Let $M$ be a compact connected manifold with boundary whose interior has dimension $n\geq 2$. Suppose that we have a map $f:M\rightarrow S^n$ which is continuous and such that the restriction of $f$ to the boundary of $M$ is injective. Suppose further that $f$ is locally injective - that is, for each $p\in M$, there is an open $U$ containing $p$ such that the restriction of $f$ to $U$ is injective.

Does it follow that $f$ is injective?

I thought of this question while I was studying some discrete geometry. For my purposes, I was able to prove and use the weaker statement that if $f:M\rightarrow \mathbb R^n$ satisfies the given conditions and has that the image of the boundary of $M$ under $f$ is the boundary of a convex set $C$, then $f$ is injective.

I proved this by first noting that the image of the interior of $M$ is a subset of the interior of $C$, which follows from the fact that the dot product $f(x)\cdot v$ may have no local minima or maxima on the interior of $M$ due to invariance of domain.

Then, I suppose that I have a pair $x,y$ of distinct points in the interior with $f(x)=f(y)$. If I choose any path $\gamma$ between them in the interior of $M$, then I would have that $\alpha=f\circ \gamma$ was a loop based at $f(x)=f(y)$. However, $\alpha$ has a homotopy $H$ to the trivial curve $H(p,t)=(1-t)\alpha(p)+tf(x)$. Then, we show that we can lift this to some homotopy $\tilde H$ such that $\tilde H(p,0)=\gamma(p)$ and $f\circ \tilde H = H$. However, this is a problem since if we define $\beta(p)=\tilde H(p,1)$, we have that $\beta(0)=x$ and $\beta(1)=y$, but $f\circ \beta$ is constant, contradicting local injectiveness.

My trouble with extending the statement is that my argument is littered with geometrical facts, and even though I suspect that similar methods should work in the more general case I ask about, I can't figure out how.

  • $\begingroup$ Well, at least in one dimension it is easy to see that this is wrong. Wind an interval which is long enough around $S^1$ several times parametrized by arclength such that the endpoints are mapped to different points. (I do assume you mean the sphere here). $\endgroup$ – Thomas Aug 1 '16 at 16:31
  • $\begingroup$ @Thomas Ah, that's a good point. I edited the question to ask that the dimension be at least $2$. $\endgroup$ – Milo Brandt Aug 1 '16 at 17:05

Indeed, such a map is injective under the following mild extra assumption:

Let $N$ denote $\partial M$. Then require that $f|N$ is tame. In other words, there is a product neighborhood $U\cong (-1,0]\times N$ of $N$ in $M$ such that the restriction $f|U$ extends to a continuous injective map $F: (-1,1)\times N \to S^n$. (There are alternative definitions of tameness, intrinsic to the restriction $f|N$, but this one is the simplest to work with.)

This tameness property is automatic if $M$ and $f$ are $C^1$-smooth. In general, however, it fails (consider a horned sphere).

Assume now that $f|N$ is tame. Then there exists a submanifold with boundary $L\subset S^n$ such that $\partial L= f(N)$ and $L$ contains $F([0,1)\times N)$ in the above description. Therefore, we can define a connected closed manifold (i.e. compact without boundary) $X$ by gluing $M$ and $L$ via the map $f^{-1}|f(N)$. Then we obtain an obvious continuous map $g: X\to S^n$ by combining $f|M$ with the identity embedding $id: L\to S^n$. One then checks that $g$ is a local homeomorphism. Since $X$ is closed compact, $g: L\to S^n$ is a covering map (by the "stack of records theorem"). However, $S^n$ is simply connected, hence, $g$ is a homeomorphism, and, thus $f$ is injective. qed

I think one can adjust this proof to make it work without tameness assumption (by replacing $M$ with a slightly smaller submanifold). Let me know if you need this. (Most people work in the smooth category anyway.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.