---
tag: QHJ0GBQE2E
kind: statement
role: theorem
name: Borsuk–Ulam theorem
msc: [55M25, 55M20]
uses: [T62369RFH6, 0VGFHTM75Y, TMFG2P912R]
about: [T62369RFH6]
sources:
  - url: https://en.wikipedia.org/wiki/Borsuk–Ulam_theorem
same_as:
  wikidata: Q848092
  wikipedia: Borsuk–Ulam theorem
---

## Statement

Let $n \ge 1$ and let $f \colon S^n \to \R^n$ be a [[continuous map|0VGFHTM75Y]] on the [[$n$-sphere|T62369RFH6]] $S^n \subset \R^{n+1}$.
Then there is a point $x \in S^n$ with $f(x) = f(-x)$.

## Remarks

An equivalent form says that there is no continuous odd map $g \colon S^n \to S^{n-1}$, that is, no map with $g(-x) = -g(x)$ for all $x$.
Indeed, if $f(x) \neq f(-x)$ for all $x$, then $g(x) = (f(x) - f(-x)) / \norm{f(x) - f(-x)}$ is such a map.
One proof uses the fact that an odd map $S^{n-1} \to S^{n-1}$ has odd [[degree|TMFG2P912R]].
The restriction of an odd map $S^n \to S^{n-1}$ to the equator is an odd map that extends over a hemisphere.
So it is null-homotopic and has degree $0$, a contradiction.

Example: for $n = 1$, let $h(x) = f(x) - f(-x)$ on the circle.
Then $h(-x) = -h(x)$, so $h$ takes values of both signs or vanishes, and by the intermediate value theorem it has a zero.

Example of use: at any moment there are two antipodal points on the Earth with the same temperature and the same air pressure, if both are continuous functions on $S^2$.
The theorem fails for maps $S^n \to \R^{n+1}$: the inclusion $S^n \subset \R^{n+1}$ takes different values at $x$ and $-x$.

## Explanations

### L1 human

If you squash a sphere flat into a space of one dimension less, two opposite points of the sphere always land on the same spot.
For a balloon pressed flat onto a table, some pair of opposite points ends up one on top of the other.

### L2 human

The theorem is the starting point of the use of topology in combinatorics and geometry.
It implies the [[ham sandwich theorem|GDXX60SNG5]], the Lusternik–Schnirelmann covering theorem, and Lovász's computation of the chromatic number of Kneser graphs.
Its combinatorial counterpart is [[Tucker's lemma|PADZ7D9KBT]], from which it can also be proved.