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Aug 22, 2011 at 17:39 history edited Hugo Chapdelaine CC BY-SA 3.0
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Aug 18, 2011 at 22:08 comment added Benjamin Steinberg In my opinion, the pullback sheaf is easiest to understand using the etale space formulation of sheaves rather than the functorial description. If $Z\rightarrow Y$ is etale, then the pullback $X\times_Y Z\rightarrow X$ is etale over X.
Aug 18, 2011 at 19:11 comment added Hugo Chapdelaine @Angelo, thanks for your very simple and instructive example!
Aug 18, 2011 at 19:04 comment added Hugo Chapdelaine I needed my map $f$ to be a closed embedding and not just proper.
Aug 18, 2011 at 19:01 history edited Hugo Chapdelaine CC BY-SA 3.0
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Aug 18, 2011 at 18:59 comment added Hugo Chapdelaine Hi @Angelo, thanks for pointing my mistake, I know what is wrong!
Aug 18, 2011 at 18:52 comment added Peter McNamara Greg Muller provides an exmaple in mathoverflow.net/questions/45212/…
Aug 18, 2011 at 18:30 comment added Angelo Take $Y$ to be a point, $X$ a non-empty Hausdorff space that is not a point, and $\mathcal F$ a non-zero sheaf: then $f'\mathcal F$ is not sheaf. This would also seem to give counterexamples to your statement, unless I misunderstand something.
Aug 18, 2011 at 18:07 history asked Hugo Chapdelaine CC BY-SA 3.0