direct limit presheaf being sheaf on Noetherian space Let $X$ be a noetherian topological space and $\{\mathcal{F}_i\}$ be a directed system. Assuming $U\mapsto \varinjlim \mathcal{F}_i(U)$ is a presheaf, I am trying to see that this is a sheaf on $X$.
Let $s,t\in \mathcal{G}(U)=\varinjlim \mathcal{F}_i(U)$. We have natural maps $\tau_i(U):\mathcal{F}_i(U)\rightarrow \mathcal{G}(U)$ for each $i$. Given $s,t\in \mathcal{G}(U)$ we have $s=\tau_i(x_i), t=\tau_j(x_j)$ for some $x_i\in \mathcal{F}_i(U)$ and $x_j\in \mathcal{F}_j(U)$.
Suppose we are given an open cover $\{U_l\}$ such that $s|_{U_l}=t|_{U_l}$ for all $l$. Then we have following commutative diagram 

from this question. So, we have  $$s|_{U_l}=\Phi_l(s)=\Phi_l(\tau_i(U))(x_i)=\tau_i(U_l)(x_i|_{U_l}).$$
So, $s|_{U_l}=t|_{U_l}$ implies $$\tau_i(U_l)(x_i|_{U_l})=\tau_j(U_l)(s_j|_{U_l})\in \mathcal{G}(U_l).$$
This, means, for some $k$ we have $$f_{ik}(U_l)(x_i|_{U_l})=f_{jk}(U_l)(x_j|_{U_l})$$
which is same as saying
$$f_{ik}(U)(x_i)|_{U_l}=f_{jk}(U)(x_j)|_{U_l}.$$
So, $f_{ik}(U)(x_i),f_{jk}(U)(x_j)\in \mathcal{F}_k(U)$ such that
$$f_{ik}(U)(x_i)|_{U_l}=f_{jk}(U)(x_j)|_{U_l}.$$
As $\mathcal{F}_k$ is a sheaf, we have $$f_{ik}(U)(x_i)=f_{jk}(U)(x_j)$$
which is same as saying $s=t$. 
I have not used that $X$ is Noetherian. What am I missing here 
 A: Identity axiom : Let $U$ be an open subset of $X$, $\{U_l\}$ be an open cover of $X$.
  Let $s,t\in \mathcal{G}(U)=\varinjlim \mathcal{F}_i(U)$ be such that $s|_{U_l}=t|_{U_l}$ for each $l$.
  As $s\in \mathcal{G}(U)$ there exists $i\in I$ such that $s=\tau_i(U)(x_i)$ for some $x_i\in \mathcal{F}_i(U)$,
  similarly $t=\tau_j(x_j)$ for some $x_j\in \mathcal{F}_j(U)$. 
To show that $s=t$ i.e., $\tau_i(U)(x_i)=\tau_j(U)(x_j)$ we show that there exists $k\in I$ with $i\leq k$ and 
  $j\leq k$ such that $f_{ik}(U)(x_i)=f_{jk}(U)(x_j)$.

So, we have $s|_{U_l}=t_{U_l}$ implies $\tau_i(U_l)(x_i|_{U_l})=\tau_j(U_l)(x_j|_{U_l})$.
Fix $l$, we have $$\tau_i(U_l)(x_i|_{U_l})=\tau_j(U_l)(x_j|_{U_l})\in \mathcal{G}(U_l).$$
So, there exists $k$ (depending on $l$) such that 
$$f_{ik}(U_l)(x_i|_{U_l})=f_{jk}(U_l)(x_j|_{U_l}).$$
Which is same as saying 
$$f_{ik}(U)(x_i)|_{U_l}=f_{jk}(U)(x_j)|_{U_l}.$$
So, we have $f_{ik}(U)(x_i),f_{jk}(U)(x_j)\in \mathcal{F}_k(U)$ such that 
$$f_{ik}(U)(x_i)|_{U_l}=f_{jk}(U)(x_j)|_{U_l}.$$
This $k$ may not work for all $l$ in the index set of the open cover $\{U_l\}$. As $X$ is noetherian,
$U$ is quasi compact, so, there exists a finite subcover $\{U_1,\cdots,U_n\}$ of $U$. For each $l\in \{1,2,\cdots,n\}$
there exists $k_l$ such that $$f_{i{k_l}}(U)(x_i)|_{U_l}=f_{j{k_l}}(U)(x_j)|_{U_l}.$$
Let $k\in I$ be such that $k_l\leq k$ for all $1\leq l\leq n$. For this $k$ and for all $l$
we have $$f_{ik}(U)(x_i)|_{U_l}=f_{jk}(U)(x_j)|_{U_l}.$$
As $\mathcal{F}_k$ is a sheaf,  we have $f_{ik}(U)(x_i)=f_{jk}(U)(x_j)$ i.e., $s=t$.
Gluing Axiom: Let $\{U_l\}$ be an open cover of $U$ and $s_l\in \mathcal{G}(U_l)$ be such that $s_l|_{U_l\cap U_m}=s_m|_{U_l\cap U_m}$.
As $X$ is noetherian, there exists a finite subcover say $\{U_1,\cdots,U_n\}$ for $U$. 
For each $1\leq p\leq n$ there exists $x_{i_p}\in \mathcal{F}_{i_p}(U_p)$ such that 
$\tau_{i_p}(U_p)(x_{i_p})=s_p$.
Fix $p,q\in \{1,2,\cdots,n\}$. We have $s_p|_{U_p\cap U_q}=s_q|_{U_p\cap U_q}$ i.e.,
$\tau_{i_p}(U_p)(x_{i_p})|_{U_p\cap U_q}=\tau_{i_q}(U_q)(x_{i_q})|_{U_p\cap U_q}$ i.e., 
$\tau_{i_p}(U_p\cap U_q)(x_{i_p}|_{U_p\cap U_q})=\tau_{i_q}(U_p\cap U_q)(x_{i_q}|_{U_p\cap U_q})$.

There exists $k_{p,q}$ with $i_p\leq k_{p,q}$ and $i_q\leq k_{p,q}$
such that $$f_{i_pk_{p,q}}(U_p\cap U_q)(x_{i_p}|_{U_p\cap U_q})=f_{i_qk_{p,q}}(U_p\cap U_q)(x_{i_q}|_{U_p\cap U_q}).$$
Which is same as saying 
$$f_{i_pk_{p,q}}(U_p)(x_{i_p})|_{U_p\cap U_q}=f_{i_qk_{p,q}}(U_q)(x_{i_q})|_{U_p\cap U_q}.$$
So, we have $f_{i_pk_{p,q}}(U_p)(s_{i_p})\in \mathcal{F}_{k_{p,q}}(U_p)$ such that 
$$f_{i_pk_{p,q}}(U_p)(x_{i_p})|_{U_p\cap U_q}=f_{i_qk_{p,q}}(U_q)(x_{i_q})|_{U_p\cap U_q}.$$
For $k\in I$ with $k_{p,q}\leq k$ for all $p, q$ we have
$$f_{i_pk}(U_p)(x_{i_p})|_{U_p\cap U_q}=f_{i_qk}(U_q)(x_{i_q})|_{U_p\cap U_q}.$$
So, there exists $s\in \mathcal{F}_k(U)$ such that 
$$s|_{U_p}=f_{i_pk}(U_p)(x_{i_p}).$$
We have $$\tau_k(U)(s)|_{U_p}=\tau_k(U_p)(s|_{U_p})=\tau_k(U_p)f_{i_pk}(U_p)(x_{i_p})
=\tau_{i_p}(U_p)(x_{i_p})=s_p.$$

So, we have $\tau_k(U)(s)\in \mathcal{G}(U)$ is such that 
$\tau_k(U)(s)|_{U_p}=s_p$ for all $1\leq p\leq n$. The same is true for any open subset in the open cover. 
Thus, gluing axiom is satsfied.
So, $U\mapsto \mathcal{G}(U)=\varinjlim \mathcal{F}_i(U)$ gives a sheaf structure on noetherian space $X$.
