $$d_0,d_1$$ are both zero as there is only one 0-cell. The Space is path connected thus $$H_0(X)=\mathbb{Z}$$. But what about $$d_2: \mathbb{Z}^3\to\mathbb{Z}^4$$? Is there a problem with the approach?

• Can you prove your space is homotopy equivalent to a wedge of two spheres and one circle?
– Pedro
Commented Aug 14, 2021 at 6:58
• Not sure, i could. I could try using the Meier-Vietoris LES with fitting subsets of X that deformation retract to spheres and their intersection to a circle. But I was wondering how to do this with cellular complexes and boundary maps since I’m having difficulties with them. @PedroTamaroff Commented Aug 14, 2021 at 7:11

The quickest way to do this is to follow the comment of @PedroTamaroff: Collapse the disks to points, then stretch one of them to an edge. Then move the end points of the edge to the other collapsed disk, to get the wedge of two spheres and a circle.

Thus we know the answer is: $$\begin{eqnarray*}H_2(X)&=&\mathbb{Z}^2\\ H_1(X)&=&\mathbb{Z}\\H_0(X)&=&\mathbb{Z}\\ \end{eqnarray*}$$

However you may not be familiar with the results that allow you to do this, so lets follow your approach and cellularize $$X$$.

In your picture you have two disks which do not touch. Thus you will need two vertices $$a,b$$ to construct $$X$$ as a cell complex.

Let $$\alpha,\beta,\gamma,\delta$$ be the $$1$$-cells as shown above, and let $$C,D,E,F$$ be the two cells, again as shown above.

We have:

$$\begin{eqnarray*} \partial_1 \alpha&=& b-a\\ \partial_1 \beta&=& a-b\\ \partial_1 \gamma&=& 0\\ \partial_1 \delta&=& 0\\\\ \partial_2 C&=& \gamma\\ \partial_2 D&=& \delta\\ \partial_2 E&=& \gamma+\delta\\ \partial_2 F&=& \gamma+\delta \end{eqnarray*}$$

Thus $$H_0(X)$$ is freely generated by $$a$$, $$H_1(X)$$ is freely generated by $$\alpha+\beta$$ and $$H_2(X)$$ is freely generated by $$E-C-D, F-C-D$$, confirming what we got by the first method.