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I'm studying Gathmann's notes (https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2021/alggeom-2021.pdf), and I'm currently stuck in Exercise 5.7:

Show:

(a) Every morphism $f:\mathbb{A}^1\setminus\{0\}\to \mathbb{P}^1$ can be extended to a morphism $\mathbb{A}^1\to \mathbb{P}^1$.

(b) Not every morphism $f:\mathbb{A}^2\setminus\{0\}\to \mathbb{P}^1$ can be extended to a morphism $\mathbb{A}^2\to \mathbb{P}^1$.

I know that questions have been asked here (see Extending morphisms between varieties, Extending a morphism $\mathbb{A}^1-0\to \mathbb{P}^1$ to $\mathbb{A}^1$). But none of them uses a construction of $\mathbb{P}^1$ as in Gathmann: In the notes, $\mathbb{P}^1$ is constructed by gluing together two copies of $\mathbb{A}^1$ along $\mathbb{A}^1\setminus\{0\}$ using the gluing isomorphism $x\mapsto \frac1x$.

My question is, how should I understand morphisms to $\mathbb{P}^1$? I know that they are continuous maps such that the pullback of a regular function is regular, but the definition of the structure sheaf on $\mathbb{P}^1$ is quite complicated here: it is defined by $$\mathcal{O}_{\mathbb{P}^1}(U)=\{\phi:U\to K:i_1^\ast\phi\in\mathcal{O}_{X_1}(i_1^{-1}(U))~\text{and}~i_2^\ast\phi\in\mathcal{O}_{X_2}(i_2^{-1}(U))\}$$ where $i_1,i_2:\mathbb{A}^1\to \mathbb{P}^1$ are two natural embeddings.

More precisely, in (b), I tried to construct a morphism $f:\mathbb{A}^2\setminus\{0\}\to\mathbb{P}^1$ via $$f(x,y)=\begin{cases}[x/y] &\text{if}~y\neq 0 \\ [y/x] &\text{if}~x\neq 0\end{cases}$$ where $[\cdot]$ is the equivalence class of $\cdot$ in $\mathbb{P}^1$. I think this is well-defined, but I can't even show this is a morphism. Would anyone be able to help and enlighten me?

(I know that homogeneous coordinates can be used here, but using this definition of $\mathbb{P}^1$ it seems to be unsuitable.)

blastzit
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  • For (b): a point in $\mathbb{P}^1$ is an equivalence class of pairs of coordinates, not of single coordinates, i.e. of the form $[(a,b)]$, but it is more commonly denoted by $(a:b)$. But you're not completely off with your attempt at an answer. – R.P. Jul 13 '22 at 12:56
  • @RP_ Thanks for the reply! I do know $(a:b)$ (or $[a:b]$) represents a point in $\mathbb{P}^1$ usually, but I don’t see why that’s the case in the definition given by Gathmann. Isn’t a point in $\mathbb{P}^1$ just a equivalent class $[x]$ which contains $x$ and $1/x$? Or did I miss something obvious? – blastzit Jul 13 '22 at 13:26
  • My apologies for the delay. I think you must misunderstand the construction, the way Gathmann defines it is not equivalence classes of elements of $\mathbb{A}^1$, but of elements of $\mathbb{A}^1 \coprod \mathbb{A}^1$, in which a non-zero element $x$ in the first copy of $\mathbb{A}^1$ is identified with the element $1/x$ in the second copy of $\mathbb{A}^1$. – R.P. Jul 13 '22 at 19:31
  • @RP_ That makes much more sense! So does $[(a,b)]$ mean the equivalence class of $a$ in the first $A^1$ and $b$ in the second $A^1$ instead? From where can I see that this $[(a,b)]$ matches with the homogeneous coordinates, so e.g. that $[(4,2)]=[(2,1)]$? – blastzit Jul 13 '22 at 19:36
  • No you have to distinguish those two forms of notation very carefully. The notation $(a,b)$ only makes sense when you're considering products, not disjoint unions. An element of a disjoint union $A \coprod B$ means picking an element of $A$ or an element of $B$, not (as in the case of a cartesian product) an element of $A$ and an element of $B$. – R.P. Jul 13 '22 at 19:39
  • @RP_ In that case, what is a representative of an equivalence class in $\mathbb{P}^1$? Would it be something like $[(x,1)]$ and $[(x,2)]$ (as in construction of disjoint union), and $[(x,1)]=[(1/x,2)]$ if $x\neq 0$? – blastzit Jul 13 '22 at 19:46
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    That would be one way to write it, yes. But I don't think it is a very helpful notation. A better way is just to observe that the second copy of $\mathbb{A}^1$ is really only adding a single point, namely $0$, because every other point $y$ corresponds to $x=1/y$ in the first copy. So for this single extra point coming from the second copy you can just write $\infty$, so that you get $\mathbb{P}^1 = \mathbb{A}^1 \coprod { \infty }$. – R.P. Jul 13 '22 at 19:50
  • I think comment 5.5(a) gives a good clue: it talks about gluing morphisms. The variety $\mathbb{P}^1$ is defined in terms of two copies of $\mathbb{A}^1$, and it easily seen that it contains these two copies of $\mathbb{A}^1$ as open subsets. So the morphism $f$ from (a) can be seen as a gluing of two morphisms $f_1 : U_1 \to \mathbb{A}^1$ and $f_2 : U_2 \to \mathbb{A}^1$ where $f_1$ is the restriction of $f$ to the first copy of $\mathbb{A}^1$ and $f_2$ is the restriction of $f$ to the second copy of $\mathbb{A}^1$ (and where $U_1$ and $U_2$ are open subsets of $\mathbb{A}^1\backslash{0}$). – R.P. Jul 13 '22 at 20:04
  • The fact that $f_1$ and $f_2$ glue together to give a morphism to $\mathbb{P}^1$ translates into the condition that, for $u \in U_1 \cap U_2$, we have $f_1(u) = f_2(u)^{-1}$. This observation brings the whole thing down to a consideration of morphisms between affine varieties, which should be a lot less confusing. – R.P. Jul 13 '22 at 20:04
  • @RP_ Thank you so much for the detailed explanation. One last question: why is $U_i$ affine? – blastzit Jul 13 '22 at 20:10
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    Oh it's actually easy to see: every open subset of the affine line is either empty, or the affine line minus a finite number of points. And given a finite subset $S$ of $\mathbb{A}^1$, there exists a polynomial $g$ with the elements of $S$ as its zeros, and then we can write $\mathbb{A}^1 - S$ as the affine variety corresponding to the ring $K[X,g^{-1}]$. (By inverting $g$, you are killing precisely the ideals $(X-\alpha)$ where $\alpha$ ranges over the zeros of $g$.) – R.P. Jul 13 '22 at 20:17
  • Which means that $f_1$ and $f_2$ correspond to two ring homomorphisms $\phi_1 : K[X] \to K[X,g_1^{-1}]$ and $\phi_2 : K[X] \to K[X,g_2^{-1}]$. And the zeros of each of the $g_i$ are exactly the points that you have deleted from the affine line. I'm going to leave it here for the moment, have to get on with other things. – R.P. Jul 13 '22 at 20:30
  • @RP_ Thank you so much for the help! I think I can try to continue from here. – blastzit Jul 13 '22 at 20:31

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