Eilenberg–Mazur swindle

In geometric topology, the addition used in the swindle is usually the connected sum of knots or manifolds.

A typical application of the Mazur swindle is the proof that the sum of two non-trivial knots ${\displaystyle A}$ and ${\displaystyle B}$ is non-trivial.^[1] For knots it is possible to take infinite sums by making the knots smaller and smaller, so if ${\displaystyle A+B}$ is trivial then

${\displaystyle A=A+(B+A)+(B+A)+\cdots =(A+B)+(A+B)+\cdots =0}$

so ${\displaystyle A}$ is trivial (and ${\displaystyle B}$ by an analogous argument). The infinite sum of knots is usually a wild knot, not a tame knot. There are many geometric examples of the swindle.^[2]

Oriented ${\displaystyle n}$-manifolds have an addition operation given by connected sum, with identity the ${\displaystyle n}$-sphere. If ${\displaystyle A+B}$ is the ${\displaystyle n}$-sphere, then ${\displaystyle A+B+A+B+\cdots }$ is Euclidean space, so the Mazur swindle shows that the connected sum of ${\displaystyle A}$ and Euclidean space is again Euclidean space. Hence ${\displaystyle A}$ is the one-point compactification of Euclidean space and therefore ${\displaystyle A}$ is homeomorphic to the ${\displaystyle n}$-sphere. (This does not show in the case of smooth manifolds that ${\displaystyle A}$ is diffeomorphic to the ${\displaystyle n}$-sphere, and in some dimensions, such as ${\displaystyle 7}$, there are examples of exotic spheres ${\displaystyle A}$ with inverses that are not diffeomorphic to the standard ${\displaystyle n}$-sphere.)

In algebra the addition used in the swindle is usually the direct sum of modules over a ring.

A typical application of the Eilenberg swindle in algebra is the proof that if ${\displaystyle A}$ is a projective module over a ring ${\displaystyle R}$ then there is a free module ${\displaystyle F}$ with ${\displaystyle A\oplus F\cong F}$.^[3][4] To see this, choose a module ${\displaystyle B}$ such that ${\displaystyle A\oplus B}$ is free, which can be done as ${\displaystyle A}$ is projective, and put

${\displaystyle F=B\oplus A\oplus B\oplus A\oplus B\oplus \cdots }$

so that

${\displaystyle A\oplus F=A\oplus (B\oplus A)\oplus (B\oplus A)\oplus \cdots =(A\oplus B)\oplus (A\oplus B)\oplus \cdots \cong F.}$

For another application of the swindle, recall that finitely-generated free modules over commutative rings have a well-defined natural number as their dimension which is additive under direct sums, and are isomorphic if and only if they have the same dimension.^[5]

This is false for some noncommutative rings, and a counterexample can be constructed using the Eilenberg swindle as follows: let ${\displaystyle X}$ be an abelian group such that ${\displaystyle X\cong X\oplus X}$ (for example the direct sum of an infinite number of copies of any nonzero abelian group), and let ${\displaystyle R}$ be the ring of endomorphisms of ${\displaystyle X}$. Then the left ${\displaystyle R}$-module ${\displaystyle R}$ is isomorphic to the left ${\displaystyle R}$-module ${\displaystyle R\oplus R}$.

As a final interesting example,^[6] if ${\displaystyle A}$ and ${\displaystyle B}$ are any groups then the Eilenberg swindle can be used to construct a ring ${\displaystyle R}$ such that the group rings ${\displaystyle R[A]}$ and ${\displaystyle R[B]}$ are isomorphic rings: take ${\displaystyle R}$ to be the group ring of the restricted direct product of infinitely many copies of ${\displaystyle A\times B}$.

The proof of the Schröder-Bernstein theorem might be seen as antecedent of the Eilenberg–Mazur swindle. In fact, the ideas are quite similar. If there are injections of sets from X to Y and from Y to X, this means that formally we have X = Y + A and Y = X + B for some sets A and B, where + means disjoint union and = means there is a bijection between two sets. Expanding the former with the latter,

X = X + A + B.

In this bijection, let Z consist of those elements of the left hand side that correspond to an element of X on the right hand side. This bijection then expands to the bijection

X = A + B + A + B + ⋯ + Z.

Substituting the right hand side for X in Y = B + X gives the bijection

Y = B + A + B + A + ⋯ + Z.

Switching every adjacent pair B + A yields

Y = A + B + A + B + ⋯ + Z.

Composing the bijection for X with the inverse of the bijection for Y then yields

X = Y.

This argument depended on the bijections A + B = B + A and A + (B + C) = (A + B) + C as well as the well-definedness of infinite disjoint union.