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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 $\mathcal F'(U)$.

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    $\begingroup$ That looks like a subpresheaf to me. $\endgroup$ – Lord Shark the Unknown Mar 30 '19 at 12:31
  • $\begingroup$ @Lord So even when we assume that $\mathcal F$ is a sheaf, is there $\mathcal F'$ that satisfies the definition in the quote, but is not a sheaf? $\endgroup$ – zxcv Mar 30 '19 at 12:41
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    $\begingroup$ Think of a presheaf $\cal P$, and its sheafification ${\cal P}^+$. There's a map ${\cal P}\to{\cal P}^+$. This is not necessarily an injection, but in many familiar examples it is. So setting ${\cal F}={\cal P}^+$ and ${\cal F}'={\cal P}$ we see that sheaves can have subpresheaves that are not sheaves. $\endgroup$ – Lord Shark the Unknown Mar 30 '19 at 12:53
  • $\begingroup$ @Lord Thanks! I'll think about an explicit example. $\endgroup$ – zxcv Mar 30 '19 at 13:01
  • $\begingroup$ To give a concrete example of Lord Shark's comment, you can consider a monomorphism of sheaves $\mathcal{F\to G}$ and consider the quotient presheaf as a subpresheaf of the quotient sheaf. In general, it's not a sheaf $\endgroup$ – Max Mar 30 '19 at 13:02
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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.

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