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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!

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up vote 1 down vote accepted

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.)

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Thanks! That was actually just a typo, thanks for pointing it out! – Rudy the Reindeer Sep 6 '11 at 9:34
If I take $X$ to be the boundary plus the line in the middle at $\frac{1}{2}$ this is homotopy equivalent to a wedge of two copies of $S^1$, so $H_n(X) = H_n(S^1) \oplus H_n(S^1) = 0$ for $n > 1$ and $\mathbb{Z} \oplus \mathbb{Z}$ for $n=1$ and $\mathbb{Z}$ for $n=0$. – Rudy the Reindeer Sep 6 '11 at 9:38
So in that case, it's only different to my other solution in the case $n=1$... – Rudy the Reindeer Sep 6 '11 at 9:38
$f$ is not an isomorphism for $n=1$: there is no zero following $H_0(U\cap V)$. Write down the groups that follow and see what happens with $f$ for $n=1$ in the examples I suggested. – a.r. Sep 7 '11 at 1:11

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