I cannot understand the proof of Proposition 3.11 in Sheaf Thoery by Tennison and Hitchin.
If $F,F'$ are subsheaves of a sheaf $G$, then $$ F \leq F' \; \Leftrightarrow \; \forall x \in X \; F_x \subseteq F'_x $$ (as subgroups of $G_x$).
$F \leq F'$ is defined in Definition 3.10 as follows:
If $F,F'$ are subsheaves of $G$, we write $F \leq F'$ iff $\exists$ a morphism $F \rightarrow F'$ (necessarily mono) such that $F \rightarrow F' \rightarrow G$ and $F \rightarrow G$ are equal.
In the above, a sentence after "such that" is my own description of the commutative diagram in the book.
I can understand the $\Rightarrow$ part of the proof.
$\Leftarrow$:Suppose $\forall x \in X \; F_x \subseteq F'_x$. Given $s \in \Gamma(U,F)$ with $U$ open in $X$, there exists a unique section of $F'$ with germs $s_x$ at each $x \in U$; hence there is a morphism : $F \rightarrow F'$ such that the maps $F \rightarrow F' \rightarrow G$ and $F \rightarrow G$ agree on all stalks. By 2.1.10 they are equal so $F \leq F'$.
Propsition 2.1.10 states that two morphisms of sheaves that agree on all stalks are equal. My questions are
1.How can I prove that there exists a unique section $t$ of $F'$ with $t_x = s_x$ at each $x \in U$ rather than sections $t(x)$ with $t(x)_x = s_x$
2.How can I construct a morphism of sheaves from a morphism of stalks. I know the converse is possible by $f_x(s_x) := f(s)_x$.
Any suggestion is appreciated.
Try to proceed in this way: $\forall x \in X$ you get sections $t(x)$, such that $t(x)_x=s_x$, as you said. Hence, on a smaller neighbourhood $U_x \subset U$ of $x$ you have $t(x)_{|U_x}=s_{|U_x}$. Now glue the compatible sections $(t(x)_{|U_x})_{x \in U}$ to get the desired $t \in F'(U)$. This should work.