I've just started studying algebraic topology this semester and I've been given some homework to do. Right now I'm struggling with the following exercise (which has already taken me too much time):
Given integers k and l, let $X(k,l)$ be the CW complex with a single 0-cell, a single 1-cell and two 2-cells attached by maps $f, g : S^1 \to S^1$ which, respectively, wrap the circle around itself k and l times.
I've managed to prove that $X(k,1)$ is homotopy equivalent to $X(k',1)$ for $k \neq k'$. Moreover, I believe I've showed that $X(2,2)$ is Not homotopy equivalent to $X(2,1)$. My question now is:
Is $X(2,2)$ homotopy equivalent to $X(2,0)$??? If I'm not mistaken both have the same fundamental group, so I can't use that. Also, I haven't learned homology, so that's not allowed too. The complex $X(2,0)$ is a wedge between the projective plane and $S^2$ (I believe), and $X(2,2)$ are two projective planes glued...(?) Any hints or suggestions will be much appreciated!