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.
Best Answer
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, $f(M) = N$.