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This question came up when I was reading over some information about sheaves; specifically, that if $\mathscr{F}$ is a sheaf on the topological space $X$, $x\in X$, and $Z\subseteq X$, then $(\mathscr{F}|_Z )_x =\mathscr{F}_x$. I don't know if this is supposed to be trivial, and while it definitely seems to be a desirable property, I didn't see it as being obvious that it is true. After thinking for a while, I came to this conclusion:

Let $\mathfrak{I}$ be a directed category, and $F:\mathfrak{I}\rightarrow\mathfrak{C}$ a covariant functor; for $i\in \mathfrak{I}$, denote $F(i)$ by $A_i$. Fix a particular $I\in\mathfrak{I}$, and let $\mathfrak{J}$ be the full subcategory of $\mathfrak{I}$ satisfying $\mathrm{obj}(\mathfrak{J})=\lbrace j\in \mathfrak{I}\mid \mathrm{Mor}_\mathfrak{I}(I,j)\neq \emptyset\rbrace$. Then $\underrightarrow{\mathrm{lim}}_{i\in \mathfrak{I}}A_i =\underrightarrow{\mathrm{lim}}_{j\in \mathfrak{J}}A_j$.

I believe that this is true (well, I wrote a proof that convinced myself, anyway). If this is correct, it would explain why the statement about the stalks of the sheaves earlier is true. I was wondering if something more general can be said about when a colimit of a subcollection is equal to the colimit of the entire collection? Is there an even more general principle at work here, other than just a property of colimit?

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If you're taking a filtered colimit (like in this case) then I think any cofinal subcollection will do? – Dylan Wilson Nov 21 '11 at 16:15
up vote 4 down vote accepted

Dylan's comment is right. More generally, if $D \colon J \to C$ is a diagram and $L \colon J' \to J$ a functor, then the colimit of D is isomorphic to the colimit of DL if and only if L is (co)final. See Mac Lane, Cats Work, section IX.3.

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Ah, yes. That's exactly the property I used in the proof I wrote. Thanks guys! (Sorry for the not-so-interesting question...) – topspin1617 Nov 21 '11 at 16:23

As for the equation $(F|_Z)_x = F_x$ for sheaves: This is an instance of the following lemma from category theory: If $\alpha,\beta$ are composable functors such that their left adjoints $\alpha^\*,beta^\*$ exist, then $(\alpha \circ \beta)^\* = \beta^\* \circ \alpha^\*$. Now apply this to the direct image functors $f_\*$ and $g_\*$ for the inclusions $f = \{x\} \to Z$ and $g : Z \to X$.

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Is $I$ is a small category then you have $I=\bigcup_{c\in \pi_0(I)} I^{(c)}$ i.e. I is the (disjoint) union of the family $\pi_0(I)$ of its connect components, (full categories) (connection is the equivalent relation on objects generated by "have some arrow betwenn them" ).

Then if in some category $\mathcal{C}$ exist a colimt $X=Colim_I X_i$ then you have that $X=\coprod_{c\in\pi_o(I)} X^{(c)}$ and your assert is flase (in general) if your category isnt connected. If $I$ is connected, then for any subcategory $J\subset I$ the canonical morphism $colim_{j\in J}X_j\to colim_{i\in I}X_i$ is a isomorphism (for any category $\mathcal{C}$ and any diagram whenever exist such colimts) $iff$ for any $i\in I$ the comma category $i\downarrow J$ is connected: this means that isnt empty i.e. exist a arrow of type $\phi: i\to j,\ j\in J$, and for two of these arrow $\phi: i\to j,\ \phi': i\to j'$ exixst a finite sequence of arrow in $J$ : $j\to j_1\leftarrow j_2\ldots j_n\to k$ and arrow's $\phi_k: i\to j_k\ 1\leq k\leq n$ such that the diagram of the $\phi, \phi', \phi_1,\ldots, \phi_n, j\to j_1\leftarrow j_2\ldots j_n\to k $ is commutative.

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Yes, if you saw the comment I wrote already on the other answer, I already realized that the original question wasn't very difficult. I just didn't realize it when I asked. I was still curious about whether things like that could happen more generally, i.e. when a limit/colimit/etc of a subset of objects is enough to determine the same on the whole set, but I'm sorry if that's a silly thing to wonder about. – topspin1617 Nov 21 '11 at 18:28
"Too elementary" really is in the eye of the beholder, isn't it? I'd say the question is fine. MO is ready-made for mathematicians asking for confirmation on graduate-level proofs outside their domain of expertise. I'd also say that not noticing a chapter with the vague title of "special limits" would have especial bearing on one's question is quite forgivable. :-) – Todd Trimble Nov 21 '11 at 18:33
I edit my answere, mine was really a wrong consideration. – Buschi Sergio Nov 21 '11 at 18:54

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