Fundamental group of a quotient of $S^2 \times I$. Let $X=S^2 \times [0,1]$ and let Y be the quotient space obtained from X identifying each point $x\in S^2 \times \{1\}$ with its antipodal in $S^2 \times \{0\}$. How can I calculate $\pi_1 (Y)$?
All I have been able to do so far is draw some loops in Y, based at the north pole of $S^2 \times \{1\}$ and passing through both poles of $S^2 \times \{1\}$. They do not seem to be homotopic, but I just don't know if there are relations between them, neither if there are more loops.
 A: Note that if you can write down your complex in a cellular form, then the fundamental group will have generators given by the 1-cells and relations given by the 2-cells.
If you prefer using the van Kampen Theorem directly without a full cellular decomposition, I think there is at least one reasonable path forward.  If you break $S^2$ into its equatorial circle and its complement, there is a corresponding decomposition of $S^2 \times I$ given by one piece $A$ being an open neighborhood of $S^1 \times I$ and another being $B = (S^2\setminus S^1) \times I$.  The intersection $A \times B$ is then a deformation retract of $(S^1 \coprod S^1) \times I$.  
This covering is not good for computing the fundamental group of $S^2 \times I$, since the intersection isn't connected, but upon taking the quotient relation the two components of the intersection get glued together.  I claim that the intersection now has the homotopy type of a torus.   I think that the image of $A$ under the gluing is a model for the Klein bottle and the image of $B$ deformation retracts to the circular path passing through the poles.  It remains to determine the corresponding maps on fundamental groups to compute the relevant amalgamated product.
