Let $\phi:{\mathbb{N}}\rightarrow {\mathbb{N}}$ be a bijection. Can we extend the notion of parity (of a finite permutation) to $\phi$ ? 


In other words, Can we define a group homomorphism $\Lambda $, between the group of all bijections of $\mathbb{N}$ to the group {1,-1,$\cdot$} such that $\Lambda(\psi)=-1$ if $\psi$ is a "transposition": 


There exists $n\_{1},n\_{2}\in \mathbb{N}$, where $n\_{1}\neq n\_2$ and $\psi(n\_{1})=n\_{2} ,\psi(n\_{2})=n\_{1}$ and ${\psi(x)=x , \forall x \neq n\_{1},n\_{2} }$.


My guess is that building such a homomorphism requires some version of the axiom of choice / ultra-filters. 



Defining parity on the set of all bijections of ${\mathbb{N}}$
It would be easier, and equivalent, to consider $\mathbb{Z}.$ But I'll stick to $\mathbb{N}.$ 


Consider $\sigma=(2\ 3)(4\ 5)(6\ 7)\cdots$ (so $0$ and $1$ are fixed points) and $(0\ 1)\sigma.$ They have the same cycle type but should not have the same parity. Note that fixed points and the particular ordering of $\mathbb{N}$ should be irrelevant.


For a similar example with conjugation consider $\tau=(5\ 7)(9\ 11)(13\ 15)\cdots$ and $\tau'=(1\ 3)(5\ 7)(9\ 11)(13\ 15)\cdots$ which should again not gave the same parity. To turn $\tau$ into $\tau',$ conjugate by this product of two infinite cycles $$(\cdots 13\ 9\ 5\ 1\ 0\ 4\ 8 \cdots) (\cdots 11\ 7\ 3\ 2\ 6\ 20 \cdots).$$


