How to embed the torus into the Klein bottle?

Question

Is there a continuous map of the torus into the Klein bottle? Can one do this so that it is locally a homeomorphism (or a complete embedding)?

My idea is to take the square $[-1,2] \times [-1,1]$ and identify $(-1,y) \sim (1,y)$ and $(x,-1) \sim (x,1)$ to create the torus. To create the Klein bottle, I take the square $[-1,1] \times [-1,1]$ and identify $(-1,y) \sim (1,y)$ and $(x,-1) \sim (-x,1)$.

Because of the difference in identification between the top/bottom torus and the top/bottom of the Klein bottle squares, I know there must be a flip involved. My idea is to use affine linear transformations. I tried various points along the bottom line of the square but everything I tried either broke the continuity of the map or did not meet the orientation of the Klein bottle. Any ideas on what to try?

As for is there an embedding, there cannot be a complete embedding as the Klein bottle is not orientable but the torus is not. However, I feel that the torus can be mapped into the Klein bottle so that it is a local homeomorphism or at least "piecewise" a local homeomorphism. I feel this can be done as above using affine linear transformations.

Answer 1

The answers above give a good explanation of a double cover from the torus to the Klein bottle which answers your questions about finding a continuous map which is a local homeomorhpism. I wanted to add that the torus cannot topologically embed in the Klein bottle $K$, which addresses the title.

More generally, if $M$ and $N$ are both closed connected topological $n$-manifolds which are not homeomorphic, then $M$ cannot embed into $N$. In fact, even more precisely, if $M$ is a closed topological $n$-manifold and $N$ is a connected $n$-manifold, then any embedding $f:M\rightarrow N$ is a homeomorphism

Here's the idea of the proof. If $f:M\rightarrow N$ is an embedding, then it is, by definition, continuous and injective. A continuous injective map from a compact space ($M$) to a Hausdorff space ($N$) is a homeomorphism onto its image. Thus, we need only show the image is all of $N$.

Well, $f$ is an open map using invariance of domain and also $f$ is a closed map since $M$ is compact. Thus, $f(M)$ is a clopen subset of $N$ and thus, since $N$ is connected, $f(M) = N$.

Answer 2

As aes and janmarqz have suggested, the torus double covers the Klein bottle. As for the map, look at their fundamental polygons. What happens when you start gluing Klein bottles side to side? Don't get too caught up starting out with the explicit map. Rather, start with the geometric intuition, then once you've convinced yourself where this map comes from, try making the map explicit.

Reference: http://www.math.cornell.edu/~hatcher/AT/ATch1.3rev.pdf

Edit: maybe looking at this might help with your question: Why this map is a covering map?

Answer 3

To do it with linear transformations, first note that the torus is the quotient of the plane by the action of a full lattice of translations - for instance the standard lattice of vectors with integer coordinates. So its fundamental group is $\mathbb{Z}\times\mathbb{Z}$.

If the torus is a two fold cover of the Klein bottle then the fundamental group of the Klein bottle must contain the lattice plus an additional transformation of the plane that has the following properties

1. Its square is in the lattice

2. Its matrix part has integer coefficients.

3. The group generated by the lattice and this extra affine transformation must be torsion free (since it is a group of covering transformations of the Klein bottle). In particular the translational part of the extra affine transformation can not be zero.

So the fundamental group of the Klein bottle must be a torsion free group extension of $\mathbb{Z}_2$ by $\mathbb{Z}\times\mathbb{Z}$.

$0\longrightarrow\mathbb{Z}\times\mathbb{Z}\longrightarrow K\longrightarrow\mathbb{Z}_2\longrightarrow0$

in which the action of $\mathbb{Z}_2$ (by conjugation) on $\mathbb{Z}\times\mathbb{Z}$ has determinant $-1$ .

It is not hard to find an affine transformation that works. In fact you can choose its translational part to be half a lattice point.