Exotic bijection from $\mathbb R$ to $\mathbb R$ Clearly there is no continuous bijections $f,g~:~\mathbb R \to \mathbb R$ such that $fg$ is a bijection from $\mathbb R$ to $\mathbb R$.
If we omit the continuity assumption, is there such an example ?
Notes: to follow from Dustan's comments:
Notes:  By definition $fg~:~x \mapsto f(x)\times g(x)$ and not $f\circ g$.
If there were continuous bijections just look at the limits of $f$ and $g$ at $+\infty$ and $-\infty$ to conclude that $fg$ can't be a bijection 

Let $f(x)=x$, and define $g$ piecewise by
$$
  g(x) =
  \begin{cases} 
    -x ,& x \in \cdots \cup (-16,-8] \cup (-4,-2] \cup \cdots
        &= \bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k) \\
    4x ,& x \in \cdots \cup (-8,-4] \cup (-2,-1] \cup \cdots
        &= \bigcup_k -[4^k, 2 \cdot 4^k) \\
    0 ,& x = 0 \\
    x ,& x \in \cdots \cup [1,2) \cup [4,8) \cup \cdots
        &= \bigcup_k [4^k, 2 \cdot 4^k) \\
    -\frac14 x ,& x \in \cdots \cup [2,4) \cup [8,16) \cup \cdots
        &= \bigcup_k [2 \cdot 4^k, 4 \cdot 4^k) \\
   \end{cases}
$$
(where $-[b,a)$ just means $(-a,-b]$).
This $g$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$, since


*

*$g$ maps $\bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k [2 \cdot 4^k, 4 \cdot 4^k)$.

*$g$ maps $\bigcup_k [2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k)$.

*$g$ maps $\bigcup_k -[4^k, 2 \cdot 4^k)$ one-to-one onto itself.

*$g$ maps $\bigcup_k [4^k, 2 \cdot 4^k)$ one-to-one onto itself.

*$g$ maps $0$ to itself.


And $fg$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$, since


*

*$fg$ maps $\bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k -[4 \cdot 16^k, 16 \cdot 16^k)$.

*$fg$ maps $\bigcup_k [2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k -[16^k, 4 \cdot 16^k)$.

*$fg$ maps $\bigcup_k -[4^k, 2 \cdot 4^k)$ one-to-one onto $\bigcup_k [4 \cdot 16^k, 16 \cdot 16^k)$.

*$fg$ maps $\bigcup_k [4^k, 2 \cdot 4^k)$ one-to-one onto $\bigcup_k [16^k, 4 \cdot 16^k)$.

*$fg$ maps $0$ to itself.

