I was reading these questions: What does it mean for a sequence of sheaves to be exact and Exactness can be checked on stalks And I am wondering if I am understanding them correctly, because I'm still new to the topic. And then by rectifications and suggestions in the comments, I will edit the post so it can maybe be a good reference for someone who is starting to learn the concepts, because at least for me, it was a slow start specially with the concepts I will refer to. And then I also have a question about the second post.
First of all, it seems to me that it is most useful to not give a precise definition of the sheafification of a presheaf, but define a sheafification of a presheaf $F$ as the inductive limit $\lim\limits_{\to}{F}$ (this limit exists and we can give a precise example of a sheafification), meaning, the inductive limit of the discrete diagram with $F$ (the diagram is just $F$ and the identity map on $F$). So, we can use the vagueness of this definition to provide a satisfying one for exact sequences of sheaves: $F \overset{\alpha}{\to} G \overset{\beta}{\to} H$. We then define the written sequence to be exact if $Im(\alpha) = Ker(\beta)$, where $Ker(\beta)$ is already a sheaf and (here are the nuances) $Im(\alpha)$ is the canonical sheafification of the presheaf $Im^{pre}(\alpha): U \subset X \mapsto \alpha_{U}(F(U))$ as a subsheaf of $G$,which can be described uniquely in the following way. It is the unique subsheaf of $G$ satisfying the commutative diagram:
$\require{AMScd}$ \begin{CD} Im^{pre}(\alpha) @>i>> G\\ @V j V V @AA k A\\ Im(\alpha) @=Im(\alpha) \end{CD}
Where all the maps are inclusions. Let us see it is well-defined. If you take a sheafification $F'$ of a presheaf $F$, it has an associated morphism $l: F \to F'$, and if we also have a sheaf $G$ and monomorphism $f: F \to G$, by the universal property of a sheafification, there is a unique $u:F' \to G$ which will be a monomorphism in this case, since $f$ is one. We then have that $u$ induces an isomorphism $F' \cong Im(k) \subset G$ which will make $Im(k) =: H$ also a sheafification of $F$. We now have the commutative diagram:
$\require{AMScd}$ \begin{CD} F @>f>> G\\ @V k V V @AA i A\\ H @=H \end{CD}
Where $i$ is the inclusion.
This $H$ is the unique subsheaf of $G$ which is a sheafification of $F$ and satisfying the commutative diagram with the inclusion mapping from $H$ to $G$. To see this take some other subsheaf of $G$, $H'$, $g: F\to H'$ and $i':H' \to G$ where $i'$ is also an inclusion mapping. By commutativity of the diagrams, $i'g = ik = f$. Since $H$ and $H'$ are both sheafifications of $F$, there is an isomorphism $\phi : H \to H'$ such that $\phi k = g$. By uniqueness of $i$ and $i'$ we must have $i' \phi = i$ which implies that $H = i(H) = i'(H') = H'$. This $H$ is then what I previously called the canonical sheafification of $Im^{pre}(\alpha)$ as a subsheaf of $G$ (recall that here $F = Im^{pre}(\alpha)$)
Notice that $H$ is not necessarily the only subsheaf of $G$ that is a sheafification of $F$. It is rather the only subsheaf of $G$ which is a sheafification of $F$ and satisfies the respective commutative diagram when the maps are all inclusions.
My questions now are:
- Is my reasoning correct?
- Does the category of sheaves admit small limits?
I know that the category of presheaves admits small limits, but if we take a limit of a diagram in the category of sheaves, can we guarantee that the limit is a sheaf? I ask this because it seems to me that in the second post Exactness can be checked on stalks it is used that the category does indeed have at least finite limits.
Thank you very much for your time. I appreciate all the effort in advance :)