Is $f:\mathbb{Z}_{\geq 0}\times \mathbb{Z} \rightarrow \mathbb{Z}$ defined as $f(m,n)=2^m(2n-1)$ bijective? Let $f:\mathbb{Z}_{\geq 0}\times \mathbb{Z} \rightarrow \mathbb{Z}$ defined as $f(m,n)=2^m(2n-1)$. Is $f$ bijective?


So every integer can be written in the form $2^i \xi$ where $\xi$ is an odd integer. And such a representation is unique. So clearly this is a bijection. Am I right?
The function is  not onto. $f(m,n)=0$ does not have a solution.
The function is not bijective, consider $f(k,0)=2^{k}(2(0)-1)$. Note injectivity is violated. 

Recall the definition of injectivity is $f(x)=f(y)$ iff $x=y$. But clearly we have $f(k,0)=f(p,0)=0$ since both $2^{p}(2(0)-1)=0$ and $2^{k}(2(0)-1)=0$. 

Surjectivty holds, because we can choose $f(0,n)=(2n-1)=r$ has a solution for every $n$