Is a continuous open map of a closed ball into Euclidean space a homeomorphism, if it restricts to an embedding of a sphere?

Question

Let $B_n = \{x \in \mathbb{R}^n : \lVert x \rVert \leq 1\}$ be the closed $n$-ball, and $S_n = \{x \in \mathbb{R}^{n + 1} : \lVert x \rVert = 1\}$ be the $n$-sphere. Let $X' \subset \mathbb{R}^m$, and $f : B_n \to X'$ be such that

• $f$ is continuous,
• $f$ is open,
• $f$ is surjective,
• $f|S_{n - 1}$ is injective (equivalently: an embedding),
• $f[S_{n - 1}] \cap f[B_n \setminus S_{n - 1}] = \emptyset$.

Note the last requirement which I was unable to fit into the title.

Problem

Is $f$ a homeomorphism?

Notes

• All topologies are subspace topologies of the Euclidean topology.
• $f$ is open to its image $X'$, not to the whole space.
• It suffices to show that $f$ is injective.
• Borsuk-Ulam theorem shows that $m \geq n$: otherwise $f|S_{n - 1}$ is not injective.
• I can show that the set of non-injectivity points $$I = \{x \in B_n : |f^{-1}[\{f(x)\}]| > 1\}$$ is open in $B_n$. This proof works for any continuous open map from a Hausdorff space to an arbitrary space.
• The problem is equivalent to: "Is $I$ closed in $B_n$?". For suppose that $I$ is also closed in $B_n$. In that case $I$ is clopen in $B_n$, but not the whole $B_n$. Since $B_n$ is connected, $I = \emptyset$. Then $f$ is injective, and hence a homeomorphism.
• The openness of $I$ is a quick way to show that simple attempts of self-intersection fail the requirements. For example, consider a path $f$ such that $I$ is finite. Then $I$ is not open, and hence does not satisfy the requirements.
• For $n = 1$ and $m = 1$, $f$ is a homeomorphism. The idea is to show that $f$ cannot have local extrema on $B_1 \setminus S_0$, because that breaks openness.

Examples

Let $f : B_1 \to [0, 1]$ be defined piecewise linearly with...

• ... $-1 \mapsto 0$, $-0.5 \mapsto 0.75$, $0.5 \mapsto 0.25$, and $1 \mapsto 1$. Then $f$ satisfies all other requirements, except that $f$ is not open.
• ... $-1 \mapsto 0$, $0 \mapsto 1$, $1 \mapsto 0$. Then $f$ satisfies all other requirements, except that $f|S_0$ is not injective. Note that here $I = B_n \setminus \{0\}$ is not closed.
• ... $-1 \mapsto 0$, $-0.5 \mapsto 1$, $0.5 \mapsto 0$, and $1 \mapsto 1$. Then $f$ satisfies all other requirements, except that $f[S_0] \cap f[B_1 \setminus S_0] \neq \emptyset$.

Answer 1

As discussed in comments, the conjecture is false for all $n=m\ge 3$: There exists an open continuous map $f: I^n\to I^n$ hich is the identity on the boundary, $f^{-1}(\partial I^n)=\partial I^n$ but $f$ is not 1-1, moreover, for all $p\in Int(I^n)$, $f^{-1}(p)$ has cardinality of continuum. See Theorem 3 in

Wilson, David, Open mappings on manifolds and a counterexample to the Whyburn conjecture, Duke Math. J. 40, 705-716 (1973). ZBL0273.54008.

For other weird examples of open continuous maps, see the references in my answers here and here.

Answer 2

The question as stated was already answered in the negative. In this answer I'll collect some additional conditions under which the claim does hold.

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Wilson, David, "Open mappings on manifolds and a counterexample to the Whyburn conjecture", Duke Math. J. 40, 705-716 (1973) contains the following theorem.

Theorem 6. Let $M^m$ be a compact connected m-manifold with boundary. Let $f$ be a discrete open mapping of $M^m$ onto $M^m$ such that $f(\textrm{Bd}(M^m)) = \textrm{Bd}(M^m)$ and $f(\textrm{Int}(M^m)) = \textrm{Int}(M^m)$. If $f|Bd(M^m)$ is a homeomorphism, then $f$ is a homeomorphism.

A mapping is discrete, if it has discrete fibers.

Note: Discreteness is not defined in the above paper, but I found its definition from

Väisälä, J. 1967. "Discrete open mappings on manifolds". Annales Fennici Mathematici. 392 (Jan. 1967).

The proof of Theorem 6 refers to the above paper, along with the McAuley's paper below.

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Louis F. McAuley, "Conditions under which light open mappings are homeomorphisms", Duke Math. J. 33(3): 445-452 (September 1966).

Above paper contains many theorems which show $f$ to be a homeomorphism given that $f$ is a light mapping together with some additional assumptions. A map is light, if it has totally disconnected fibers.

Note 1: I'm not sure whether totally disconnected here refers to trivial components or trivial quasicomponents (totally separated).

Note 2: Some other papers define a light mapping as having 0-dimensional fibers. Every 0-dimensional space is totally disconnected, but not vice versa.

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Morris L. Marx, "Whyburn’s conjecture for C2 maps", Proc. Amer. Math. Soc. 19 (1968), 660-661

The above paper shows the following:

Theorem. Let $F: D^n \to D^n$ be an open $C^2$ mapping such that $f^{-1}(S^{n - 1}) = S^{n - 1}$ and $f|S^{n - 1}$ is a homeomorphism. Then $f$ is a homeomorphism.

Here $D^n$ is the n-ball and $S^n$ is the n-sphere. Regarding the definition of a $C^2$ mapping:

By $f \in C^n$ on $D^n$ we mean that there exists an open set $U \supset D^n$ and a function $F: U \to E^n$ such that $F|D^n = f$ and $F$ has continuous second order partial derivatives in $U$.