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I have done the following problem but I wonder whether there is a generalization of the statement for the usage of direct limit. I will denote $lim$ as direct limit with the open set containing $p$.

Let $F'\to F$ be an injective sheaf morphism. Then I have $(F/F')_p\cong (F_p/F'_p)$.

I do not care for the sheafification at the moment as there is no difference to distinguish the stalk on sheaf or pre-sheaf as they are the same through connoical isomorphism on stalk level.

Q: Is there a generalization of about direct limit usage? (i.e. $lim(\frac{A}{B})=\frac{lim A}{lim B}$ obvious? Sounds like quotient of the limit exists if each limit exists. However it does not make sense unless I can make sure the sequence of B is always a subset of any sequence of $A$. In particular, I wish to say if I have another direct limit system $lim'$ such that $lim'B\cong lim B$, then I conclude $\frac{lim A}{lim'B}\cong\frac{lim A}{lim B}\cong lim(\frac{A}{B})\cong lim'(\frac{A}{B})$. This might be the another version of freshman's dream? )

I am too lazy to draw a lot of diagram. I want to think an intuitive way of direct limit for this. I hope there is an obvious way to see this.

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  • $\begingroup$ Taking stalks is exact by definition, hence this holds. $\endgroup$
    – MooS
    Sep 27 '17 at 18:27
  • $\begingroup$ @MooS The direct limit functor is exact. That is known. But I do not think this answers my question in a direct manner. You have to realize exact sequence first and then take the direct limit. What if I replace the limit with some other limit and suppose the answer is the same, how do I realize such an incidence. There should be a more obvious and general way to interpret the statement. $\endgroup$
    – user45765
    Sep 27 '17 at 21:16
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    $\begingroup$ Whenever a functor on an abelian category is exact, it commutes with taking quotients. It is as simple as that. $\endgroup$
    – MooS
    Sep 28 '17 at 6:54
  • $\begingroup$ @MooS I see. Thanks a lot. $\endgroup$
    – user45765
    Sep 28 '17 at 17:00
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Let $I$ be a directed set, and let $\{A_i\}, \{B_i\}, \{C_i\}$ be directed systems of abelian groups indexed by $i$. For each $i \in I$, let

$$\{A_i\} \rightarrow \{B_i\} \rightarrow \{C_i\} $$

be an exact sequence of directed systems. By this I mean a collection of exact sequences $ A_i \rightarrow B_i \rightarrow C_i$ indexed by $I$, which are compatible with the homomorphisms $A_i \rightarrow A_j, B_i \rightarrow B_j,C_i \rightarrow C_j$ for all $i \leq j$.

The universal property of direct limits induces a sequence of abelian groups

$$\varinjlim A_i \rightarrow \varinjlim B_i \rightarrow \varinjlim C_i$$

which by a standard exercise is exact e.g. see here Why do direct limits preserve exactness? .

In particular, if you started with a short exact sequence

$$0 \rightarrow \{A_i\} \rightarrow \{B_i\} \rightarrow \{C_i\} \rightarrow 0$$

then for each $i$, you have $B_i/A_i \cong C_i$, and the above results lets you say that

$$\varinjlim B_i / \varinjlim A_i \cong \varinjlim C_i \cong \varinjlim B_i/A_i$$

In your case, $I$ is the collection of open neighborhoods of $p$ ordered by reverse inclusion ($U \leq V$ if and only if $U \supseteq V$).

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  • $\begingroup$ The direct limit is exact functor. I am aware of that and I knew that proof as well. Should not there be a more natural obvious construction other than exact sequence? In addition to this, the quotient of direct limit is upto isomorphism. What if I take a different limit with respect to $A_i$? The final answer is the same. But why and will you be offering me a natural reason? $\endgroup$
    – user45765
    Sep 27 '17 at 21:14
  • $\begingroup$ What is more natural than an exact sequence? Saying a sequence is short exact is literally saying the thing in the middle mod the thing on the left is the thing on the right. $\endgroup$
    – D_S
    Sep 28 '17 at 21:01
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    $\begingroup$ Do you understand the principle of of a direct limit being "unique up to unique isomorphism?" This is what makes taking a different direct limit of the same directed system not really change anything. If you draw a diagram and work with different direct limits $$(\varinjlim A_i)', (\varinjlim B_i)', (\varinjlim C_i)'$$ then you are immediately in possession of a corresponding exact sequence $$(\varinjlim A_i)' \rightarrow (\varinjlim B_i)' \rightarrow (\varinjlim C_i)'$$ $\endgroup$
    – D_S
    Sep 28 '17 at 21:06
  • $\begingroup$ OK. I see your point. I am more leaning towards direct construction of a directed system and show it is indeed the direct limit though I am using two different directed limits. $\endgroup$
    – user45765
    Sep 28 '17 at 23:10

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