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I see this sentence in Hartshorne, exercices 3.6 in chapter 1. And he build a counter-example, namely $\mathbb{A}^2 \backslash \{(0,0)\}.$ But for me, this sentence is absolutely trivial.

Indeed, an affine variety is closed and a quasi-affine variety is dense open. So if a variety is affine and quasi-affine, it is the whole space $\mathbb{A}^n .$ There is something that I'm probably missing here. Thanks for any helpful comment.

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The subtle thing here is that we call closed subsets of $\mathbb A^n$ affine varieties and then define affine varieities to be closed under isomorphisms. Hence the quasi-affine variety $\mathbb A^1 - \{0\}$ is also an affine variety, since it is isomorphic to the closed subset $V(xy-1) \subset \mathbb A^2$.

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    So $f(x) = (x,1/x)$ is an isomorphism $\mathbb{A}^1 - \{0\} \to V(xy-1)$ – 2017-01-30
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    Yes, that is correct. – 2017-01-30
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    Sorry, I'd like to understand $f^{-1}$ and how [it acts on $k[x,y]/(xy-1)$](https://en.wikipedia.org/wiki/Morphism_of_algebraic_varieties#Definition). Is it $f^{-1}(x) = z, f^{-1}(y) = 1/z$ inducing an isomorphism $ k[x,y]/(xy-1) \to k[z,z^{-1}]$ ? – 2017-01-30
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    Oh ok, I understand. But this is was never said anywhere in the book. Thank you. – 2017-01-30
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    The inverse map $f^{-1}$ is just the projection $(a,b) \mapsto a$. The ring isomorphism induced by $f$ is given by $x \mapsto z, y \mapsto z^{-1}$. – 2017-01-30
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    @C.Dubussy This is mentioned on p. 25 of Hartshorne, but it is pretty cryptic: "We say loosely that a variety is *affine* if it is isomorphic to an affine variety." MooS's explanation would have been much better! – 2017-01-30