Do Darboux real valued functions form a sheaf? Do real valued Darboux functions   satisfy the glueing and locality axioms?
 A: Your statement would be: if a function $f\colon D \to \mathbb{R}$ ( $D$ open subset of $\mathbb{R}$) is locally Darboux then it is a Darboux function. That is true.
We have to show that the image of any interval $J \subset D$ is connected. Now there exists a cover of $D$ by open intervals $(I_i)_i$ such that $f_{|I_i}$ is Darboux for every $i$. Consider only those $i$ such that $J\cap I_i\ne \emptyset$. The sets $f(J\cap I_i)$ are intervals and $f(J) = \cup_i f(J\cap I_i)$. Assume that $f(J)$ is not an interval, that is, $f(J)$ is not connected. Since $f(J \cap I_i)$ are intervals, it follows that there exists a partition into two parts of the indexes $i$ so that for $i$, $i'$ in different parts the segments $f(J \cap I_i)$, $f(J\cap I_{i'})$ do not intersect. We conclude that $J\cap I_i$, $J \cap I_{i'}$ do not intersect. It follows that $J$ is not connected, contradiction. 
Note: considering $J$ a closed interval, we can reduce to a finite cover of $J$ that is nice. In other words, we can find a sequence of intermediate points 
$a= a_0 < a_1< \ldots < a_n = b$ and open intervals $I_i \supset [a_i, a_{i+1}]$ such that $f_{|I_i}$ is Darboux. Take now $c$ between $f(a) = f(a_0)$ and $f(b) = f(a_n)$. There exists $i$ so that $f(a_i) \le c < f(a_{i+1})$ ( consider the largest $i$ such that $f(a_i) \le c$). Then $c$ is achieved between $a_i$ and $a_{i+1}$.
