I'm trying to prove that in a noetherian topological space the following property is satisfied:
Consider a direct system of sheaves and morphisms $\{ \cal{F}_t, f_{ij} \}_t$. Consider the presheaf given by $$ U \mapsto \varinjlim \cal{F}_t (U)$$ I want to prove that this presheaf is a sheaf. I managed to prove that the first axiom os sheaves is satisfied, using the idea given in the comment of my previous question (First axiom of sheaves: in noetherian topological spaces the direct limit presheaf is a sheaf.). I thought that the second axiom would be similar, but when I tried to write it using similar argument I got stuck.
These are my attemps and my ideas:
Let $U$ an open set and $\{ V_j \}_j$ an open covering of $U$. Being $X$ noetherian it is compact, so we can suppose $\{ V_j \}_j = \{ V_1, \dots , V_n \}$. For each $j$ consider $s_j \in \varinjlim \cal{F}_t (V_j)$. Suppose they have the glueing property: ${s_i}_{|V_i \cup V_j}={s_j}_{|V_i \cup V_j} \quad \forall i, j$. We want to show that there exists $s \in \varinjlim \cal{F}_t (U)$ such that $s_{|V_j}=s_j$.
For each $j \quad \exists k_j, s'_{k_j} \in \cal{F}_{k_j} \quad s.t. f_{k_j}(V_j)(s'_{k_j})=s_{k_j}$.
We have ${s_i}_{|V_i \cup V_j}={s_j}_{|V_i \cup V_j}$ so $f_{k_i}(V_i \cup V_j)({s'_{k_i}}_{V_i \cup V_j})=f_{k_j}(V_i \cup V_j)({s'_{k_j}}_{V_i \cup V_j})$.
If I could define $s''_j \in \cal{F}_n(V_j)$ such that $s_j=f_n(V_j)(s''_j)$ and verifying the glueing property, then I could work with the $s''_j$ and the $s_j$ and prove the glueing property for the $s_j$.
EDIT: The difference between the proof of the second axiom of sheaves and the first one is the following: in the first axiom one can use that, if $f_i$ is a morphism of the direct limit such that $f_i(x_i)=0$, then there exists some $j \geq i$ such that $f_{ij}(x_i)=0$, being $f_{ij}$ a morphism of the direct system. This property allows to put together the information of the representative elements of the original $s_{V_i}$, using that the original cover of the open set $U$ can be considered finite, since $X$ is compact.