# Homology groups of unit square with parts removed

I did exercise 19 in Hatcher on page 132 and I was wondering if anyone could tell me if this is right:

19. Compute the homology groups of the subspace of $I \times I$ consisting of the four boundary edges plus all points in the interior whose first coordinate is rational.

My solution:

(i) The homology group of homotopy equivalent spaces is the same.

(ii) if $A \subset X$ is contractible then $X$ is homotopy equivalent to $X/A$

So by (i) and (ii) I can quotient the space by the top boundary union the bottom boundary union any one of the vertical lines. Doing this yields a wedge of circles $X/A = \vee_i S^1$ where $i \in \mathbb{Q} \cap [0,1] - \{ * \}$.

Let $*$ denote the point where all the $S^1$ are attached to each other. Then $(S^1, \{*\})$ is a good pair i.e. there exists a neighbourhood of $\{*\}$ that deformation retracts onto $*$ (in fact every neighbourhood does).

(iii) if $(X,x_0)$ and $(Y,y_0)$ are good pairs then $H_k(X \vee Y) = H_k(X) \oplus H_k(Y)$

Using (iii), $H_k(X/A) = \oplus_i H_k(S^1)$ where $i \in \mathbb{Q} \cap [0,1] - \{ * \}$.

Now:

$k=0$: $H_0(X) = \mathbb{Z}$ because $X$ is path connected.

$k=1$: $H_1(X) = H_1(X/A) = \oplus_i H_1(S^1) = \oplus_i \mathbb{Z}$

$k>1$: $H_k(X) = \oplus_i H_k(S^1) = \oplus_i \{ 0 \}$ because $S^1 = e^0 \cup e^n$ i.e. the union of a zero cell and an $n$ cell. But in a cell complex $C$ of finite dimension $n$, $H_k(C) = \{ 0 \}$ for $k > n$.

There is a problem in the reasoning. When you take $X/A$ and try to view it as a wedge of circles, you have changed the topology. There is a continuous bijection from the wedge of circles to $X/A$, but it is not a homeomorphism: taking a generic point in the wedge of circles, a small neighborhood of that point will only contain points from that single circle, but in $X/A$, the neighborhood will contain points from infinitely many circles. That said, I believe that your final answer is probably correct. – Aaron May 15 '11 at 15:57
@Aaron: in your first comment you argue that the wedge is not homeomorphic to $X/A$ but you also mention that there is a continuous bijection. But that means they are homotopy equivalent and therefore have the same homology groups. Or is this wrong reasoning? Not that it makes any difference but I'd like to know anyway. – Matt N. May 16 '11 at 5:52