Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be an irreducible variety and let $U \subset X$ be a proper open set. Question : can there be a morphism $f : X \rightarrow U$ such that the composition $U \hookrightarrow X \stackrel{f}{\rightarrow} U$ is the identity?

My guess is that the answer is "no", but I can't seem to prove it.

One observation is that if $f$ exists, then it would give a procedure for taking a regular function $\phi : U \rightarrow k$ and extending it to all of $X$. But this can definitely be done in some cases; for instance, if $X = \mathbb{A}^2$ and $U = \mathbb{A}^2 \setminus \{(0,0)\}$.

share|cite|improve this question
The title seems to ask a different question than the body. The question in the body asks whether a proper open subset can be a retract. – Benjamin Steinberg Feb 10 '12 at 23:00
The question in the title is much more interesting than the question in the question. For the question in the question, regard $f$ as a function from $X$ to $X$, and observe that $f$ restricts to the identity on $U$. Thus $f$ must be the identity on the closure of $U$, which is all of $X$ since $X$ is irreducible. – Andy Putman Feb 11 '12 at 4:35
The retraction question is slightly more subtle than it appears since the Zariski topology is not Hausdorff. One needs to use that varieties are separated. – Benjamin Steinberg Feb 11 '12 at 13:11

1 Answer 1

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.