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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?

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  • $\begingroup$ I should write an answer later, but the sheafification as a subsheaf here is extremely unique. $\endgroup$
    – Hoot
    Dec 5, 2015 at 20:39
  • $\begingroup$ I would love it if you also wrote an answer. $\endgroup$
    – D_S
    Dec 6, 2015 at 1:20

2 Answers 2

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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 $i(Im(\alpha))=i'(Im(\alpha '))$.

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  • $\begingroup$ 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, 2015 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, 2015 at 19:48
  • $\begingroup$ Thank you for the excellent answer. $\endgroup$
    – D_S
    Dec 6, 2015 at 1:02
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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).

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    $\begingroup$ I don't see how your answer try to answer the O.P's question. Can you elaborate? $\endgroup$
    – Babai
    Dec 6, 2015 at 7:09

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