# How to make sense of this induced map on homology

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.

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 $$H_1(K,\Bbb Z_2) \cong\Bbb Z_2^2$$.
• In that case, the map $\rho$ is orientation preserving on the $2$-cell. But again, that doesn't matter, since $H_2(K,\Bbb Z)=0$. @Daven – Lord Shark the Unknown Jun 12 '19 at 18:27