Equivalence of gluability axiom for sheaves Suppose $F$ is a sheaf on a topological space X. Suppose we have an open set $U\subseteq X$. For simplicity, suppose we have an open cover $U=U_1\cup U_2$. The gluability axiom states that given $f_1\in F(U_1)$ and $f_2\in F(U_2)$ such that $f_1|_{U_1\cap U_2}=f_2|_{U_1\cap U_2}$, then there is some $f\in F(U)$ such that $f|_{U_1}=f_1$ and $f|_{U_2}=f_2$.
I am trying to understand why this statement of the gluability axiom is equivalent to the following sequence being an equalizer: $F(U)\rightarrow F(U_1)\times F(U_2){{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} F(U_1\cap U_2)\times F(U_1\cap U_2)$. So to show that this implies the above axiom, suppose we have $f_1\in F(U_1)$ and $f_2\in F(U_2)$. But don't we needs to know the first map is surjective? What am I missing?
 A: Why do you think the first map needs to be surjective? This is easily seen to be false by example: consider holomorphic functions on $\mathbb C$. The natural map $\mathcal O_{\mathbb C}(\mathbb C) \to \mathcal O_{\mathbb C}(\mathbb C \setminus \left\{0\right\})$ is not surjective since $f(z) = 1/z$ is not in its image, so pairing with the restriction map to any other open set $U \not \ni 0$ you get a map of this type which is not surjective.
The hypothesis (existence of two sections that agree on $U \cap V$) is that we have a pair $(f_u,f_v)$ in the middle spot which equalizes the double arrows (i.e. lies in the difference kernel of the two arrows, if I'm getting my terminology correct). Now the sheaf axiom is equivalent to exactness, i.e. the difference kernel of the double arrows is equal to the image of the first arrow. In other words, the conclusion (existence of a section that glues $f_u$ and $f_v$) is that we can lift the pair to a section on $X$, i.e. the pair is in the image of the first map.
