Isomorphism between $\mathbb R^2$ and $\mathbb R$

Based on this question:

Is it true that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as abelian groups?.

I’m trying to prove that this map $\varphi:\mathbb{R}\to\mathbb{R}^2$ defined by its action on element of Hamel basis of $\mathbb R$: $\varphi(e_\lambda)= (e_{i(\lambda)_1},e_{i(\lambda)_2})$, where $i:\Lambda\to\Lambda\times\Lambda$ is a set theoretic bijection, is an isomorphism between $\mathbb R$ and $\mathbb R^2$ as $\mathbb Q$-vector spaces.

The author were focusing in another aspects of the question and leave this proof unsolved. I’m trying to understand why this is an isomorphism, I have a basic knowledge of abstract algebra, I need help.

Thanks

Solutions Collecting From Web of "Isomorphism between $\mathbb R^2$ and $\mathbb R$"

I’m not sure what you mean by $\Lambda$, but it appears that the map $\phi$ gives a bijection between a $\bf Q$-basis of $\bf R$ and a $\bf Q$-basis of ${\bf R}^2$, right? Now you extend $\phi$ by linearity to be a map from $\bf R$ to ${\bf R}^2$. Since you have extended it by linearity, it is a linear transformation. To show it’s a $\bf Q$-vector space isomorphism, you have to show it’s one-one and onto; equivalently, you have to show it’s invertible.

So, given an element of ${\bf R}^2$, you can express it as a finite linear combination of the basis vectors; then can you see how to use $\phi$ to show it is the image of a unique element of $\bf R$?