---
tag: VK39X07EYK
kind: conjecture
name: Bing–Borsuk conjecture
msc: [57P05, 55M15]
uses: [DK3MT3HYX5, 0QT00PB5G2, A3W7RGXXK9, QQQ9B1DK4P]
about: [DK3MT3HYX5, QQQ9B1DK4P]
sources:
  - url: https://en.wikipedia.org/wiki/Bing%E2%80%93Borsuk_conjecture
same_as:
  wikidata: Q55605472
  wikipedia: Bing–Borsuk conjecture
history:
  posed_by: "R. H. Bing and Karol Borsuk"
  first_stated: "1965"
---

## Statement

Let $n \ge 1$ and let $X$ be an [[absolute neighborhood retract|DK3MT3HYX5]] of [[covering dimension|A3W7RGXXK9]] $n$.
Suppose that $X$ is homogeneous: for all points $x, y \in X$ there is a [[homeomorphism|0QT00PB5G2]] $h \colon X \to X$ with $h(x) = y$.
Then $X$ is a [[topological manifold|QQQ9B1DK4P]] of dimension $n$.

## Remarks

Every connected topological manifold is a homogeneous absolute neighborhood retract, so the conjecture asks for a converse.
The conjecture is known to hold for $n = 1$ and $n = 2$.
In dimension $3$, it is known to imply the Poincaré conjecture.
In general it is open.
The finite dimension is needed: the Hilbert cube is a homogeneous compact absolute retract of infinite dimension that is not a manifold of any finite dimension.

## Explanations

### L1 human

A manifold looks the same at every point, in the sense that any point can be moved to any other by a self-homeomorphism.
It is also a well-behaved space in the sense of homotopy theory.
The conjecture says that these two properties, in finite dimension, are enough to force a space to be a manifold.