This seems to be a rather simple (stupid?:)) question; yet I was not able to find an answer quickly.

For a morphism $f:X\to Y$ of schemes (or topological spaces) and an (etale or topological) sheaf $f$ (of abelian groups) on $Y$ one defines the inverse image $f^*(S)$ (or is $f^{-1}$ more standard?) as the sheafification of the inverse image of $S$ considered as a presheaf.

Now suppose that $f$ is a (closed) embedding. Is the sheafification necessary here? It seems that the answer is no, since any covering of an open $U\subset X$ could be presented as a 'limit' of coverings of open $V\subset Y$, $V\supset f(U)$ (in the topogical context); so the 'presheaf inverse image' of a sheaf is a sheaf also. This seems to be easy, as well as carrying this argument over to the etale context; yet I would be deeply grateful for a definite answer and a reference for it. Also, does the situation change when one passes to simplicial schemes and sheaves? Is this fact 'standard' (if it is true)?

  • $\begingroup$ I would have thought that the answer would be "not necessary", rather, for an open embedding, no? (Or étale morphism?) $\endgroup$ Sep 16, 2011 at 12:10
  • $\begingroup$ It seems that you are not quite right. Certainly, everything is ok for an open embedding. My point is that one can present a closed subset as a 'limit' of open ones. On the other hand, let $f$ be the map from a (finite) collection of points with the discrete topology to a single point; certainly, such an $f$ is etale. Then the pullback of a constant presheaf (which is a certainly a sheaf) is a constant presheaf; one certainly has to sheafify here in order to obtain a sheaf. $\endgroup$ Sep 16, 2011 at 13:23
  • $\begingroup$ Also, do you have a counterexample to my claim? $\endgroup$ Sep 16, 2011 at 13:24
  • $\begingroup$ sorry, I missed the point. I don't have a counterexample, either. (You probably mean to write $V\subset Y$ above.) $\endgroup$ Sep 16, 2011 at 14:12
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    $\begingroup$ In the topological case, you need some additional hypotheses, otherwise you get a counterexample from $Y=\{g,s,t\}$ with $U\subseteq Y$ open iff $U$ empty or $g\in U$, so $X=\{s,t\}$ is a closed subspace, but $\Gamma(X,f^{-1}\mathbf Z_Y)=\mathbf Z^2\ne\mathbf Z=\varinjlim_{V\supseteq X}\Gamma(V,\mathbf Z_Y)$. $\endgroup$
    – user2035
    Sep 17, 2011 at 9:39

1 Answer 1


Perhaps I'm thinking of sheaves a bit differently from you here, but as long as we think of sheaves as \'{e}tale maps this holds for arbitrary continuous maps $f:X\to Y$ and not just open or closed embeddings. The reason for this is that, \'{e}tale maps are stable under pullback along arbitrary continuous maps. See, e.g., Mac Lane and Moerdijk (Sheaves in Geometry and Logic) Lemma 1 (II.9, page 99).

I guess that you know this though and are asking about the the case where we are not regarding sheaves as \'{e}tale maps. In this case you are right that some sheafification is required (since the construction of the inverse image along $f$ involves taking a colimit in $\text{Sh}(X)$). (I had thought that this was incorrect, but I had misunderstood precisely how you were claiming one should compute the inverse image in this case and so have removed my counterexample since it did not quite address your question.)


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