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Let $K$ denote an algebraically closed field. Define that an algebraic variety over $K$ is a ringed space that can be covered by open sets, each of which is isomorphic to an affine algebraic variety, possibly subject to some further conditions if it helps.

Question. Let $(f,\alpha) : A \rightarrow B$ and $(f,\beta) : A \rightarrow B$ be two morphisms of locally ringed spaces. Are $\alpha$ and $\beta$ necessarily equal as natural transformations? If not, is this at least true for "nice" spaces and/or "nice" fields? If not, how is it possible to define a function between algebraic varieties and then "prove" that this function is a morphism, when the corresponding natural transformation isn't unique?

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2 Answers 2

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No, as the following example proves.

Example. Let $\mathbb{K}=\overline{\mathbb{Z}_p}$ be the algebraic closure of the field of integer numbers modulo $p\in\mathbb{P}$, and let $X=Y=\mathbb{A}^1_{\mathbb{K}}$ as locally ringed space; considering the following maps of $X$ in itself: \begin{gather} O:x\in X\to 0\in X,\\ F:x\in X\to x^p-x\in X; \end{gather} one has that as maps $O=F$, but as regular maps $O\neq F$ because $O^{*}$ is the zero endomorphism of $\mathbb{K}[t]$ and $F^{*}$ is the Frobenius endomorphism of $\mathbb{K}[t]$. $\triangle$

And if one fixes a continuous map $f:(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ then the pull-back $f^{*}:\mathcal{O}_Y\to f_{*}\mathcal{O}_X$ is well-defined ever; and if $f^{*}$ is a morphism of sheaves then $(f,f^{*})$ is a morphism of locally ringed spaces.

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  • $\begingroup$ Consider $\mathbb{F}_p$ to refer to the field of $p$ elements, since its unambiguous; $\mathbb{Z}_p$ is also used to to the ring of $p$-adic integers. $\endgroup$
    – user14972
    Sep 15, 2018 at 11:01
  • $\begingroup$ Yes, I know; indeed I specify what is $\mathbb{Z}_p$. ;) $\endgroup$ Sep 15, 2018 at 11:03
  • $\begingroup$ So is it correct to say that each continuous map $f$ between ringed spaces induces a canonical morphism of ringed spaces, namely $(f,f^*)$? $\endgroup$ Sep 15, 2018 at 11:40
  • $\begingroup$ No, it is not! For example: any permutation of $\mathbb{A}^1_{\mathbb{K}}$ with Zariski topology is a continuous map, but not any permutation is a morphism of locally ringed spaces. $\endgroup$ Sep 15, 2018 at 11:49
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    $\begingroup$ Wait why is $O=F$? We don't have $$X^p=X$$ for all $$X \in \overline{\mathbb{F}_P}$$ right.. $\endgroup$
    – M. Van
    Mar 1, 2022 at 14:08
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Here's another example (I have problems with the other answer given, see my comments below that): take the algebraic closure $k$ of $\mathbb{F}_p$ as your ground field and consider $X= \text{Spec}(k)$. Then any two distinct automorphisms of $k$ give the same map on the level of sets, since $X$ consists of only one point. You can take the frobenius map and the identity map for example.

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