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Is the direct image of a constant sheaf a constant sheaf? I'm not an expert on sheaf theory and can't find this anywhere

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Notation: $f:X \to Y$ is the map we're pushing forward along, and $F$ is our sheaf on $X$. In general the stalks of $f_*F$ at different points will not be isomorphic. For instance if $f$ misses the point $y \in Y$ and your space is sufficiently separated then the stalk of $f_*F$ at $y$ will be 0 while it will be nonzero for points in the image.

An extreme case is when the map has image a point. Then you get a skyscraper sheaf, which is very far from constant on most spaces and most points (Note: if you're hitting the generic point of $Y$ then the direct image will in fact be constant).

Edit: Another extreme case is when $X$ is a large discrete space. Then one can get direct image sheaves where no stalk is isomorphic to any other stalk. For instance this happens if all the fibers of $f$ have different cardinalities. I think you would usually need the axiom of choice to even define such a map.

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Not in general. For example take the double cover of the unit circle $S^1$ by the connected helix, say $H$, and call the projection $\pi:H\rightarrow S^1$. Then the sections of the direct image of the constant sheaf $\mathcal A$ over any connected open set $U$ of $S^1$ will give the direct product of two copies of the group $A$, while the constant sheaf over $U$ would give only one copy of $A$ as its group of sections.

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  • $\begingroup$ The point is that for U=S^1 you get only one copy of A but for, say connected nonempty U you get two copies of A. $\endgroup$
    – Jan Weidner
    Commented Apr 13, 2011 at 11:16
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As HNuers example shows, the pushforward of a constant sheaf isn't constant anymore. However one also doesn't get arbitrary sheaves this way. The direct image is still a constructible sheaf, which means that your space is a finite disjoint union of locally closed pieces on which the sheaf is locally constant. For example the sheaf in NHuers example is locally constant (but not constant!) on the whole space. The pushforward of a vector space on a point to some space in the example of K.J. Moi would be constant on the point and its complement.

Edit: Of course one needs assumptions here. For example everything works for pushforward along morphisms of complex algebraic varieties (sheaves considered in analytic topology!).

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When the map is well behaved, say a locally trivial fibration over a base $B$ with fiber $F$, then the direct image of a constant sheaf will be locally constant (all of the fibers of this sheaf will be equal to the cohomology of the constant sheaf on $F$), but won't be constant in general. The obstruction is the monodromy, the defect of compatibility between all local trivializations along a loop in the base space.

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  • $\begingroup$ What is a constant if not the sheaf of locally constant functions? $\endgroup$
    – bavajee
    Commented Apr 13, 2011 at 16:52
  • $\begingroup$ Sorry, I meant to write: What is a constant sheaf, if not a sheaf of locally constant functions? $\endgroup$
    – bavajee
    Commented Apr 13, 2011 at 16:53
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    $\begingroup$ @bavajee: "Locally constant sheaf" is not the same as "sheaf of locally constant functions", in the same way that a covering space is a local homeomorphism but not a homeomorphism on connected components. Actually, the second is an example of the first, if you construct the sheaf of maps from the base to the covering space. This is, basically, what ACL is saying in his answer. $\endgroup$
    – Ryan Reich
    Commented Apr 13, 2011 at 17:32
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    $\begingroup$ @Ryan: I see. Constant sheaf = sheaf associated to constant presheaf. Locally constant sheaf = sheaf that is locally isomorphic to a constant sheaf. Thanks. $\endgroup$
    – bavajee
    Commented Apr 13, 2011 at 20:02
  • $\begingroup$ Does this result also hold for direct image with compact support? More specifically: if $f:X\to Y$ is a locally trivial fibration between locally compact spaces, is it true that $R^q f_! A_X$ is a local system on Y? The stalks seem to suggest that this is the case, since we know $(R^q f_! A_X)_y = H^q_c(f^{-1}(y), A) = H^q_c(F, A)$ for all $y \in Y$. $\endgroup$
    – Eduardo de Lorenzo
    Commented Jul 26, 2021 at 13:50

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