Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f\colon X \rightarrow Y$ be a map between topological spaces. Let $\mathfrak{O}(X)$(resp. $\mathfrak{O}(Y))$ be the set of open subsets of $X$(resp. $Y$). Suppose the application $U \rightarrow f^{-1}(U)$ is a bijection $\mathfrak{O}(Y) \rightarrow \mathfrak{O}(X)$. Then $f$ is called a quasi-homeomorphism.

Now we use the definitions of this question. Let $k$ be a field and $X$ be an affine $k$-variety. Let $A = \Gamma(X, \mathcal{O}_X)$. Let $x \in X$. We denote $\{f \in A|\ f(x) = 0\}$ by $\mathfrak{p}_x$. It is a prime ideal of $A$. Hence we get a map $\psi\colon X \rightarrow Spec(A)$ by defining $\psi(x) = \mathfrak{p}_x$.

Let $Y = Spec(A)$. Let $f \in A$. Let $h \in A_f = \Gamma(D(f), \mathcal{O}_Y)$. Then $h$ is identified with a regular function on $\psi^{-1}(D(f))$ in $Y$. Hence we get a canonical map $\psi_U^{\#}\colon \Gamma(U, \mathcal{O}_Y) \rightarrow \Gamma(\psi^{-1}(U),\mathcal{O}_X)$ for any open subset $U$ of $Y$. Are the following assertions true?

(1) $\psi$ is a quasi-homeomorphism.

(2) $\psi_U^{\#}\colon \Gamma(U, \mathcal{O}_Y) \rightarrow \Gamma(\psi^{-1}(U),\mathcal{O}_X)$ is an isomorphism of $k$-algebras for any open subset $U$ of $Y$.

share|cite|improve this question
Probably you mean that $V \mapsto f^{-1}(V)$ is a bijection from $\mathfrak{O}(Y) \to \mathfrak{O}(X)$. – Adeel Dec 1 '12 at 17:02
@Adeel Right. Thanks. – Makoto Kato Dec 1 '12 at 17:45
What is the definition of quasi-homeomorphism? A homeomorphism outside a closed set? – Fredrik Meyer Dec 1 '12 at 17:52
@FredrikMeyer A quasi-homeomorphism is a map $f\colon X \rightarrow Y$ of topological spaces satisfying the above condition. – Makoto Kato Dec 1 '12 at 17:58

Your Answer


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

Browse other questions tagged or ask your own question.