Let $U$ be a connected open subset in $\mathbb{R^n}$. Let $f: U \rightarrow U$ be a differentiable projection, i.e. $f\circ f = f$. It's well-known that $f(U)$ is a submanifold of $U$ (Henri Cartan, Sur les rétractions d’une variété.(1986). My question is

"Under which condition on $f$ (and possibly on $U$) we have $Cl_{\mathbb{R}^n} f(U) \cap \partial_{\mathbb{R}^n} U \neq \emptyset$?"

Here the notation $Cl_{\mathbb{R}^n}$ stands for the closure when considering as a subset of $\mathbb{R}^n$ and the notation $\partial_{\mathbb{R}^n}$ stands for the topological boundary as a subset of $\mathbb{R}^n$

For example, if we take $U= \{ x \in \mathbb{R}^n | \|x \| <1\}$ the open solid sphere and the map $f(x_1,\ldots,x_n)=(0, x_2,\ldots,x_n)$ the projection onto the plane $\{x_1=0\}$. Then $f(U)$ is the unit disk in the plane $\{x_1=0\}$ and its closure as a subset of $\mathbb{R}^n$ is $$\{x=(x_1,\ldots,x_n) \in \mathbb{R}^n |x_1=0, \sum\limits_{i=2}^n x_i^2 \le1\}$$ indeed has non empty intersection with the boundary of $U$.

Further more, I am wondering if we could replace the hypothesis by $U$ is in $\mathbb{C}^n$ and the map $f$ is holomorphic retraction. Is that more helpful if we add condition on the boundary of $U$ (a submanifold, a algebraic subset)?

  • $\begingroup$ Well, $Cl_{\mathbb{R}^n} f(U) \cap \partial_{\mathbb{R}^n} U \neq \emptyset$ happens $\endgroup$
    – js21
    May 16, 2017 at 10:52
  • $\begingroup$ Yes. So is that always happens ? Is this just a particular example ? $\endgroup$
    – Curiosity
    May 16, 2017 at 10:55
  • 2
    $\begingroup$ This does not always happen, take a constant function with image not in the boundary of $U$. $\endgroup$
    – Nick L
    May 16, 2017 at 11:26
  • $\begingroup$ Then my question is under which condition that the closure of f(U) has nonempty intersect with the boundary of U. Do you have any example of non constant project that the closure of f(U) stay strictly inside U like the constant case ? $\endgroup$
    – Curiosity
    May 16, 2017 at 11:50
  • $\begingroup$ Take a solid torus (also called an anchor ring), and map to the central circle by the obvious (rotation equivariant) projection. The image is the central circle, not touching the boundary. $\endgroup$
    – Ben McKay
    May 16, 2017 at 12:00

2 Answers 2


In the real case, take $U = \mathbb{R}^2 \setminus 0$. Then the projection to the unit circle $Re^{i \theta} \mapsto e^{i \theta}$ would be a example of a non-constant smooth projection where the intersection you define is empty. There is generalisation in all dimensions, so I would be surprised if under some reasonable assumption we can make to make the intersection non-empty, (although one that comes to mind is to assume $U$ simply connected?? -I can't prove this).

  • $\begingroup$ Actually, I guess my guess is wrong for the reason that we can take $U$ with empty boundary. I mean, standard projection $\mathbb{R}^2 \rightarrow \mathbb{R}^1$ would have been an easier counter-example. (Maybe also could assume $U$ has non-empty boundary...). $\endgroup$
    – Nick L
    May 16, 2017 at 13:25
  • $\begingroup$ So the second hypothesis beside that $f$ is non constant that is $U$ is bounded. $\endgroup$
    – Curiosity
    May 16, 2017 at 15:22
  • $\begingroup$ For this hypothesis let $f$ be the projection to the circle and $U$ an open disk of radius more than $1$ (minus the origin!). $\endgroup$
    – Nick L
    May 16, 2017 at 15:51

The only non trivial case such that the closure of $f(U)$ has non empty intersect with the boundary of $U$ is when $U$ is bounded open subset of $\mathbb{C}^n$ and $f:U \rightarrow U$ is non constant projection. Details can be found in this link, Corollary 2.1.32, p122



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.