Homology groups of unit square with parts removed -- revisited I had another go at an exercise that I tried some time ago, the question I asked here. Can you tell me if this is right:
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:
Using Mayer Vietoris. The two open sets $U$, $V$ can be the upper three quarters and the lower three quarters of the space $X$. Then $U,V$ are both homotopy equivalent to a point. Their intersection is homotopy equivalent to $\mathbb{Q} \cap [0,1]$. This gives me the following sequence:
$$ 0 \rightarrow H_n(X) \xrightarrow{f} H_{n-1}(U \cap V) \rightarrow 0$$
where $f$ is an isomorphism and $ H_n(U \cap V) =  H_n(\mathbb{Q} \cap [0,1]) = 0$ for $n > 0$ so $H_n(X) = 0$ for $n > 1$ and $\mathbb{Z}$ for $n=0$.
Do you agree with this? Many thanks for your help!
Edit:
So the only remaining case is $n=1$:
$$ H_1(X) = H_0(U \cap V) = H_0(\mathbb{Q} \cap [0,1]) = \oplus_{q \in \mathbb{Q} \cap [0,1]} \mathbb{Z}$$
Is that correct? Many thanks for your help!

What if you begin with $X$ equal to the four boundary edges plus all points in the interior whose first coordinate is $\frac{1}{2}$? Then I would continue with $X$ equal to... whose first coordinate is $\frac{1}{4}, \frac{1}{2}, \frac{3}{4}$. Does the result agree with yours? If not: why?
(Hint: you have written incorrectly the MVS.)
