# Prove that there is a two-sheeted covering of the Klein bottle by the torus

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 $q:S^1 \times S^1 \to K$?

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.

• In spite of the upvotes and the acceptance, I'm afraid this answer is not correct. See my answer below. Oct 5 '18 at 12:41
• To augment the comment of @LukasLewark, I have corrected this answer. See my edit summary for details of what was done. Jun 16 '19 at 18:57
• @LeeMosher $\phi$ is not continuous, nor it is well-defined at $(1/2,y)$ for $y\not=1/2$ Mar 13 '20 at 1:37

It is my impression that the Banana-image can be understood to give a correct covering - however, the map $$f:T\to K$$ from the Torus $$T$$ to the Klein bottle $$X$$ described in the accepted answer is not a covering map, but merely a continuous map such that all points in the Klein bottle have exactly two preimages.

Note that this is necessary, but not a sufficient condition for $$f$$ to be a covering. For $$f$$ to be a double covering map, every point $$x\in K$$ must have an open neighborhood $$U$$ such that there exist a homeomorphism $$h: U\times \{0,1\}\to f^{-1}(U)$$ with (*) $$f\circ h|_{U\times\{i\}}$$ a homeomorphism from $$U\times\{i\}$$ to $$U$$ for $$i\in\{1,2\}$$.

For the given map, everything is satisfied except condition (*). Thus it is a quite cool example of something that is nearly a covering map!

If you fold along the middle line, then a point $$x\in K$$ that's on the image of the fold line does have a small open neighborhood $$U$$ such that $$f^{-1}(U)$$ is homeomorphic to two copies of $$U$$. However, restricted to one of these copies, $$f$$ is not a homeomorphism, because it is two-to-one for points in $$U$$ not lying on the fold.

It's quite subtle! Another way to see that this is not a covering map goes as follows. If it were, because it is a double cover, there would be exactly one non-identical deck transformation $$g: T\to T$$, which maps every $$y\in T$$ to the unique $$g(y) \neq y$$ with $$f(g(y)) = f(y)$$. For $$y$$ in the interior of the square, and not on the fold, $$g(y)$$ is the reflection of $$y$$ across the fold. So when you pick a sequence of $$y_i$$'s approaching a point $$y$$ on the fold, then $$y_i$$ and $$g(y_i)$$ approach one another. By continuity of $$g$$, we must have $$g(y) = y$$! But for $$y$$ on the fold, $$g(y)$$ is on the boundary of the square.

If you have trouble seeing the mistake, here's a simpler map that fail to be a covering in the same way: let $$k: S^1\to S^1$$ be defined as $$k(z) = z^2$$ for $$\text{Im} z\geq 0$$, and $$k(z) = z^{-2}$$ for $$\text{Im} z \leq 0$$. Every point has two preimages, but it's not a covering...

Finally, here's how to construct a map $$m: T\to K$$ that is a double covering map. Instead of folding along the line (identifying $$(x,y)$$ with $$(1-x, y)$$), take the glide reflection (identifying $$(x,y)$$ with $$(1-x, y + \frac{1}{2} \pmod{1})$$. I'll let you work out the details...

Note that this fits in well with Amitai Yuval's answer.

I realize my answer is some years late, but I didn't want to leave this uncommented.

One way to define the torus is as the quotient $T=\mathbb{R}^2/\mathbb{Z}^2$, where $\mathbb{Z}^2$ acts on $\mathbb{R}^2$ by translations. Thus, for any space $X$, specifying a map $f:T\to X$ is equivalent to specifying a map $\overline{f}:\mathbb{R}^2\to X$, which satisfies $\overline{f}\circ g=\overline{f}$ for any $g\in\mathbb{Z}^2$.

One way to define the Klein bottle is as the quotient $K=\mathbb{R}^2/G$, where $G$ is a group of symmetries which contains $\mathbb{Z}^2$. Thus, the natural projection $\pi:\mathbb{R}^2\to K$ descends to the desired double cover $p:T\to K$.

• what would be that $G$ exactly? Sep 9 '19 at 1:33
• @rmdmc89 It's the group of homeomorphisms from $\mathbb{C}$ to $\mathbb{C}$ generated by $a: z\mapsto z+i$ and $b: z\mapsto \overline{z}+\frac{1}{2}+i$. See Exercise 17.9 parts (e), (f), (g) and (i) in Czes Kosniowski's $\textit{A First Course in Algebraic Topology}$.
– WLOG
May 7 '20 at 1:52

I would like to give my formula. Let us think of $$T^2$$ as $$S^1\times S^1$$ and $$S^1$$ as the unit circle in $$\mathbb{C}$$. Let $$f$$ act on $$T^2$$ by $$f(z,w)=(-z, \bar{w})$$, then $$f^2$$ is the identity. $$f(z,w)\sim(z,w)$$ gives an equivalence relation on $$T^2$$ and $$T^2/\sim$$ provides the desired two-fold covering map from $$T^2$$ to the Klein bottle, as illustraded in the figure by OP.