TITLE: Computation of Betti numbers of a given space
QUESTION [5 upvotes]: Trying to verify my computations below. I don't have much intuition for homology or the Betti numbers computation. It's a simple case, yet somehow intuitively I'm surprised to get $\beta_1(W) = 3$. 
Is my computation wrong?  


Here is how I construct $W$. 
$W$ is constructed by connecting two (hollow) cylinders by drilling a hole in the two cylinders and connecting them along the hole. 
So you get a space $W = X \cup Y$ of the form
$\quad$$\quad$ $\quad$ 
Where the spaces $X$ and $Y$ are essentially identical 
$\quad$$\quad$ $\quad$  
And $W$ is obtained by gluing $X$ and $Y$ so that their intersection is a circle: 
$\quad$ $\quad$ $\quad$ 


I do not want to use the Mayer-Vietoris sequence. Instead, I want to use Euler characteristics $\chi(W)$ and recover $\beta_i$ from 
$$\chi(W) = \beta_0(W)- \beta_1(W)+\beta_2(W)$$
I notice that $\beta_0(W)=1$ ($W$ has one connected component) and $\beta_2(W) = 0$ (no enclosed void in $W$). Next, I want to use additivity of the Euler characteristic:
$$
\chi(X\cup Y) = \chi(X) + \chi(Y) - \chi(X \cap Y)
$$
Using the fact that $X\cap Y$ is a circle so its Euler characteristic is zero. 

Because $X$ and $Y$ are identical, it is enough to compute $\chi(X)$. Now, the space $X$ is equivalent to a cylinder with an open disk removed. I know that the Euler characteristic of a cylinder is zero. Consequently, removing a disk from a cylinder will yield a space whose Euler characteristic is $-1$, yielding $\chi(X) = \chi(Y) = -1$. 
Overall, I obtain 
$$
\chi(X\cup Y) = - 1 + -1  - 0 = -2,
$$ 
yielding 
$$
-2 = \chi(X\cup Y) =   \chi(W)  = \beta_0(W) - \beta_1(W) + \beta_2(W) = 1 - \beta_1(W) + 0,
$$ 
it follows that $\beta_1(W) = 3$. 

Intuitively I would expect to have a higher $\beta_1$ for $W$. Did I miss something or this calculation is correct? Any comments would be appreciated.

REPLY [5 votes]: That looks reasonable to me.  Another way to think about it is that $W$ is homotopic to a sphere with four holes punched in it.  A loop around each hole is a generator of $H_1(W)$, but the sum of all of those loops is homologous to a loop around all the holes, and that's contractible by going around the other end of the sphere.  So the sum of all four generators is zero in homology.