Edit: Nevermind, I believe I found all of my mistakes, so I just need to redo the computation.
I am working on Hatcher 2.2.9 c)
He asks us to find the homology groups of 2-disc with two circle cut out and their clockwise oriented boundaries identified...
I call this space X, and the boundaries A. Then I can create a long exact sequence of homology groups between $X, A, X/A$.
$X/A$ is homotopy equivalent to $S^1 \vee S^1 \vee S^2$, and $A$ is the disjoint union of three circles. Their homology groups are easy to determine, and so I am trying to use the long exact sequence of the pair. I identify $H_n(X,A) = \tilde H_n(X/A)$.
The long (reduced homology) exact sequence of the pair gives me, after some diagram chasing ...
$0 \to H_2(X) \to H_2(X,A) = Z \xrightarrow{\delta} H_1(A) = Z^3 \to H_1(X) \to 0$.
So it's clear that understanding the boundary map will tell me everything about my homology.
I know that $\delta [\alpha] = [\partial \alpha]$, for $\alpha$ a relative 2-cycle in $X$. However, I doesn't seem like any of the 1 boundaries in $A$ are boundaries of relative cycles in $A$. I'm not sure how to argue this more rigorously. Moreover, this gives me a different answer than my other approach - since this would tell me that $H_2(X) = Z$, which seems wrong.
Another approach I am trying -
I am visualizing the space X by gluing tubes instead of making identifications. This gives me a space like a house with two rooms, except that each room is has only a circular column in it, and not the divider. (So it is not contractible.)
My naive intuition says that this space should have no second homology, since each 2-dimensional hole has a hole poked into it. This tells me that $H_2(X) = 0$, so $H_1(X) = Z^2$ from the above diagram.
Still, this feels awfully handwaivy - especially since I am getting two different answers. What should I be doing differently?