Computing the homology groups of the twice-punctured disk Let $X$ be space obtained by first removing the the interior of two disjoint closed disks from the unit closed disk in $\mathbb R^{2}$ and then identifying their boundaries clockwise. Compute the homology of this space.
My idea is to do this using cellular homology: We can have cell complex structure on $X$: one $0$-cell, one $1$-cell and one $2$-cell.
Attaching the $2$-cell to the $1$-skeleton by first diving the $S^{1}$ into $3$ parts, then mapping these parts to the $1$-skeleton in the same direction.
Thus the cellular boundary map $d_2$ will be multiplication by $3$ and we have the homology groups $H_{0}(X)=\mathbb Z$ and $H_{1}(X)=\mathbb Z_{3}$ and $H_{i}(X)=0$, otherwise.
Please check the calculations and share some ideas for such questions.
Thanks in advance!
 A: Many thanks to Steve D, user17786, and Dave Hartman for their helpful corrections.
First, I put a cell structure on the twice-punctured disk with 3 0-cells, 5 1-cells, and 1 2-cell:

Note that the boundary of the 2-cell $D$ is
$$d_2D=\alpha+\beta+\gamma-\beta+\delta+\epsilon-\delta=\alpha+\gamma+\epsilon,$$
and that the boundaries of the 1-cells are
$$\begin{align}
d_1\alpha&=0\\
d_1\beta&=y-x\\
d_1\gamma&=0\\
d_1\delta&=z-x\\
d_1\epsilon&=0
\end{align}
$$
Now, we identify $y$ with $z$, and $\gamma$ with $\epsilon$, to produce a cell structure on $X$:

For $X$, the chain groups are
$$\begin{align}
C_0(X)&=\langle x,y\rangle\\
C_1(X)&=\langle \alpha,\beta,\gamma,\delta\rangle\\
C_2(X)&=\langle D\rangle
\end{align}$$
where $D$ is our 2-cell, and we have
$$\begin{align}
d_1\alpha&=0\\
d_1\beta&=y-x\\
d_1\gamma&=0\\
d_1\delta&=y-x
\end{align}
$$
$$d_2D=\alpha+\beta+\gamma-\beta+\delta+\gamma-\delta=\alpha+2\gamma.$$
Thus,
$$H_0(X)=\ker(d_0)/\mathrm{im}(d_1)=\langle x,y\rangle/\langle y-x\rangle=\left\langle\overline{x}\right\rangle\cong\mathbb{Z}$$
$$H_1(X)=\ker(d_1)/\mathrm{im}(d_2)=\langle \alpha,\gamma,\beta-\delta\rangle/\langle \alpha+2\gamma\rangle=\left\langle\overline{\gamma},\overline{\beta-\delta}\right\rangle\cong\mathbb{Z}^2$$
$$H_2(X)=\ker(d_2)/\mathrm{im}(d_3)=0/0\cong 0.$$
A: $X$ as mentioned earlier is a punctured Klein Bottle, hence deformation retracts onto wedge of two circles. So $H_1(X) = \mathbb{Z} \oplus \mathbb{Z}$.
