Consider the representation of the Klein bottle $K$ as a square $[0, 1]$ $X$ $[0, 1]$, with the top and bottom edges identified by $(x, 1)$~$(x, 0)$, and the left and right edges identified by $(0, x)$ ~ $(1, 1-x)$. We have $\mathbb{Z}_2$ homology groups given by $\mathbb{Z}_2$, $\mathbb{Z}_2 \bigoplus \mathbb{Z}_2$, and $\mathbb{Z}_2$ in dimensions $0$, $1$ and $2$ respectively.
I want to find the map induced on homology by the map that rotates $K$ $180$ degrees about its midpoint. I calculated the homology by using a CW-complex, and so I am confused as to how to interpret the rotation. I have a hunch that the map on homology will be the identity map but I have little to back it up.
We have a class $\alpha$ that corresponds to a simplex that maps to the top edge of the square. When rotated, it maps to the bottom edge of the square in the opposite direction. This induces a map $1 \rightarrow -1 = 1 \in \mathbb{Z}_2$ in homology, since the composition of $\alpha$ with the reversed chain of $\alpha$ is clearly a boundary. How can we interpret the action of the rotation on the single $2$-cell of $K$? I have no idea how to deal with this.
Best Answer
You can give this a CW-complex structure with one $0$-cell, two $1$-cells and a $2$-cell. Over $\Bbb Z_2$ the boundary maps are all zero, so the homology groups over $\Bbb Z_2$ have dimensions $1$, $2$ and $1$.
You needn't worry about the effect of your map $\rho$ on your two-cell. It's an automorphism, so can only act trivially on $H_2(K,\Bbb Z_2)\cong\Bbb Z_2$.
The only thing you care about is the action on $H_1$. But that is generated by the homology classes of the upper and left edges of your square. But each of these is taken by $\rho$ either to itself or its negative. Over $\Bbb Z_2$ that is the same homology class. So $\rho$ acts trivially on $H_1(K,\Bbb Z_2) \cong\Bbb Z_2^2$.