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Suppose $X$ is the four edges of the unit square, together with the vertical line segments $L_q=\{(q,y):y\in I\}$ for $q\in\mathbb{Q}\cap I$.

I'm trying to compute $H_1(X)$ in particular. I think of setting $U$ to be the top three-fourths of $X$, and $V$ to be the bottom three fourths. These are both open in $X$, and their union is $X$. Moreover, $U$ and $V$ are contractible, and $U\cap V$ is homotopic to a countable set of points, $\mathbb{Q}\cap I$. Mayer-Vietoris implies $$ H_1(U)\oplus H_1(V)\to H_1(X)\to H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X) $$ From the above observations, this gives the sequence $$ 0\to H_1(X)\to\oplus_{q\in\mathbb{Q}\cap I}\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z} $$ since $H_0(X)\simeq\mathbb{Z}$ as it is clearly path connected. This sequence is not short exact unfortunately, but I think at least $$ H_1(X)\simeq\ker(\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}) $$

This map is induced by the sum of the inclusions of $U\cap V$ into $U$ and $V$. So a basis for the kernel of this map would be the elements$[q]-[u]$, where $[q]$ ranges over the generators of $H_0(U\cap V)$ and $[u]$ is a fixed single generator for both $H_0(U)$ and $H_0(V)$, since any class of a point in $U$ and $V$ is a generator as any two points are homologous as $U$ and $V$ are connected. This basis is countable, so I think $H_1(X)\simeq\oplus\mathbb{Z}$.

Is this valid reasoning?

Geovanna Anthony
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    If you consider reduced homology $\tilde H_*$, then the sequence becomes exact, with $Z\oplus Z$ vanishing. That doesn't change the result since $H_1=\tilde H_1$. – Stefan Hamcke May 06 '14 at 11:16
  • Isn't this space homotopy equivalent to a countable infinite bouquet of circles? Just collapse top and the bottom edges to points. On the other hand, MVP computation that you are doing is pretty much clear too. – Moishe Kohan May 06 '14 at 15:38
  • @StefanHamcke If I use MVS for reduced homology, wouldn't $\tilde{H}_0(U\cap V)\simeq\oplus\tilde{H}_0(pt)=0$ as well? Implying $\tilde H_1(X)=0$, but I don't think that's correct. Have I applied it wrong? – Geovanna Anthony May 07 '14 at 00:12
  • Oh, it seems $\tilde{H}_0$ is a little different from $H_0$ and is free abelian of rank one less than the number of path components, so it'd be the same here. – Geovanna Anthony May 07 '14 at 00:18

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