6
$\begingroup$

Let $F, G, H$ be sheaves on a topological space $X$, and let $$F \xrightarrow{\alpha} G \xrightarrow{\beta} H $$ be morphisms of sheaves. By definition, $\textrm{Ker } \beta$ is the subpresheaf $U \mapsto \textrm{Ker}(\beta(U))$ of $G$, and it is in fact a sheaf. If $O$ is the subpresheaf of $G$ given by $U \mapsto \textrm{Image}(\alpha(U))$, then $\textrm{Im } \alpha$ is defined to be any sheafification of $O$. Let $\theta: O \rightarrow \textrm{Im } \alpha$ be a morphism of presheaves such that the pair $(\textrm{Im } \alpha, \theta)$ satisfies the universal property for sheafification.

I've seen the definition that the sequence above is exact if $\textrm{Ker } \beta = \textrm{Im } \alpha$, but I'm confused as to what this really means. I am aware that the sheafification $(\textrm{Im } \alpha, \theta)$ may be chosen to literally be a subsheaf of $G$ (from the fact that $\theta$ preserves an isomorphism on the stalks, and a morphism of sheaves is injective on the sections if and only if it is injective on the stalks). But even so, any sheafification of $O$ which is a subsheaf of $G$ need not be uniquely determined (as far as I can see). Can anyone clarify what it means for a sequence of sheaves to be exact?

$\endgroup$
  • $\begingroup$ I should write an answer later, but the sheafification as a subsheaf here is extremely unique. $\endgroup$ – Hoot Dec 5 '15 at 20:39
  • $\begingroup$ I would love it if you also wrote an answer. $\endgroup$ – D_S Dec 6 '15 at 1:20
3
$\begingroup$

The image of a map is defined to be the sheafification of $Im^{pre}$. By the universal property of the sheafification this sheaf is unique up to unique isomorphism. Now as you already mentioned there is a natural injection $i: Im(\alpha) \rightarrow G$. We can now define the above sequence to be exact if $i(Im(\alpha))=Ker(\beta)$.

Edit: To see why this works assume the two sheafs $Im(\alpha)$ and $Im(\alpha)^{'}$ are both sheafifications of the sheaf $Im^{pre}(\alpha)$(with natural maps $\theta$ and $\theta ^{'}$). If $j:Im^{pre}(\alpha)\rightarrow G$ is the inclusion map and $\phi: Im(\alpha)) \rightarrow Im(\alpha))'$ the unique isomprhism, then $i \theta=j$. But since $i^{'} \theta ^{'}=j$ and $\theta ^{'}=\phi \theta$, by uniqueness we have $i{'} \phi=i$. Thus $i(Im(\alpha))=i'(Im(\alpha '))$.

$\endgroup$
  • $\begingroup$ And $i(Im(\alpha))$ does not depend on the choice of original sheafification $(Im(\alpha), \theta)$ by the universal property. So in other words, I can say that there is a canonical sheafification of $O$ which is a subsheaf of $G$? (which is not the same thing as saying there is a unique subsheaf of $G$ which is a sheafification of $O$, I don't expect that to be true). $\endgroup$ – D_S Dec 5 '15 at 18:50
  • $\begingroup$ Exactly. I edited the answer, it now contains the purely formal argument that shows that the images under the inclusion maps dont depend on the given sheafification. $\endgroup$ – math635 Dec 5 '15 at 19:48
  • $\begingroup$ Thank you for the excellent answer. $\endgroup$ – D_S Dec 6 '15 at 1:02
3
$\begingroup$

Let's take an example, a piece of the Poincare sequence on a smooth manifold $M$

$$ \Omega^k_M \overset{d_k}{\to}\Omega^{k+1}_M \overset{d_{k+1}}{\to} \Omega^{k+2}_M$$

It is exact at the term $\Omega^{k+1}(M)$. This means two things:

First, $d_{k+1}\circ d_k=0$.

Second, for every open subset $U$ of $M$ and $\omega \in \Omega^{k+1}_M(U)$ a $k+1$-differential form with $d \omega = 0$, there exists a covering $V_i$ of $U$ and $\eta_i \in \Omega^k_M(V_i)$ so that $d_k(\eta_i)= \omega_{|V_i}$ ( any closed form is ${\it locally}$ exact).

$\endgroup$
  • $\begingroup$ I don't see how your answer try to answer the O.P's question. Can you elaborate? $\endgroup$ – Babai Dec 6 '15 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.