Embedding $\mathbb A^2-(0,0)$ into $k^n$, the image is not closed

Question

For an arbitrary embedding $\mathbb A^2-(0,0)$ into $k^n$, must the image be not closed?

In Mumford's Redbook p25, he argued as every coordinate function can be extended to $\mathbb A^2$, so the morphism can be extended to $\mathbb A^2$, so the image would not be closed.

If the extended morphism is injective, we delete the point corresponding to $(0,0)$, the image cannot be closed since the image of $\mathbb A^2$ is irreducible.

I don't know how to show the extended morphism is also injective or does it hold?

Answer 1

If $\mathbb{A}^2−(0,0)$ admits a closed embedding into $\mathbb{A}^n$, then $\mathbb{A}^2−(0,0)$ is a closed subscheme of an affine scheme, hence $\mathbb{A}^2−(0,0)$ is affine, but $\mathbb{A}^2−(0,0)$ is not affine as Mumford explained.