# 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 $$\mathcal F'(U)$$.

• That looks like a subpresheaf to me. – Lord Shark the Unknown Mar 30 '19 at 12:31
• @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? – zxcv Mar 30 '19 at 12:41
• 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. – Lord Shark the Unknown Mar 30 '19 at 12:53
• @Lord Thanks! I'll think about an explicit example. – zxcv Mar 30 '19 at 13:01
• 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 – Max Mar 30 '19 at 13:02

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 $$\mathcal F$$.
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.