Presheaves Sheaves Stalks and Sheafification - Examples

Restriction of a continuous function is continuous, so $U\mapsto C^0(U,\mathbb{R})$ is a presheaf. If two continuous functions on $U$ agree on every $U_i$, then they agree pointwise on all of $U$ because the $U_i$ cover $U$.

Let $f_i\in C^0(U_i,\mathbb{R})$ be compatible on overlaps. Define $f:U\to \mathbb{R}$ by $f(x)=f_i(x)$ for any $i$ with $x\in U_i$. Compatibility makes this well defined, and continuity is local on the open cover because $f{\mathrm{\mid }}_{U_i}=f_i$ is continuous for each $i$. Thus [L1] holds.

Restriction preserves local constancy. If two locally constant functions on $U$ agree on an open cover, then they agree pointwise on $U$. If locally constant functions $f_i:U_i\to A$ are compatible on an open cover of $U$, the pointwise glued function $f$ is well defined and locally constant because near any point it agrees with one of the local functions $f_i$. Hence [L1] holds and ${\underset{‾}{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ is a sheaf.

Define ${ϵ}_x:{\underset{‾}{A}}_{\mathrm{l}\mathrm{o}\mathrm{c},x}\to A$ by ${ϵ}_x([U,f])=f(x)$. This is well defined because equal germs agree on some neighbourhood of $x$, hence have the same value at $x$. Every $a\in A$ is the value at $x$ of the constant function $a$ on any neighbourhood of $x$, so ${ϵ}_x$ is surjective. If ${ϵ}_x([U,f])={ϵ}_x([V,g])$, then $f(x)=g(x)$. Since both functions are locally constant, there is a neighbourhood $W$ of $x$ on which $f$ and $g$ are both constantly this common value, so $[U,f]=[V,g]$. Therefore ${ϵ}_x$ is bijective.

For each integer $n\ge 1$, let $U_n:=(-n,n)$ and let $s_n:U_n\to \mathbb{R}$ be the identity function $s_n(x)=x$. Each $s_n$ lies in $C_b^0(U_n,\mathbb{R})$ because $U_n$ is bounded.

If $m\le n$, then $s_n{\mathrm{\mid }}_{U_m}=s_m$, so the family $(s_n)$ is compatible on the open cover $\mathbb{R}={\bigcup }_{n\ge 1}{U}_n$.

Any glued section on $\mathbb{R}$ would have to equal the identity function $x\mapsto x$, because it agrees with each $s_n$ on $U_n$. But $x\mapsto x$ is not bounded on $\mathbb{R}$, so it does not lie in $C_b^0(\mathbb{R},\mathbb{R})$. This violates the gluing requirement in [L1]. Therefore the stated presheaf is not a sheaf.

By [F1], ${\underset{‾}{A}}_{\mathrm{p}\mathrm{r}\mathrm{e}}$ is a presheaf. On the disjoint cover $U=U_1\cup U_2$, choose the local sections $s_1:=a\in {\underset{‾}{A}}_{\mathrm{p}\mathrm{r}\mathrm{e}}(U_1)$ and $s_2:=b\in {\underset{‾}{A}}_{\mathrm{p}\mathrm{r}\mathrm{e}}(U_2)$. Because $U_1\cap U_2=\mathrm{\varnothing }$, there is no overlap condition to check, so the pair is compatible.

A glued section over $U$ would have to be an element $s\in A$ whose restriction to $U_1$ is $a$ and to $U_2$ is $b$. But every restriction map is the identity, so this would force $s=a=b$, contradicting the choice $a\mathrm{\ne }b$. Therefore ${\underset{‾}{A}}_{\mathrm{p}\mathrm{r}\mathrm{e}}$ is not a sheaf.

The sheafification of ${\underset{‾}{A}}_{\mathrm{p}\mathrm{r}\mathrm{e}}$ records exactly the data obtained by gluing constant local sections on an open cover, which is the same as an $A$-valued locally constant function. By [L1], that sheaf is ${\underset{‾}{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$, so [F2] identifies the locally constant sheaf as the sheafification of the constant presheaf.

If an open set $U$ does not contain $x_0$, then every section of ${\mathcal{S}}_{x_0,A}(U)$ is the unique element $\ast$, so locality and gluing are trivial. If $x_0\in U$ and $U={\bigcup }_i{U}_i$, then at least one $U_i$ contains $x_0$. Compatibility forces all sections on such $U_i$ to be the same element of $A$, and every $U_j$ not containing $x_0$ contributes only the unique section $\ast$. Thus there is a unique glued section on $U$. Therefore [L1] holds and ${\mathcal{S}}_{x_0,A}$ is a sheaf.

For the stalk at $x_0$, every neighbourhood of $x_0$ has section set $A$ and every transition map is the identity on $A$. Hence the colimit in [F1] is canonically $A$.

If $y$ has an open neighbourhood $V$ with $x_0\mathrm{\notin }V$, then every smaller neighbourhood of $y$ inside $V$ also has section set $\{\ast \}$. Hence the stalk diagram is eventually constant at $\{\ast \}$, so [F1] gives $({\mathcal{S}}_{x_0,A}{)}_y\cong \{\ast \}$.

If $V\subseteq W$, then every $V_i\subseteq W$ has the unique section $\ast$, and the only possible glued section on $V$ is again $\ast$. If $V\subseteq ̸W$, choose $x\in V\setminus W$. Some cover member $V_i$ contains $x$, so $V_i\subseteq ̸W$ and ${\mathcal{E}}_W(V_i)=\mathrm{\varnothing }$. Hence there is no compatible family of local sections on this cover. In either case the gluing and uniqueness clauses in [L1] are satisfied.

Restriction maps are forced: from $\{\ast \}$ to $\{\ast \}$ they are the identity, and into $\mathrm{\varnothing }$ there is only the empty function from $\mathrm{\varnothing }$. Thus ${\mathcal{E}}_W$ is a presheaf, and step 1.1 shows it is a sheaf.

Let $U_1=S^1\setminus \{(-1,0)\}$ and $U_2=S^1\setminus \{(1,0)\}$. These open arcs cover $S^1$. On each $U_i$ choose a continuous argument ${\theta }_i:U_i\to \mathbb{R}$ with $e^{2\pi i{\theta }_i(z)}=z$. Therefore the identity map ${\mathrm{id}}_{S^1}$ restricts to sections in the objectwise image presheaf on both $U_1$ and $U_2$.

On the overlap $U_1\cap U_2$, both local sections are equal to the same target section ${\mathrm{id}}_{S^1}{\mathrm{\mid }}_{U_1\cap U_2}$, so they are compatible.

Suppose ${\mathrm{id}}_{S^1}$ lay in the global objectwise image. Then there would be a continuous $f:S^1\to \mathbb{R}$ with $e^{2\pi if(z)}=z$ for every $z\in S^1$. Writing $z=e^{2\pi it}$ with $t\in [0,1]$, the function $g(t):=f(e^{2\pi it})-t$ is continuous and integer valued, hence constant. So $f(e^{2\pi it})=t+n$ for some fixed integer $n$. Evaluating at $t=0$ and $t=1$ gives two values of $f$ at the same point $1\in S^1$, namely $n$ and $n+1$, a contradiction. Thus ${\mathrm{id}}_{S^1}$ is not in the global objectwise image.

Steps 1.1 to 3.1 give compatible local sections in the image presheaf that do not glue globally, so the objectwise image is not a sheaf and hence not a subsheaf by [F1]. By [L1], its sheafification is the correct image sheaf.

The function $g$ vanishes on the neighbourhood $(-1,1)$ of $0$, so its germ at $0$ equals the germ of the zero function by [F1]. But $g(2)=1$, so the two functions are not equal globally.

If $f_x=g_x$ for every $x\in U$, then [F1] gives for each $x$ an open neighbourhood $V_x\subseteq U$ on which $f{\mathrm{\mid }}_{V_x}=g{\mathrm{\mid }}_{V_x}$. The sets $V_x$ cover $U$, so $f$ and $g$ agree at every point of $U$ and hence are equal.

This pointwise-germ criterion is the section-level instance behind [L1]: one stalk does not determine a section, but all stalks together do.

By [F1], there is no open set to consider except $\mathrm{\varnothing }$, so the whole presheaf data of $\mathcal{F}$ is just the set $\mathcal{F}(\mathrm{\varnothing })$ together with its identity restriction map.

By [L1], this set $\mathcal{F}(\mathrm{\varnothing })$ is a singleton. Therefore every sheaf on the empty space has exactly one section over its only open set.