Different ways of identifying the boundaries of a disc with two holes I am reading Hatcher's Algebraic Topology and I have encountered a problem with exercise 1.2.13.  The question goes as follows:
Let $Y$ be the space obtained from a disc with two holes by identifying each of the three boundary circles.  There are only two essentially different ways of doing this each with a non-isomorphic fundamental group.
I understand that two different spaces come from whether you reverse the orientation of one of the identifications.  However, my problem is, is there a nice way to realize these distinct spaces as CW complexes, preferably with a diagram?
Other questions regarding this exercise have been asked here and here.
 A: Begin by drawing the disk with two holes cut out. Give this a cell complex structure with three 0-cells, five 1-cells, and one 2-cell. Going from this space to the space $Y,$ you must identify the three 0-cells and two of the five 1-cells. Let $a$ be the 0-cell of $Y$; let $c,$ $\lambda,$ and $\ell$ be the 1-cells of $Y$; and let $P$ be the 2-cell of $Y.$ Our diagram should look similar to the following.

Computing the cellular homology of $Y$ now amounts to simply writing down the matrices of the cellular boundary maps with 0s, 1s, and -1s, and calculating kernels and cokernels.
A: We have basically two options after the identification of any of the two circles : a torus with a hole or a Klein bottle with a hole. After the final identification with the third circle these will correspond to $N = \langle aba^{-1}b^{-1}cb^{\epsilon}c^{-1}\rangle$ or $N = \langle abab^{-1}cb^{\epsilon}c^{-1} \rangle $. After the Abelianization we got $\mathbb{Z}\times \mathbb{Z}$ for $Y$ and $\mathbb{Z}_2\times \mathbb{Z}$ for $Z$. Hence they are not isomorphic.  Yet I am not so sure. Is my answer correct?:) 
