Is a subsheaf indeed a sheaf? I'm reading Qing Liu's "Algebraic Geometry and Arithmetic Curves." In the book p. 34, it gives the definition of subsheaf.

There is a natural notion of subsheaf $\mathcal F'$ of $\mathcal F$ : $\mathcal F'(U)$ is a subgroup of $\mathcal F(U)$, and the restriction $\rho'_{UV}$ is induced by $\rho_{UV}$.

I see that a subsheaf of a sheaf is a presheaf. But I can't prove that it's a sheaf. I proved the uniqueness condition of sheaf, but I couldn't prove the condition of glueing local sections.
Here's my attempt for proving the glueing local sections condition.
Let $X$ be the given topological space. Let $U$ be an open subset of $X$, $\{U_i\}_{i\in I}$ a covering of $U$ by open subsets $U_i$. Let $s_i\in\mathcal F'(U_i)$, $i\in I$, be sections such that $s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}$. I should prove that there is $s\in\mathcal F'(U)$ such that $s|U_i = s_i$ for each $i\in I$. Since $\mathcal F$ is a sheaf, there exists a section $s\in\mathcal F(U)$ such that $s|U_i = s_i$. But the problem is that $s$ may not be in $\mathcal F'(U)$.
 A: As mentioned in the comments, it is not true that that $\mathcal F'$ is necessarily a subsheaf: the definition in your book only defines a separated subPREsheaf of $\mathcal F$.
Counterexample:
The sheaf $\mathcal C$ on $\mathbb R$ of continuous real-valued functions has as subpresheaf $\mathcal C_b \subset \mathcal C$ the preshaf of bounded continuous functions defined by the requirement that for $U\subset \mathbb R$ open  $$\mathcal C_b(U)=\{f:U\to \mathbb R\vert f \operatorname {is continuous and bounded}\}$$
This is not a sheaf because if we consider the open covering $U_i=(-i,i) (i=1,2,3,\dots)$ of $ \mathbb R$, the functions $f_i\in \mathcal C_b(U_i)$ defined by $f_i(x)=x$ are of course compatibly defined but do not glue to a bounded function in $\mathcal C_b (\mathbb R)$.
Of course they do glue to the continuous (unbounded!) function  $f\in \mathcal C(\mathbb R)$ defined by $f(x)=x$, as they should since $\mathcal C$ is a sheaf.
A: Here is another counterexample, on the discrete space $X=\{a,b\}$. Let $\mathcal F$ be the constant presheaf $U\mapsto\mathbb Z$ if $U\ne\emptyset$ and $\emptyset\mapsto 0$. Here, $\mathcal F$ is not a sheaf, since $0\in\mathcal F(\{a\})$ and $1\in\mathcal F(\{b\})$ both restricts to $0\in\mathcal F(\emptyset)=0$, but they are not a common restriction of anything in $\mathcal F(X)=\mathbb Z$.
Then, the sheafification $\widetilde{\mathcal F}$ is given $\emptyset\mapsto 0$, $\{a\},\{b\}\mapsto\mathbb Z$, and $X\mapsto\mathbb Z^2$, with the natural projection as the restriction. We see that $\mathcal F$ is a sub-presheaf of $\widetilde{\mathcal F}$, but is not a sheaf.
