Linear mapping between two different countably infinite sets

Question

Take the natural numbers $\mathbb{N}_0 = \{0, 1,2,3,\dots\}$ and the integers $\mathbb{Z} = \{\dots, -1, 0, 1, \dots\}$.

Can we come up with a bijective linear mapping $T: \mathbb{N} \rightarrow \mathbb{Z}$?

Consider the mapping $0 \mapsto 0$, $1 \mapsto 1$, $2\mapsto -1$, $3\mapsto 2$, $4 \mapsto -2$, etc. This can be summarized by the function $$T(x) = \begin{cases} -\frac{x}{2} \text{ if } x \text{ even}, \\ \frac{x+1}{2} \text{ if } x \text{ odd}. \end{cases} $$

$T$ is not linear: if $x$ is even and $y$ is odd, then $x+y$ is odd and so $T(x+y) = \frac{x+y+1}{2}$, but $T(x)+T(y) = \frac{-x+y+1}{2}$.

Is it possible to come up with a linear mapping? Intuitively, a linear mapping should stretch, rotate, or translate a set, rather than "zig-zagging". I don't know if there is a way to do this to turn $\mathbb{N}$ into $\mathbb{Z}$?

In general: is there some explanation of when a linear mapping exists between two countably infinite sets, and when it doesn't exist?

Edit: it seems from some comments that a linear mapping exists between two sets when those sets are both vector spaces. Is this the answer?

Answer 1

Suppose $T(0)=a$ and $T(1)=b$. Then $T^{-1}(a)=0$ and $T^{-1}(b)=1$, so by linearity $T^{-1}(a-b)=-1$, which is not in $\Bbb N$.

Answer 2

The term "linear mapping" is generally defined on vector spaces, and neither $\mathbb{N}$ nor $\mathbb{Z}$ is a vector space, so I'm going to try to rephrase your question to avoid that term:

Rephrased Question: Does there exist a bijective function $T:\mathbb{N} \rightarrow \mathbb{Z}$ such that for all $x,y \in \mathbb{N}$, $T(x+y) = T(x) + T(y)$?

The answer to this question is No. To prove this, we'll prove an even stronger statement:

Claim: Any $T:\mathbb{N} \rightarrow \mathbb{Z}$ such that $T(x+y) = T(x) + T(y)$ is of the form $T(n) = nz$, for some fixed $z \in \mathbb{Z}$.

Proof of Claim: First we see that $T(0) = T(0+0) = T(0) + T(0)$, so $T(0) = 2T(0)$. $0$ is the only integer equal to its double, therefore $T(0) = 0$.

Next, for any $n > 0 \in \mathbb{N}$, we can express it as the sum of $n$ copies of $1$. Thus:

$T(n) = T(1 + \dots+1) = T(1) + \dots + T(1) = nT(1)$.

Setting $z = T(1)$ proves the claim.

The claim implies the image of $T$ is either completely nonnegative or completely nonpositive, meaning it cannot be bijective.