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How to construct a bijection $f: A\to \Bbb{Z}$ where $A$ is the set of odd natural numbers?

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    Find a method that would list all the numbers in each set. – 2017-01-21
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    Have you ever heard of Cantor's diagonal argument? that should give you some inspiration. – 2017-01-21

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An odd natural number can be written as $2n-1$, $n \in \mathbf N$, hence $$ A \to \mathbf N, \quad 2n-1\mapsto n $$ is a bijection. Now we map $\mathbf N$ to $\mathbf Z$. Let $n$ a natural number, if it is even, say $n = 2k$, we map it to $k-1$, if it is odd, say $n = 2k-1$, we map it to $-k$. This gives the bijection $$ \mathbf N \to \mathbf Z, \quad n \mapsto \begin{cases}  k-1 & n = 2k\\ -k & n = 2k-1 \end{cases}  $$ Concatenating both maps, we have the bijection $A \to \mathbf Z$ $$ 2n-1 \mapsto \begin{cases}  k-1 & n = 2k\\ -k & n = 2k-1 \end{cases}  $$