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Jun 12, 2014 at 4:36 history edited bananastack CC BY-SA 3.0
fixed link to sard's theorem
Sep 1, 2011 at 11:25 comment added David E Speyer Your right that I am glossing over steps. In more (but not complete detail): $X$ is closed in $U \times \mathbb{CP}^4$ (exercise). Thus the map $X \to U$ is proper. Proper plus finite fibers means a finite map. Finite map with all fibers the same cardinality is etale (in characteristic $0$). Etale and finite is a covering map.
Sep 1, 2011 at 10:42 comment added Bruno Martelli Nice. How does the fact that the fiber of $X \to U$ has constant cardinality $9^4$ imply that it is a covering?
Sep 1, 2011 at 3:29 history edited Charles Staats CC BY-SA 3.0
added 1 characters in body
Sep 1, 2011 at 2:49 comment added Steven Landsburg (PS---"Exactly what I had in mind" should not be misread to mean that I had this example in mind. This is, instead, exactly the kind of thing I was looking for.
Sep 1, 2011 at 2:48 comment added Steven Landsburg Terrific --- exactly what I had in mind, except perhaps for the easily-explainable to-undergraduates part. Thank you!
Sep 1, 2011 at 2:34 history answered David E Speyer CC BY-SA 3.0