example of a presheaf
Firstly I should warn you that I am just learning about sheaves and presheaves for the first time at the time of writing this answer.
With that out of the way, lets start off with a couple of definitions :
Definition (presheaf) For $X$ a topological space, a presheaf (of groups) $\mathcal{F}$ on $X$ is function such that:
- for every open set $U\subseteq X$, there is a corrisponding group $\mathcal{F}(U)$,
- for every inclusion $V\subseteq U$ of open sets in $X$, there is a group hommomorphism $$\rho_{_{U,V}}:\mathcal{F}(U)\to\mathcal{F}(V),$$
such that
- $\mathcal{F}(\emptyset)=0$,
- $\rho_{_{U,U}}$ is the identity map $\mathcal{F}(U)\to\mathcal{F}(U)$,
- if we have the inclusion $W\subseteq V\subseteq U$ of open sets of $X$, then $\rho_{_{U,W}}=\rho_{_{V,W}}\circ \rho_{_{U,V}}$.
$\color{white}{hi}$
Definition (sheaf) A sheaf on a topological space $X$ is a presheaf that satisfies the following:
- if $\{U_i\}$ is an open cover of an open subset $U\subseteq X$, and if $s\in \mathcal{F}(U)$ such that $s_i|_{U_{i}}=0$ for each $i$, then $s=0$,
- if $\{U_i\}$ is an open cover of an open subset $U\subseteq X$, and if we have $s_i\in\mathcal{F}(U_i)$ for each $i$, with the property that for each $i,j$, we have $s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j}$, then there exists an $s\in \mathcal{F}(U)$ such that $s|_{U_i}=s_i$ for all $i$.
As an example we can consider two different presheaves of groups on $X$, where $X$ is the two-point topological space $\{a,b\}$ with the discrete topology. One of these presheaves will be a sheaf, and the other will not .
We can define one presheaf on $X$ as follows: $$\mathcal{F_1}(\emptyset)=0,\ \mathcal{F_1}(\{a\})=\mathbb{Z},\ \mathcal{F_1}(\{b\})=\mathbb{Z},\ \text{and } \mathcal{F_1}(\{a,b\})=\mathbb{Z}\times\mathbb{Z}, $$ with the following restriction maps \begin{align*} \begin{array}{l l} &\rho_{_{X,X}}:\mathcal{F_1}(\{a,b\})\to\mathcal{F_1}(X)\ &\text{ by }\ (z_1,z_2)\to (z_1,z_2)\\ &\rho_{_{X,\{a\}}}:\mathcal{F_1}(\{a,b\})\to\mathcal{F_1}(\{a\})\ &\text{ by }\ (z_1,z_2)\to (z_1,0)\\ &\rho_{_{X,\{b\}}}:\mathcal{F_1}(\{a,b\})\to\mathcal{F_1}(\{b\})\ &\text{ by }\ (z_1,z_2)\to (0,z_2)\\ &\rho_{_{X,\emptyset}}:\mathcal{F_1}(\{a,b\})\to\mathcal{F_1}(\emptyset)\ &\text{ by }\ (z_1,z_2)\to (0,0)\\ & \rho_{_{\{a\},\{a\}}}:\mathcal{F_1}(\{a\})\to\mathcal{F_1}(\{a\})\ & \text{ by }\ (z_1,0)\to (z_1,0)\\ &\rho_{_{\{a\},\emptyset}}:\mathcal{F_1}(\{a\})\to\mathcal{F_1}(\emptyset)\ & \text{ by }\ (z_1,0)\to (0,0)\\ &\rho_{_{\{b\},\{b\}}}:\mathcal{F_1}(\{b\})\to\mathcal{F_1}(\{b\})\ & \text{ by }\ (0,z_2)\to (0,z_2)\\ &\rho_{_{\{b\},\emptyset}}:\mathcal{F_1}(\{b\})\to\mathcal{F_1}(\emptyset)\ & \text{ by }\ (0,z_2)\to (0,0)\\ &\rho_{_{\emptyset,\emptyset}}:\mathcal{F_1}(\emptyset)\to\mathcal{F_1}(\emptyset)\ & \text{ by }\ (0,0)\to (0,0). \end{array} \end{align*} It turns out that this is a presheaf (which can be seen by just checking each part of the definition). It also turns out that this is a sheaf (we now check the two parts of the sheaf definition):
We will just verify part (1.) of the sheaf definition for a single open cover (of an open set), however all the other open covers are just as straightforward to check. We consider the open cover $\{\{a\}, \{b\}\}$ of $X$. Now for any $(z_1,z_2)\in\mathcal{F_1}(X)=\mathbb{Z}\times\mathbb{Z}$, if $(z_1,z_2)|_{\{a\}}=z_1=0$ then we must have $z_1=0$. Similarly if $(z_1,z_2)|_{\{b\}}=z_2=0$ then we must have $z_2=0$. Therefore in order for $(z_1,z_2)|_{\{a\}}=0$ and $(z_1,z_2)|_{\{b\}}=0$ we must have $(z_1,z_2)=(0,0)$, as desired.
For part two of the sheaf definition we again just check a single example, but all other cases are similar. If we consider our open cover $\{\{a\}, \{b\}\}$ of $X$, then for any $(z_1,0)\in \mathcal{F_1}(\{a\})\cong\mathbb{Z}$ and any $(0,z_2)\in \mathcal{F_1}(\{b\})\cong\mathbb{Z}$ we have $$ (z_1,0)|_{\{a\}\cap\{b\}}=(z_1,0)|_{\emptyset}=0=(0,z_2)|_{\emptyset}=(z_1,0)|_{\{a\}\cap\{b\}},$$ and we see that there is an element $(z_1,z_2)\in\mathcal{F_1}(X)$ such that $$(z_1,z_2)|_{\{a\}}=(z_1,0)\ \text{ and }\ (z_1,z_2)|_{\{b\}}=(0,z_2),$$ as desired.
We now give an example of a presheaf of groups on $X$ which is NOT a sheaf. This time we define a presheaf on $X$ as follows: $$\mathcal{F_2}(\emptyset)=0,\ \mathcal{F_2}(\{a\})=\mathbb{Z},\ \mathcal{F_2}(\{b\})=\mathbb{Z},\ \text{and } \mathcal{F_2}(\{a,b\})=\mathbb{Z}\times\mathbb{Z}\times \mathbb{Z}, $$ with the following restriction maps \begin{align*} \begin{array}{l l} &\rho_{_{X,X}}:\mathcal{F_2}(\{a,b\})\to\mathcal{F_2}(X)\ &\text{ by }\ (z_1,z_2,z_3)\to (z_1,z_2,z_3)\\ &\rho_{_{X,\{a\}}}:\mathcal{F_2}(\{a,b\})\to\mathcal{F_2}(\{a\})\ &\text{ by }\ (z_1,z_2,z_3)\to (z_1,0,0)\\ &\rho_{_{X,\{b\}}}:\mathcal{F_2}(\{a,b\})\to\mathcal{F_2}(\{b\})\ &\text{ by }\ (z_1,z_2,z_3)\to (0,z_2,0)\\ &\rho_{_{X,\emptyset}}:\mathcal{F_2}(\{a,b\})\to\mathcal{F_2}(\emptyset)\ &\text{ by }\ (z_1,z_2,z_3)\to (0,0,0)\\ & \rho_{_{\{a\},\{a\}}}:\mathcal{F_2}(\{a\})\to\mathcal{F_2}(\{a\})\ & \text{ by }\ (z_1,0,0)\to (z_1,0,0)\\ &\rho_{_{\{a\},\emptyset}}:\mathcal{F_2}(\{a\})\to\mathcal{F_2}(\emptyset)\ & \text{ by }\ (z_1,0,0)\to (0,0,0)\\ &\rho_{_{\{b\},\{b\}}}:\mathcal{F_2}(\{b\})\to\mathcal{F_2}(\{b\})\ & \text{ by }\ (0,z_2,0)\to (0,z_2,0)\\ &\rho_{_{\{b\},\emptyset}}:\mathcal{F_2}(\{b\})\to\mathcal{F_2}(\emptyset)\ & \text{ by }\ (0,z_2,0)\to (0,0,0)\\ &\rho_{_{\emptyset,\emptyset}}:\mathcal{F_2}(\emptyset)\to\mathcal{F_2}(\emptyset)\ & \text{ by }\ (0,0,0)\to (0,0,0). \end{array} \end{align*} Again one can check that this is a presheaf (just use the definition). However, if we try to show that this is a sheaf, then we run into trouble with part (1.) of the sheaf definition (even tho part (2.) actually ends up holding).
To see that part (1.) of the sheaf definition is not true for this presheaf, consider the open cover $\{\{a\}, \{b\}\}$ of $X$. We have that $(0,0,1)\in \mathcal{F_2}(X)$, and that $$(0,0,1)|_{\{a\}}=(0,0,0)=(0,0,1)|_{\{b\}}, $$ however $(0,0,1)\neq (0,0,0)$, which contradicts part (1.) of the sheaf definition.
Finally I should note that a similar example of a presheaf that is not a sheaf is given on the Sheaf Wikipedia page, and I also mention that these same ideas should carry over to when $X$ is a three-point topological space with the discrete topology.
Consider the space $\mathbb R$ and the sheaf of sets $F$ such that for each open set $U$ the set $F(U)$ is the set of constant functions whose integral is at most $1$.
Is it a sheaf?