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Sep 18, 2010 at 21:07 vote accept Martin Brandenburg
Sep 18, 2010 at 13:14 answer added Qing Liu timeline score: 8
Sep 18, 2010 at 12:03 comment added Martin Brandenburg I don't understand. There are morphisms $f : X \to Y$, which are not quasi-compact, although $X$ is affine and $Y$ is quasi-compact. For example, take some affine $X$ which has an open subset $U$ which is not quasi-compact, and glue two copies of $X$ along $U$ to get $Y$. Then the, say first, inclusion $X \to Y$ is not quasi-compact.
Sep 18, 2010 at 11:46 comment added Matthieu Romagny Well, if you take for $U$ any open affine neighbourhood of $x$ then $f(U)$ is open because $f$ is, and $U\to f(U)$ is quasi-compact because $U$ is...
Sep 18, 2010 at 10:54 history asked Martin Brandenburg CC BY-SA 2.5