algebraic-topologycovering-spacesgeneral-topologyklein-bottle
Prove that there is a two-sheeted covering of the Klein bottle by the torus.
OK, so we take the the polygonal representation of the torus and draw a line in the middle as follows:
Then there are two Klein bottles in there, but how do I write down the actual covering map $q:S^1 \times S^1 \to K$?
Consider the fundamental group of the Klein bottle, $K=\langle x,y\mid y^{-1}xy=x^{-1}\rangle$.
You are looking for a finite index subgroup of $K$ (call it $H$), that is abelian and non-normal. Note that any such covering factors through the orientable double cover, corresponding to the subgroup $\langle x, y^2\rangle$.
Consider $H=\langle xy^{-2}, y^6\rangle$. This is abelian, non-normal, and of index $6$ in $K$. It can be described by the diagram below:
To see that this covering is non-normal, here is a picture of the resulting torus (right before the final gluing): the different colored regions are left invariant by the deck transformations (you can also see there are two distinct paths traced by the black arrows, which border the different colors):
I'm not sure where you get $3$ circles from. The model only has two $1$-cells after identifying - horizontal and vertical edges. This coincides with the usual $1$-skeleton for the Klein bottle given by the wedge of two circles.
The picture you should have drawn is a rectangle with $7$ edges; four are vertical, all labelled $a$ and pointing in the same direction, the last three are horizontal, all labelled $b$ and pointing right, left, right as we go from top to bottom of the rectangle.
There are two $2$-cells, both labelled $A$, the top one oriented clockwise, the bottom one oriented anti-clockwise.
Each half is a fundamental domain equal to the usual model of the Klein bottle, and if you remove the central horizontal $1$-cell, gluing the two $2$-cells together along the edge, and forget about identifying the vertical edges with their diagonal partners (so only identifying top-left with top-right, and bottom-left with bottom-right), then this model is the usual model for the torus.
Best Answer
Most topologists would be happy just drawing the diagram you've drawn (though the topologists I know prefer to draw on apples), but if you want to do it explicitly then you can as well.
As you know, the torus $S^1\times S^1$ is homeomorphic to $[0,1]\times [0,1]/\equiv$, where $\equiv$ identifies the edges of the square by $(x,0)\equiv(x,1)$ and $(0,y)\equiv(1,y)$. We also define the Klein bottle to be $K=[0,1]\times [0,1]/\sim$, where $\sim$ identifies the edges of the square by $(x,0)\sim(x,1)$ and $(0,y)\sim(1,1-y)$.
For the torus, we have an explicit continuous surjection $$ \pi:[0,1]\times[0,1]\to S^1\times S^1: (x,y)\mapsto\left(e^{i\pi x},e^{i\pi y}\right) $$ using the standard identification of $S^1$ with the unit circle in the complex plane (more a notational convenience than anything else). Note that we now have: $$ (x_1,y_1)\equiv(x_2,y_2)\Longleftrightarrow \pi(x_1,y_1)=\pi(x_2,y_2) $$ In other words, $\pi$ induces a well-defined homeomorphism $([0,1]\times[0,1]/\equiv)\to S^1\times S^1$.
The next step is to interpret your diagram as a map $[0,1]^2\to[0,1]^2$. This map is then going to induce the two-sheeted covering we want. Explicitly, we have: $$ \phi:[0,1]\times[0,1]\to[0,1]\times[0,1]: (x,y)\mapsto \begin{cases} (2x,y) &\mbox{if } x\le\frac12 \\ (2x-1,1-y) & \mbox{if } x\ge\frac12. \end{cases} $$ Composing this map $\phi$ with the projection $\pi_\sim:[0,1]\times[0,1]\to K$, we get a map $\pi_\sim\circ\phi : [0,1]\times[0,1] \to K$.
We claim that this map $\pi_\sim\circ\phi$ induces a two-to-one covering map $$\psi : S^1 \times S^1 \,\,\, = \,\,\, [0,1]\times[0,1]/\equiv \,\,\,\to\,\,\,[0,1] \times [0,1] / \sim \,\,\,= \,\,\,K $$ Proving that $\psi$ is two-to-one means checking $$ |(\psi^{-1}(\{q\})/\equiv)|=2 $$ for each $q \in K$. And to prove that $\psi$ is a covering map it suffices to check that $\psi$ is a local homeomorphism at $p \in S^1 \times S^1$ (ordinarily this is not enough for checking that something is a covering map, but it suffices when the domain and range are compact manifolds). So one has to check something for the points in $[0,1] \times [0,1]$ that form the equivalence class of the relation $\equiv$ corresponding to $p$: the four corner points; or a pair of opposite side points; or an interior point. Namely one must find neighborhoods of those points which, when fitted together under $\equiv$, form an open neighborhood of $p$ that maps homeomorphically onto an open neighborhood of $q=\psi(p)$. Checking these things is the real content of the proof, and I'll leave them as exercises. It's basically what your diagram is telling you.
Now we have a double-cover by $[0,1]\times[0,1]/\equiv$ of $K$. We already remarked that there is a homeomorphism between $S^1\times S^1$ and $[0,1]\times[0,1]/\equiv$; putting these together gives us a double cover of $K$ by $S^1\times S^1$.
I should stress - there's very little content in any of this, and it really is just a way of making your diagram 'rigorous' in some sense. It's good to work thorugh a few examples like this one explicitly, but you'd be bananas to try and be completely rigorous all the time in topology.