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Let $X$ be a variety and $Z\subset X$ be a closed subvariety. Denote by $Z^c$ the set-theoretic complement in $X$ together with the structure as subvariety of $X$. Assume that there are morphisms $f: Z\to Y$, $f^c: Z^c\to Y$ where $Y$ is another variety. Then define $$F: X\to Y, x\mapsto\begin{cases}f(x),& x\in Z,\\ f^c(x),& x\in Z^c.\end{cases}$$

I am searching for a proof or reference of the following: if $F$ is continuous then it is a morphism. This should hold true but I was unable to find it in my textbooks.

Two brief remarks:
1) If we drop the assumption that $f$ should be continuous, the assertion is false for sure.
2) If we would know that $X$ is the coproduct of $Z$ and $Z^c$, the assertion would follow. But I don't expect this to be true in general.

This seems to be related to morphism of ringed spaces glue but I'm a little confused as in this question, continuity always seems to hold. But it there is also assumed that $Z$ and $Z^c$ should be open so it can't be directly applied to my question.

References and hints to textbooks/articles are very welcome.

sampleuser
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  • What is your definition of a variety? If, for example, you mean scheme of finite type over a field, this should probably be false via some nilpotency trick (I think Georges Elencwajg wrote an answer to this effect at one point, but I can't find it right now). – KReiser Mar 13 '20 at 08:16
  • @KReiser In my situation, varieties are defined over $\mathbb C$ and always reduced, so no nilpotent elements. However, if you find the answer you mentioned, please provide a link. – sampleuser Mar 13 '20 at 08:49
  • So your definition of variety is a reduced scheme of finite type over $\Bbb C$? I'll see if I can't dig up the answer or remember exactly what the counterexample was. – KReiser Mar 13 '20 at 09:05
  • @KReiser I think so, yes. – sampleuser Mar 13 '20 at 09:07

1 Answers1

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You won't find a reference for your statement, because it is false !
Indeed let $X\subset \mathbb A^2_k$ be the curve of equation $v^2-u^3=0$.
Let $P$ be the point with coordinates $u=v=0$, and take $Z=\{ P\}$ .
Then define $F:X\to \mathbb A^1_k$ by $F(P)= 0$ and $F(u,v)=\frac vu$ for $(u,v)\in Z^c$, i.e. $(u,v)\neq P=(0,0)$.
That function $F$ is continuous everywhere and its restrictions $f$ to $Z$ and $f^c$ to $Z^c$ are regular.
However $F$ is not a morphism of algebraic varieties.

Georges Elencwajg
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