Fundamental group computation

I am trying to compute the homology groups and the fundamental group of the space $X$ obtained as the disjoint union of a circle and a cylinder $S^1\times I$ by attaching the cylinder along its boundary to the circle with the attaching map on $S^1\times\{0\}$ having degree 2 and on $S^1\times\{1\}$ degree 3.

I am thinking of this space as a square in which I identify one pair of opposite sides (call it a), and then I glue the other pair via a map of degree 6 (if I glued with a map of degree 1 I would get the torus).

From this and our knowledge of the behavior of the fundamental group after attaching 2-cells we see that $$\pi_1(X)=\langle a,b|aba^{-1}b^{-6}\rangle$$

Regarding the homology groups, I consider the cellular complex $$0\rightarrow \mathbb{Z}[A] \stackrel{d_2}{\rightarrow} \mathbb{Z}[a]\oplus \mathbb{Z}[b] \stackrel{d_1}{\rightarrow} \mathbb{Z}[p] \rightarrow 0$$ where $p$ is any point, $a$ and $b$ are 1-cells and $A$ is the 2-cell.

It is clear that $d_1=0$ since $X$ is connected and $d_2(A)=a+b-a-6b=-5b$ so we see that $H_1(X)=\mathbb{Z}[a]\oplus\mathbb{Z}_5[b]$ and $H_2(X)=0$ (since $d_2$ is injective).

Am I thinking of $X$ in the right away or am I totally off?

• The complexes defined in your first and second paragraphs are not the same. They do not even have isomorphic fundamental groups. The one described in your first paragraph has fundamental group $\langle a,b \mid a b^2 a^{-1} b^{-3} \rangle$. – Lee Mosher Jul 30 '14 at 20:39
• Okay, I see, thank you. – user54631 Jul 30 '14 at 22:15

But your construction of the CW complex for $X$ is wrong, which led you to the wrong presentation for its fundamental group.
Let $b$ be the circle to which the two components of the boundary of $S^1 \times [0,1]$ are attached. We can assume there is a point $p \in S^1$ such that $p \times 0$ and $p \times 1$ are attached to the same base point in $b$. Let $a$ be the circle obtained from $p \times [0,1]$ by identifying the two points $p \times 0$ and $p \times 1$ to form the base point of $a$. The 1-skeleton of $X$ is the union of $a$ and $b$, identifying their base points, to make a figure 8 graph. When you slit $S^1 \times [0,1]$ along $p \times [0,1]$ you get a square which is attached to that figure 8 as follows: going around the four sides of the square in order, one side is attached to $a$, the next to $b^2$, the next to $a^{-1}$, and the last to $b^{-3}$. Hence the presentation $$\langle a,b \mid a b^2 a^{-1} b^{-3} \rangle$$