A quotient space of a closed annulus is homeomorphic to a Klein bottle I was told that a quotient space of a closed annulus centered at the origin obtained with a relation $x \sim - x$ for $x$ in the boundary is homeomorphic to a Klein bottle, which is a connected sum of two real projective planes.
So if my quotient map is $q$ and $U$ is the annulus without the inner boundary and $V$ is the annulus without the outer boundary, I was told $q(U)$ and $q(V)$ are projective planes missing a point, where the missing point is $q(\text{missing boundary})$.
However, I do not see this directly, and not sure how to show this.
Also, once I show this, I can see what is the fundamental group of the space by the consequence of the Van Kampen theorem, but what would be the generators of the fundamental group?
 A: If you can appeal to the classification of connected, compact surfaces, then you can look at a cell decomposition: vertices are the 4 (really only 2) points where the boundaries meet the $x$-axis, edges are the 4 arcs (really only 2) plus the 2 segments along the $x$-axis, and 2-cells are the upper and lower regions these enclose.  You can compute the Euler characteristic and can directly find an embedded copy of a Möbius band.
For the statements about the quotient maps, it might be easier to understand what's going on if you replace "are" by "are homeomorphic to".  A surface minus a point is homeomorphic to a surface minus a closed disk centered at this point.  This point of view reveals the boundary.
For the fundamental group, since you are thinking about the Van Kampen theorem, a natural choice would be to look at a generator for the fundamental group of the intersection of $U$ and $V$ and generators for the fundamental groups of $U$ and $V$.  This will give three generators, but you can eliminate one using the relations that arise from Van Kampen's Theorem.  A little bit of care is needed in choosing a base point.
