Local isometry between simply connected manifolds

Question

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?

Answer 1

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.

Answer 2

Everything here is motivated by Lee Mosher's post.

I think of the "pretzel" he mentions as flattened pretzel dough: Start with with a long thin open rectangular strip of TWO-dimensional dough. Then mold it, passing over and under in ${\mathbb R}^3$, into a pretzel-shaped configuration, but don't press any part of the dough against any other part. The result is NOT a three-dimensional pretzel (handlebody), nor is it the boundary of a three-dimensional pretzel (higher genus surface). It is a subset of ${\mathbb R}^3$ that, with its relative topology inherited from ${\mathbb R}^3$, is homeomorphic to an open rectangle in ${\mathbb R}^2$.

I think of it as living above the $xy$-plane, and, with sun directly overhead, it casts a shadow on the $xy$-plane. The shape of the dough results in three sunny holes inside the shadow. Then the dough is stretched and pulled in such a way that the shadow loses its holes, and becomes simply-connected, even though, up above, no part of the dough gets pressed against any other part, and so the dough continues to be homeomorphic to an open rectangle in ${\mathbb R}^2$.

In the following, we describe a variant of this construction.

For all $t\in{\mathbb R}$, let $p_t:=(\cos t,\sin t,t)$, so $p$ is a parametric spiral in ${\mathbb R}^3$.

For all $t\in{\mathbb R}$, let $q_t:=(0,0,t)$, so $q$ parametrizes the $z$-axis.

Let $D:=\{(x,y)\in{\mathbb R}^2\,|\,x^2+y^2<1\}$, so $D$ is the open unit-radius disk in the $xy$-plane centered at $(0,0)$. Let $D^\times:=D\backslash\{(0,0)\}$, so $D^\times$ is the result of puncturing $D$ at its center.

Define $\Pi:{\mathbb R}^3\to{\mathbb R}^2$ by: $\forall x,y,z\in{\mathbb R}$, $\Pi(x,y,z)=(x,y)$, so $\Pi$ is orthogonal projection onto the $xy$-plane.

For any distinct $v,w\in{\mathbb R}^3$, let $(v|w)$ denote the line segment from $v$ to $w$ defined by $(v|w)=\{(1-s)v+sw\,|\,0<s<1\}$.

Let $R$ be the union, over $0<t<2\pi$, of $(p_t|q_t)$.

I think of $R$ as being a spiral ramp whose points are all above the points of $D^\times$. It lives between the plane $z=0$ and the plane $z=2\pi$. A driver who drives a car up this ramp would be spiraling once around the vertical line segment $\{(0,0,z)\,|\,0<z<2\pi\}$.

For any distinct $v,w\in{\mathbb R}^2$, let $(v|w)$ denote the line segment from $v$ to $w$ defined by $(v|w)=\{(1-s)v+sw\,|\,0<s<1\}$.

Let $a:=(0,0)$, $b:=(1,0)$. Let $L:=(a|b)$. Then $L\subseteq D^\times$.

Every subset of ${\mathbb R}^2$ is given its relative topology, inherited from the standard topology on ${\mathbb R}^2$.

Every subset of ${\mathbb R}^3$ is given its relative topology, inherited from the standard topology on ${\mathbb R}^3$.

The map $\Pi|R$ is a homeomorphism from $R$ onto $D^\times\backslash L$.

For any subset $A$ of ${\mathbb R}^2$, for any $v\in{\mathbb R}^2$, let $A+v:=\{a+v\,|\,a\in A\}$.

Let $F:=D+(0,0.5)$. Let $F^\times:=D^\times+(0,0.5)$.

Then $F$ is the open unit-radius disk in the $xy$-plane, centered at $(0,0.5)$, while $F^\times$ is the result of puncturing $F$ at its center.

For any subset $A$ of ${\mathbb R}^3$, for any $v\in{\mathbb R}^3$, let $A+v:=\{a+v\,|\,a\in A\}$.

Let $T:=R+(0,0.5,2\pi)$.

I think of $T$ as being a spiral ramp whose points are all above the points of $F^\times$. It lives between the plane $z=2\pi$ and the plane $z=4\pi$. A driver who drives a car up this ramp would be spiraling once around the vertical line segment $\{(0,0.5,z)\,|\,2\pi<z<4\pi\}$.

Let $N:=L+(0,0.5)$. Then $N\subseteq F^\times$.

The map $\Pi|T$ is a homeomorphism from $T$ onto $F^\times\backslash N$.

For any distinct $v,w\in{\mathbb R}$, let $(v|w)$ denote the open interval between $v$ and $w$ defined by $(v|w)=\{(1-s)v+sw\,|\,0<s<1\}$.

For any $v,w\in{\mathbb R}$, let $[v|w]$ denote the closed interval between $v$ and $w$ defined by $[v|w]=\{(1-s)v+sw\,|\,0\le s\le1\}$.

Let $E:=(0|1)\times[0|0.5]$. So $E$ is a rectangle in the $xy$-plane.

Let $S:=(0|1)\times[0|0.5]\times\{2\pi\}$. So $S$ is a rectangle in the plane $z=2\pi$. It lies above $E$. I think of it as connecting the top of the ramp $R$ to the bottom of the ramp $T$.

The map $\Pi|S$ is a homeomorphism from $S$ onto $E$.

Let $X:=R\cup S\cup T$.

As I think of it, $X$ is a ramp that mostly spirals upward, but has a small horizontal strip in it. It spirals upward around $\{(0,0,z)\,|\,0<z<2\pi\}$) for one full turn. Then it is horizontal, moving in the positive $y$-direction inside the plane $z=2\pi$. Then it spirals upward around $\{(0,0.5,z)\,|\,2\pi<z<4\pi\}$ for one turn.

Let $Y:=(D^\times\backslash L)\cup E\cup(F^\times\backslash N)$. Since $L\subseteq E$ and $N\subseteq E$, we get: $Y=D^\times\cup E\cup F^\times$. Since $(0,0)\in F^\times$ and $(0,0.5)\in D^\times$, we get: $Y=D\cup E\cup F$.

The map $\Pi|X$ is a local-homeomorphism from $X$ onto $Y$.

For every open subset $U$ in $X$, let $V_U:=\Pi(U)$ be the image of $U$ under $\Pi$. Let ${\cal U}$ be the set of all nonempty open subsets $U$ in $X$ s.t. $\Pi|U$ is a homeomorphism from $U$ onto $V_U$.

Then $\{\Pi|U\,\hbox{s.t.}\,U\in{\cal U}\}$ is a $C^\infty$-atlas for $X$; in fact, all of the overlap maps are identity maps. Give $X$ the $C^\infty$-manifold structure from that atlas. Give ${\mathbb R^2}$ its standard $C^\infty$-manifold structure. Give $Y$ the $C^\infty$-manifold structure that it obtains as an open subset of the manifold ${\mathbb R}^2$.

Then $\Pi|X$ is a $C^\infty$-local-diffeomorphism from $X$ onto $Y$. Also, $X$ and $Y$ are $C^\infty$-diffeomorphic to ${\mathbb R}^2$. In particular, they are both contractible. It follows that $X$ is connected and simply-connected, and that $Y$ is connected and simply-connected.

Let $\theta:=\pi/4$. Let $x:=(\cos\theta)/2$, $y:=(\sin\theta)/2$. Then $(x,y,\pi/4)\in R$ and $(x,y,2\pi)\in S$. Since $$(\Pi|X)(x,y,\pi/4)=(x,y)=(\Pi|X)(x,y,2\pi),$$ we see that $\Pi|X$ is not $1$-to-$1$. However any covering map whose domain is connected and whose image is simply-connected must be a $1$-to-$1$ map. It follows that $\Pi|X$ is not a covering map.