Is $\Bbb R^2 \subset P^2$ by isomorphism?

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According to page 10 of Construction of Number Systems by Prof. Kumar, we define a map $f:\Bbb N \rightarrow \Bbb Z$ by $f(k)=[(a+k,a)]$ for some $a\in\Bbb N$. Then we show this function $f$ is one-to-one and $f(a+b)=f(a)\oplus f(b)$ and $f(ab)=f(a)\otimes f(b)$. Here $\oplus$ and $\otimes$ are addition and multiplication on $\Bbb Z$, as opposed to $+$ and $*$ on $\Bbb N$. Prof. Kumar writes that this implies $\Bbb N \subset \Bbb Z$.

I suppose this is because $f$ is an isomorphism from $\Bbb N$ to a subset of $\Bbb Z$. I have only a basic knowledge of isomorphisms from undergraduate linear algebra.

Following this reasoning, consider $\Bbb R^2$ and $P^1$ (the set of all polynomials degree 1 or less). $\Bbb R^2$ is isomorphic to $P^1$ by the function: $f((a,b))=a+bx$. Since $P^1 \subset P^2$, does this mean that $\Bbb R^2 \subset P^2$?

Thank you for your help in advance.

EDIT: I am not familiar with abstract algebra, so I ask that you kindly explain any abstract algebra terms.