example of a presheaf

Question

I'm now trying to understand some of category theory, I think I can understand the concept of sheaf but I can not understand the difference between sheaves and presheaves.

I asked someones about it and they told me that it could be helpful if I can find an example of a presheaf that does not be a sheaf in a topology space of 3 points with the discrete topology.

But I have not idea of what this could be. So any help will be very appreciated.

Thanks.

Answer 1

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:

1. for every open set $U\subseteq X$, there is a corrisponding group $\mathcal{F}(U)$,
2. 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

1. $\mathcal{F}(\emptyset)=0$,
2. $\rho_{_{U,U}}$ is the identity map $\mathcal{F}(U)\to\mathcal{F}(U)$,
3. 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:

1. 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 $s=0$,
2. 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):

1. 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.

2. 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.

Answer 2

You are looking for presheaves that are not sheaves, it seems.

Where gluing fails.
One class of presheaves that do not satisfy the gluing axiom (but satisfy the locality axiom) can suggestively be summarized as "presheaves of functions where the functions are required to satisfy some finiteness property." Finiteness can mean bounded, or compact support, or finite support or … Mariano Suárez-Álvarez example is of this type.

Let me give a specific (rather involved) example of this kind: The presheaf of bounded holomorphic functions on a Riemann surface. Namely, let $X$ be a Riemann surface. For an open set $U\subseteq X$, let $\mathcal{B}(U)$ be the $\mathbb{C}$-algebra of bounded holomorphic functions $U\rightarrow \mathbb{C}$. For $U\subseteq V$, consider the usual restriction morphisms $\mathcal{B}(V)\rightarrow\mathcal{B}(U)$. This gives a presheaf which satisfies the locality axiom. This presheaf, however, does not satisfy the gluing axiom.

To see this, let $x\in X$. Take a chart $\phi\colon U \rightarrow V$ around $x$ which is centered at $x$. Since the map $\mathbb{C}\rightarrow \mathbb{C};z\rightarrow \alpha z$ is a biholomorphism iff $\alpha\in\mathbb{C}\setminus\{0\}$, we can without loss of generality assume that the open unit disc $D=\{z\in\mathbb{C}\ \colon \vert z \vert <1\}$ is contained in $V$. Without loss of generality further assume that $D=V$. (If this is not the case, consider $\phi^{-1}(D)$ instead of $U$ and restrict $\phi$ to $\phi^{-1}(D)$.) Next, define the function $ f\colon D\rightarrow \mathbb{C};z\mapsto \frac{1}{1-z}$. Now, let $\{V_i\}$ be an open cover of $D$ such that each $V_i$ is a contained in a centered open disc of radius strictly smaller than $1$. (For instance, for every $n\in\mathbb{N}_{>1}$ you could take any open cover of the open disc of radius $1-\frac{1}{n}$ and then consider the union of all these covers over $\mathbb{N}_{>1}.$) Since $\phi$ is a homeomorphism, the open cover $\{V_i\}$ of $D$ induces an open cover $\{U_i\}$ of $U$.

Now, for each $i$, define the restricted functions $f_i\colon=(f\circ\phi)\restriction_{U_i}$. Since $f$, as well as every transition map of $X$, is holomorphic, all the $f_i$ are holomorphic. By construction of the sets $U_i$, since $f$ is bounded on any disc of radius $<1$, all the $f_i$ are bounded. Again by construction, all the $f_i$ are additionally compatible, i.e. they agree on the overlap of their domains.

If the presheaf $\mathcal{B}$ satisfied the gluing axiom, there would be a section $\hat{f}\in \mathcal{B}(U)$ such that $\hat{f}\restriction_{U_i}=f_i$ for all $i$. By construction, $\hat{f}$ would have to be equal to $f\circ \phi$. However, $f\circ\phi$ is unbounded, since $f$ is.

With slight modifications, we could have replaced $X$ by any finite-dimensional real manifold $M$ and $\mathcal{B}$ by the presheaf of bounded (infinitely) real-differentiable functions on $M$.

Where locality fails.
Looking for presheaves that do not satisfy the locality axiom by considering presheaves of functions seems like a bad idea. If the restriction morphisms are restrictions of functions, locality is essentially guaranteed.

One example of a presheaf that does not satisfy the locality axiom is the following presheaf of abelian groups.

Consider the sheaf $\mathcal{O}^{\ast}_{\mathbb{C}}$ of non-vanishing holomorphic functions on $\mathbb{C}$ which assigns to an open set $U\subseteq \mathbb{C}$ the abelian group of non-vanishing holomorphic functions $U\rightarrow \mathbb{C}$. Similarly, define the sheaf $\mathcal{O}_{\mathbb{C}}$ of holomorphic functions on $\mathbb{C}$. For every open set $U\subseteq X$, consider the group homomorphism $\operatorname{exp}\colon \mathcal{O}_{\mathbb{C}}(U)\rightarrow \mathcal{O}^\ast_{\mathbb{C}}(U); f \mapsto {\operatorname{exp}} \circ f$. Now, consider the presheaf of abelian groups that assigns to an open set $U\subseteq X$ the quotient group $\mathcal{O}_{\mathbb{C}}^\ast(U)/\operatorname{exp}(\mathcal{O}_{\mathbb{C}}(U))$. The restriction morphisms of this presheaf are those induced by the restriction morphisms of $\mathcal{O}_{\mathbb{C}}^\ast$. That this presheaf does not satisfy the locality axiom can be proven by using the fact that a complex logarithm cannot be defined on the whole complex plane.

Answer 3

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?