Q:

Is $f(m,n)=2^m\cdot(2n+1)$ a bijection between $\Bbb{Z_{\geq0}\times Z\to Z}$?

Let $\mathbb Z$  denote the set of integers and $\mathbb Z_{\ge 0}$
  denote the set $\{0,1,2,3,...\}$. Consider the map $f:\mathbb Z_{\ge
 0}\times \mathbb Z \to \mathbb Z$ given by $f(m,n)=2^m\cdot(2n+1)$. Then
  the map $f$ is
(A)injective but not surjective.
(B)surjective but not injective.
(C)injective and surjective.
(D)neither injective nor surjective.

For injectivity,
$$2^{m_1-m_2}(2n_1+1)=(1)(2n_2+1)$$
$$2^{m_1-m_2}(2n_1+1)=2^0(2n_2+1)$$ 
$$m_1=m_2 \land n_1=n_2   $$
For surjectivity,
$m=0$, $f$ maps to odd integers.Similarly, I am getting pre-image for even integers also.
So, (C) is the correct answer. Am I correct? But, solution manual gives (A) as the correct one. Who is correct? Please help me.

A:

Note that $f(m,n) = 2^m\cdot(2n+1)$ can never be zero, since neither factor can be zero. Therefore the mapping is not surjective.

