Given a presheaf $\mathcal{F}$ on a space $X$ and a map $f: X \rightarrow Y$, when does $f_* A(\mathcal{F}) = A(f_* \mathcal{F})$, where $A$ is the associated sheaf/sheafification functor?

Since sheafification is a left adjoint and pushforward is a right adjoint, I don't expect these to always commute. What are some sufficient conditions on $f$ and $\mathcal{F}$ to make this true? (for example, $\mathcal{F}$ quasicoherent, $f$ separated, etc.)

I'd like to construct a map from the tensor product of two quasicoherent sheaves to the direct image of another tensor product by working with tensor products of modules.

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    $\begingroup$ If you want to define a map out from a tensor product, use the universal property. It also holds for sheaves of modules in the obvious way. One never needs to use the explicit constructions of associated sheaves and of tensor products of sheaves. Your question regarding commutation of associated sheaves and direct images is nevertheless interesting. $\endgroup$ Commented Apr 8, 2015 at 0:12

1 Answer 1


The question reduces to a question about sheaves of sets, so I will just work with those (instead of modules or whatever).

The point is that we have the following commutative diagram of right adjoint functors, $$\require{AMScd} \begin{CD} \mathbf{Sh}(X) @>>> \mathbf{Sh}(Y) \\ @VVV @VVV \\ \mathbf{Psh}(X) @>>> \mathbf{Psh}(Y) \end{CD}$$ and you are asking what happens when you take the left adjoints of only the vertical arrows. Well, in that case, we get a canonical natural transformation as below, $$\begin{CD} \mathbf{Sh}(X) @>>> \mathbf{Sh}(Y) \\ @AAA \Uparrow @AAA \\ \mathbf{Psh}(X) @>>> \mathbf{Psh}(Y) \end{CD}$$ whose component at a presheaf $\mathscr{F}$ on $X$ is the morphism $a f_* \mathscr{F} \to f_* a \mathscr{F}$ induced by the universal property of $a f_* \mathscr{F}$ applied to the direct image of the universal morphism $\mathscr{F} \to a \mathscr{F}$. In particular, $a f_* \mathscr{F} \to f_* a \mathscr{F}$ is automatically an isomorphism if $\mathscr{F}$ is a sheaf on $X$, as you would expect.

Now, consider a sieve $R$ in $X$, i.e. a collection of open subspaces of $X$ that is downward-closed, i.e. if $U' \subseteq U$ and $U \in R$ then $U' \in \mathfrak{U}$ as well. Let $\hat{U} = \bigcup_{U \in R} U$. We can think of $R$ as a presheaf on $X$: $R (U) = 1$ if $U \in R$ and $R (U) = \emptyset$ otherwise. The sheafification of $R$ is easy to compute: it is the sheaf $a R$ such that $(a R) (U) = 1$ if $U \subseteq \hat{U}$ and $(a R) (U) = \emptyset$ otherwise. The direct image $f_* R$ is also a sieve, namely the collection of all open subspaces $V \subseteq Y$ such that $f^{-1} V \in R$. Let $\hat{V} = \bigcup_{V \in f_* R} V$.

Notice that $f^{-1} \hat{V} = \bigcup_{V \in f_* R} f^{-1} V \subseteq \bigcup_{U \in R} U = \hat{U}$, so $a f_* R \to f_* a R$ is an isomorphism if and only if the following condition is satisfied:

  • For every open subspace $V \subseteq Y$, $f^{-1} V \subseteq \hat{U}$ if and only if $V \subseteq \hat{V}$.

The above condition being satisfied for all sieves $R$ in $X$ is something like the topology of $X$ being induced by the topology of $Y$, but it is not really the same. Certainly, if $f : X \to Y$ is the inclusion of a subspace (not necessarily open or closed), then $a f_* R \to f_* a R$ is an isomorphism for all sieves $R$. This also happens if $X$ is the spectrum of a discrete valuation ring and $Y$ is the point – even though the topology of $X$ is not induced by the topology of $Y$ in this case. And, for example, if $f : X \to Y$ is the codiagonal/fold map $Y \amalg Y \to Y$, then one can easily find a sieve $R$ such that $a f_* R \to f_* a R$ is not an isomorphism.

We still haven't really addressed the general case of a presheaf instead of a sieve. Things are more complicated here, but what is still true is that if $f : X \to Y$ is the inclusion of an open subspace then $a f_* \mathscr{F} \to f_* a \mathscr{F}$ is always an isomorphism – this is more or less obvious. On the other hand, bad things can happen if $f : X \to Y$ is the inclusion of a non-open subspace: for example, if $f : X \to Y$ is the inclusion of a point and $\mathscr{F}$ is a constant presheaf on $X$, then $a f_* \mathscr{F}$ is the sheafification of a constant presheaf on $Y$ while $f_* a \mathscr{F}$ is a skyscraper sheaf.

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    $\begingroup$ Thanks for the answer. The example is beautiful! Do you know of any conditions that we may be able to put on the presheaf $\mathcal{F}$ so that we get an isomorphism for some topologically "nice" class of maps? (I'm thinking of separated, quasi compact, etc.) What about the case of an open map which is not an immersion? $\endgroup$
    – Dorebell
    Commented Aug 25, 2015 at 5:16
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    $\begingroup$ There are already counterexamples to your suggestions in the answer. $\endgroup$
    – Zhen Lin
    Commented Aug 25, 2015 at 7:03

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