# Why isn't this space a surface?

Let $X$ be the space obtained from an annulus by identifying antipodal points on the outer circle, and points $2\pi/3$ apart on the inner circle. I've computed that $H_1(X) = \mathbb{Z}$ using cellular homology. According to this page, if Y is a compact non-orientable surface, then $H_1(Y)$ has torsion. Since $X$ is a quotient of a compact space, it's compact. So, assuming my calculation of $H_1$ is correct, I conclude that X cannot be a surface. Is there an elementary way to see this?

As a side note, if we take an annulus and identify antipodal points on both circles, we get the connected sum of two projective planes, which is the Klein bottle $K$. In this case, $H_1(K) = \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (and $K$ is a surface).

The points on the inner circle of the annulus have neighborhoods which look like $Y×(0,1)$ where $Y$ is a space that looks like the letter Y (sans serif).