# Sheafification and the unique decomposition of morphisms

My question is an attempt to understand the proof of lemma 6.17.3 (page 164) in the Stacks Project: http://www.math.columbia.edu/algebraic_geometry/stacks-git/book.pdf

This lemma states:

Let $F$ be a presheaf of sets on $X$. Any map $F\to G$ into a sheaf of sets factors uniquely as $F \to F^{S} \to G$ (when $F^{S}$ denotes the sheafification of $F$).

The part that I'd like to understand is: why is the above map $F^{S} \to G$ unique?

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A sheaf map is determined by the map on each stalks, and the stalk maps induced by $F \to G$ are the same as those of $F^S \to G$. –  Soarer Jul 25 '11 at 17:46

The key point is that another morphism of sheaves $F^S \longrightarrow G$ such that composed with the universal map $F\longrightarrow F^S$ gives you your original $F \longrightarrow G$, would induce the same morphism on all stalks $F_x = F^S_x \longrightarrow G_x$ for every point $x\in X$. Then, as morphisms of sheaves are determined by their induced maps on stalks (Iversen, lemma 2.2 (i)), they would agree.
Choose an open set $U \subset X$, and write $f:F\rightarrow G$. If we have some element $a\in F(U)$, then its image $f_U(a)\in G(U)$ is determined by its stalks since $G$ is a sheaf. So $f$ is determined by its action on stalks. Since $F^S$ is a sheaf, we then get a unique map $F^S \rightarrow G$ determined by that action.