Suppose we're given sheaves $F,G$ on a space $X$ and a morphism of sheaves (of abelian groups) $\phi:F\to G$.

I want to prove two things :

  1. the presheaf $\ker \phi$, defined by $(\ker\phi)(U):=\ker(\phi_U:FU\to GU)$ is also a sheaf.
  2. $(\ker\phi)_p \simeq \ker(\phi_p)$

I can do 1. by actually procing the axioms for a sheaf with the presheaf $\ker \phi$. But I know this should follow formally from the fact sheafification $\tilde{\,}:Psh(X)\to Sh(X)$ and the forgetful functor or inclusion $i:Sh(X)\to Psh(X)$ form a pair of adjoint functors, $\tilde{}$ being left adjoint to $i$. Because then $i$ preserves limits, and kernels are limits but the precise way to conclude here isn't clear, writing it down is a bit awkward. So $\ker \phi$ is the limit of $F\rightrightarrows G$ (taken in the category of presheaves?) where the arrows are given by $\phi$ and $0$. Saying $i$ commutes with limits, would mean $i (\lim F\rightrightarrows G )\simeq \lim iF\rightrightarrows iG$ but I don't know à priori that this limit is a sheaf, in fact this is what I want to prove so writing $i(\lim...)$ just doesn't make sense, this is I what I mean when I say the setup feels awkard.

I have a similar issue with 2. : writing down an actual map giving the isomorphism is ok, but I feel like this should formally follow from the fact that the functor $_p$ "stalk at $p$" is a filtered colimit and $\ker$ is a limit, these should commute. But I ran into similar issues, when writing it down.

Any help would be deeply appreciated.


1 Answer 1


For 1., there is indeed a general approach: given a category $\mathcal{C}$ and a (full) reflective subcategory $\mathcal{D}$ of it (i.e. the inclusion $i\colon \mathcal{D}\to\mathcal{C}$ is fully faithful and a right adjoint functor), it holds that $\mathcal{D}$ is complete if $\mathcal{C}$ is, and limits in $\mathcal{D}$ are computed in $\mathcal{C}$ (and will then automatically lie in $\mathcal{D}$). For a proof, see e.g. here. This shows that all presheaf limits of sheaves are actually sheaves and compute the sheaf limits.

You can argue the same with a bit less general nonsense, using that limits commute with limits. I will spell out this in extra detail, so I'm sorry if it is overly formal. We need to show that the presheaf kernel $\mathrm{ker}\,\varphi$ of $\varphi\colon F\to G$ satisifies the sheaf condition. The sheaf condition of a sheaf $H$ reads that $$ H(U)\to \prod_{i\in I} H(U_i)\rightrightarrows\prod_{i,j\in I}H(U_i\times_U U_j) $$ is an equalizer diagram for any open cover $U=\bigcup_{i\in I} U_i$. Call this cover $\mathcal{U}$, and call the diagram $\mathrm{D}_\mathcal{U}(H)$, which is a functor $\mathcal{I}\to\mathsf{Ab}$, where $\mathcal{I}$ is the category $$ 0\to1\rightrightarrows2 $$ The morphism $\varphi$ induces a natural transformation $\varphi_*\colon\mathrm{D}_\mathcal{U}(F)\to\mathrm{D}_\mathcal{U}(G)$ and we get even a diagram $$ \mathrm{D}_\mathcal{U}(\mathrm{ker}\,\varphi)\to\mathrm{D}_\mathcal{U}(F)\rightrightarrows\mathrm{D}_\mathcal{U}(G) $$ where on the right the top arrow is $\varphi_*$ and the bottom arrow is the zero morphism. We can consider this latter diagram as a certain functor $A\colon\mathcal{I}\times\mathcal{I}\to\mathsf{Ab}$. Here, we consider for instance $A(0,\bullet)$ to be the diagram $\mathrm{D}_\mathcal{U}(\mathrm{ker}\,\varphi)$, so the first copy of $\mathcal{I}$ encodes the ''horizontal'' arrows in the diagram above, while the second copy of $\mathcal{I}$ encodes the sheaf condition diagrams hidden inside $\mathrm{D}_\mathcal{U}(-)$.

Let $\mathcal{I}_{\geq 1}$ be the category $$1\rightrightarrows 2,$$ and write $A_j=\mathrm{lim}_{\mathcal{I}_{\geq 1}\times\{j\}} A_{\bullet,j}$ for $j\in \mathcal{I}_{\geq 1}$ and $_{j}{A}=\mathrm{lim}_{\{j\}\times\mathcal{I}_{\geq 1}} A_{j,\bullet}$ for $j\in \mathcal{I}_{\geq 1}$. This defines two functors $A_\bullet\colon\mathcal{I}_{\geq 1}\to\mathsf{Ab}$ and $_\bullet A\colon\mathcal{I}_{\geq 1}\to\mathsf{Ab}$. To say that limits commute with limits is to say that $\mathrm{lim}_{\mathcal{I}_{\geq 1}}\, A_\bullet\cong\mathrm{lim}_{\mathcal{I}_{\geq 1}}\, {_\bullet} A$.

We know that $_1 A\cong F(U)$ and $_2 A\cong G(U)$ as $F$ and $G$ are sheaves, so $\mathrm{lim}_{\mathcal{I}_{\geq 1}} {_\bullet} A\cong(\mathrm{ker}\,\varphi)(U)$. We also know that $A_j\cong A(0,j)$ for $j\geq 1$ because kernels are, well, kernels. The statement that $\mathrm{lim}_{\mathcal{I}_{\geq 1}} A_\bullet\cong\mathrm{lim}_{\mathcal{I}_{\geq 1}} {_\bullet} A\cong(\mathrm{ker}\,\varphi)(U)$ is therefore exactly the statement that $$ (\mathrm{ker}\,\varphi)(U)\to \prod_{i\in I} (\mathrm{ker}\,\varphi)(U_i)\rightrightarrows\prod_{i,j\in I}(\mathrm{ker}\,\varphi)(U_i\times_U U_j) $$ is an equalizer diagram, and this shows that $\mathrm{ker}\,\varphi$ is a sheaf.

The argument that $(\mathrm{ker}\,\varphi)_p\cong (\mathrm{ker}\,\varphi_p)$ is analogous, and we will be quicker about it. This time we write $\mathcal{J}$ for the category with opens around $p$ as objects, and maps $U\to V$ are inclusions. Given a presheaf of abelian groups $H$ on $X$, we have $H_p=\mathrm{colim}_{U\in\mathcal{J}^\mathrm{op}} H(U)$. Write $\mathrm{C}_p(H)\colon\mathcal{J}^\mathrm{op}\to\mathsf{Ab}$ for this diagram $\mathcal{J}^\mathcal{op}\to\mathsf{Ab}, U\mapsto H(U)$ that we take the colimit over, and recall our notation $\mathcal{I}$ from above. We have a diagram $$ C_p(\mathrm{ker}\,\varphi)\to C_p(F)\rightrightarrows C_p(G) $$ where one of the right arrows is induced by $\varphi$ and the other is the zero map. Write $A\colon\mathcal{I}\times\mathcal{J}^\mathrm{op}\to\mathsf{Ab}$ for this latter diagram. That filtered colimits commute with finite limits gives us that $$(\mathrm{ker}\,\varphi)_p\cong\mathrm{colim}_{j\in\mathcal{J}^\mathrm{op}}\,(\mathrm{ker}\,\varphi)(j)\cong\mathrm{colim}_{j\in\mathcal{J}^\mathrm{op}}\,\mathrm{lim}_{i\in \mathcal{I}_{\geq 1}}\, A(i,j)\cong \mathrm{lim}_{i\in \mathcal{I}_{\geq 1}}\,\mathrm{colim}_{j\in\mathcal{J}^\mathrm{op}}\, A(i,j)\cong\mathrm{ker}(F_p\rightrightarrows G_p)\cong \mathrm{ker}\,\varphi_p,$$ and this finishes the proof.


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