# Local isometry between simply connected manifolds

Suppose $D:\tilde{M}\rightarrow N$ is a local diffeomorphism between two simply connected smooth manifolds $\tilde{M}$ and $\tilde{N}$. $D$ is onto. In the case of $D$ being a covering map, it follows that $D$ also must be bijective. Are there counterexamples to this when $D$ is only a local diffeomorphism, but not a covering?

• D is onto. What I meant was if it were possible that D is not injective. Or do I miss something? – Vertex Feb 13 '16 at 0:40
• Sorry, missed the onto condition. – Dan Rust Feb 13 '16 at 0:43

Start with a pretzel, like this one, but projected to the plane, and you get a local diffeomorphism of the open disc to a subset of the plane which is homeomorphic to an open disc with three disjoint closed discs removed.

But now imagine that the projected pretzel gets just a little bit wider along some of the edges of its projected boundary. Doing this carefully, the three holes get filled in, and you still have a local diffeomorphism of the open disc, but now the image is diffeomorphic to the open disc. And it is not one-to-one.

Here's a bit more detail. Let $B$ be the closed unit disc, whose interior is denoted $\text{int}(B)$. The pretzel represents a smooth immersion $f : B \hookrightarrow \mathbb{R}^2$ with the following properties:

1. $f(\partial B)$ is self-transverse,
2. There exist $K \ge 1$ and $D_1,...,D_K \subset D \subset \mathbb{R}^2$ such that $D,D_1,...,D_K$ are all homeomorphic to closed discs and their boundaries are all piecewise smooth, $D_1,...,D_K$ are pairwise disjoint and are contained in the interior of $D$, $$f(B) = D - (\text{int}(D_1) \cup \cdots \cup \text{int}(D_K))$$ and $$f(\text{int}(B)) = \text{int}(D) - (D_1 \cup \cdots \cup D_K)$$
3. Each $\partial D_k$ is a concatenation of two or more smooth segments of $f(\partial B)$.

In the pretzel picture linked above, there are three macroscopic discs $D_1,D_2,D_3$. To be fair, the pretzel picture appears to be have one violation of self-transversality where the two pieces of pretzel twist around each other, and so that portion of the immersion should be perturbed to restore self-transversality. Depending on how that perturbation works out, there may be one more microscopic disc $D_4$.

For each $D_k$, pick $\alpha_k$ to be one of the smooth segments of $f(\partial B)$ comprising $\partial D_k$.

Now, smoothly homotope the immersion $f$ by independently pulling $\alpha_k$ across $D_k$, like the edge of a blanket pulled over a bed. At the end of the homotopy one obtains a smooth immersion $f' : B \to \mathbb{R}^2$ such that $$f'(\text{int}(B)) = f(\text{int}(B)) \cup (D_1 \cup\cdots\cup D_K) = \text{int}(D)$$ which is diffeomorphic to an open disc.

• I don't fully follow this. Could you be a bit more detailed? – Vertex Feb 13 '16 at 1:08
• I added some detail. I only wish I could add a better picture... – Lee Mosher Feb 13 '16 at 2:01