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Let $\{ \cal{F}_i, \mu_{ij}\}$ a direct system of sheaves and morphisms on a topological space $X$. Define the direct limit os the system $\{ \cal{F}_i, \mu_{ij}\}$ as the sheaf associated to the presheaf given by $U \mapsto lim_\rightarrow \cal{F}_i(U) $. I want to show that it has the following universal property:

Given a sheaf $\cal{G}$ and a collection of morphisms $g_j: \cal{F}_i \longrightarrow \cal{G}$ compatible with the maps of the direct system, then there exists a unique map $\hat f:lim_\rightarrow \cal{F}_i \longrightarrow \cal{G}$ s.t. $\forall i \quad g_i=\mu_i \circ \hat f $, being $\mu_i: \cal{F}_i \longrightarrow lim_\rightarrow \cal{F}_i$ the map associated to the direct limit of the direct system $\{ \cal{F}_i, \mu_{ij}\}$.

What I have thought: the universal property of direct limits is know for abelian groups, so we have that, $\forall U \quad \text{open set}$, $$\quad \exists ! f(U): lim_\rightarrow \cal{F}_i(U) \longrightarrow \cal{G(U)} \quad \text{s.t.} \quad \forall i \quad g_i(U)=\mu_i(U) \circ f(U) $$

If I could conclude from this property the analogue for the associate presheaves, using sheafification the problem would be solved.

My doubt: I want to show that the homomorphisms $f(U)$ conmute with the restrictions $\rho_{|UV}$ of $lim_\rightarrow \cal{F}_i(U)$, but I don't see how to do it. How can I write these restrictions in terms of the restrictions of the abelian groups $\cal{F}_i(U)$?

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Consider a restriction $\mathcal G(U) \to \mathcal G(V)$. The map $$(\varinjlim\mathcal F_\bullet)(U) \to (\varinjlim\mathcal F_\bullet)(V) \to \mathcal G(V)$$ is given by sheafification of the presheaf map $$\varinjlim(\mathcal F_\bullet(U)) \to \varinjlim(\mathcal F_\bullet(V)) \to \mathcal G(V)$$ and this is the unique map given by the morphisms $$\mathcal F_i(U) \to \mathcal F_i(V) \to \mathcal G(V).$$ Similarly the map $$(\varinjlim\mathcal F_\bullet)(U) \to \mathcal G(U) \to \mathcal G(V)$$ is given by sheafification of the presheaf map $$\varinjlim(\mathcal F_\bullet(U)) \to \mathcal G(U) \to \mathcal G(V)$$ and this is the unique map given by the morphisms $$\mathcal F_i(U) \to \mathcal G(U) \to \mathcal G(V).$$ But $$\mathcal F_i(U) \to \mathcal F_i(V) \to \mathcal G(V) \qquad \text{and} \qquad \mathcal F_i(U) \to \mathcal G(U) \to \mathcal G(V)$$ are the same map due to your compatibility conditions.

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  • $\begingroup$ I still can't see it. Let's call $\rho^i_{UV}: \cal{F}_i(U) \longrightarrow \cal{F}_i(V)$ the restrictions associated to the sheafs $\cal{F}_i$ and $\rho^G_{UV}: \cal{G}(U) \longrightarrow \cal{G}(V)$ the restrictions of $\cal{G}$. The applications $g_i$ are morphisms of sheaves, so we have $\rho^G_{UV} \circ g_i(U)= g_i(V) \circ \rho^i _{UV}$. How can I write explicitly the restrictions $\rho_{UV}$ associated to the direct limit sheaf ? $\endgroup$ – user210089 Apr 9 '15 at 20:01
  • $\begingroup$ You can't write them explicitly, they are given by sheafification of a map that's given by a universal property. You'll have to use universal properties to prove that the two compositions in your commutative diagram are identical. $\endgroup$ – Jim Apr 9 '15 at 20:04

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