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A trivial question, but my lack of working experience in algebraic geometry is a hurdle.

Show that every morphism from $\mathbb{A}^1-\{0\}$ to $\mathbb{P}^1$ extends to a morphism from $\mathbb{A}^1$ to $\mathbb{P}$, but it is not true for morphisms from $\mathbb{A}^2-\{0\}$ to $\mathbb{P}^1$.

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    $\begingroup$ For the second part, I suggest concentrating on the standard morphism $\mathbb{A}^2 - \{0\} \to \mathbb{P}^1$ sending $(x, y) \mapsto [x, y]$. $\endgroup$
    – Hoot
    Jun 5, 2015 at 21:36
  • $\begingroup$ Following up: The first part could be a little trickier. What you sort of suspect is that the morphism is given by $[f, g]$ where $f, g \in k[t,1/t]$ have no common zeros on $\mathbb{A}^1 - \{0\}$. Then the thing to do would be to multiply these guys by an appropriate power (maybe negative) of $t$ so that it makes sense to evaluate at $t=0$. $\endgroup$
    – Hoot
    Jun 6, 2015 at 0:37

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In general, if X is normal of dimension 1 and Y proper, U an open subset of X, then $f: U \rightarrow Y$ extent uniquely to a morphism from X to Y.

See Corollary 4.1.17 of Liu Qing's book.

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