Are presheaves of constant functions sheaves? - MathOverflow

Are presheaves of constant functions sheaves?

2011-08-26T15:40:37Z

Hey there, I was reading 'An introduction to homological algebra' by Rotman, and on page 279 in the section about sheaves, example 5.64, Rotman gives an example of a constant presheaf $\mathcal{P}$ that's not sheaf, the presheaf of constant real-valued functions on $\mathbb{R}^{2}$. Let the topological space $X = \mathbb{R}^{2}$ and for each $U\subseteq\mathbb{R}^{2}$ define

$\mathcal{P}(U) = $ {$f:U\rightarrow\mathbb{R}\mid$ $f$ is constant}.

If $U\subseteq V$, $\rho_{U}^{V}:\mathcal{P}(V)\rightarrow\mathcal{P}(U)$ is the restriction map $\sigma\mapsto\sigma\mid U$. Now for example let $U=U_{1}\bigcup U_{2}$, where $U_{1}$ and $U_{2}$ are disjoint nonempty sets, define $\sigma_{1}\in P(U_{1})$ by $\sigma_{1}(u_{1})=0$ for all $u_{1}\in U_{1}$, and $\sigma_{2}\in P(U_{2})$ by $\sigma_{2}(u_{2})=5$ for all $u_{2}\in U_{2}$. The overlap condition is vacuous since $U_{1}\bigcap U_{2}=\textrm{Ø}$, but there is no constant function $\sigma\in P(U)$ such that $\sigma\mid U_{i}=\sigma_{i}$, for $i=1,2$ (aka the gluing condition is not satisfied), therefore $\mathcal{P}$ is not a sheaf.

My confusion was with the overlap. How can he apply the gluing condition ff the sets $U_{1}$ and $U_{2}$ don't even overlap? In such a case wouldn't $\mathcal{P}$ satisfy the gluing condition and be a sheaf?

Answer by Ryan Reich for Are presheaves of constant functions sheaves?

2011-08-26T15:59:55Z

As you say yourself, the overlap condition is vacuous and thus automatically true. However, the sheaf condition for presheaves has two parts:

The overlap condition: that $\sigma_i | (U_i \cap U_j) = \sigma_j | (U_i \cap U_j)$. This is true, since the intersection is empty.
The gluing condition: that there is some $\sigma$ such that $\sigma | U_i = \sigma_i$. This is false, since $\sigma_1 = 0$ and $\sigma_2 = 5$, but $\sigma$ must be constant.

The fact that the overlap condition is vacuous means that you can proceed directly to checking the gluing condition, but the gluing condition itself is rarely vacuous.

Basically, you want to avoid thinking that "the sets don't even overlap". Statements in mathematical logic are always either true or false, but not void: if the intersection is empty, then it is still meaningful to discuss it; the sentence doesn't just go away.

Answer by Neil Strickland for Are presheaves of constant functions sheaves?

2011-08-26T16:29:21Z

If we take the question as written then $\mathcal{P}(\emptyset)$ is a singleton and Ryan's answer is valid. However, with that definition I would not call $\mathcal{P}$ the constant presheaf. Instead we should have $\mathcal{P}(U)=\mathbb{R}$ for all $U$. With this definition the case $U=\emptyset$ means that we cannot think of $\mathcal{P}(U)$ as a set of functions. Now the problem is more subtle. Suppose that $U$ is the union of a family of open subsets $U_i$, and we have a family of elements $p_i\in\mathcal{P}(U_i)$ for all $i$, and that $p_i$ and $p_j$ always have the same image in $\mathcal{P}(U_i\cap U_j)$. This just means that the $p_i$ are real numbers, and that $p_i=p_j$ for all $i$ and $j$, so there is a single real number $p\in\mathcal{P}(U)$ that maps to $p_i$ for all $i$. This means that $\mathcal{P}$ has unique gluing and so is a sheaf ... except that we have missed one special case.

The empty set can be covered by the empty family of subsets. (It can also be covered by the family consisting of one subset, namely the empty set, but that is different.) For this family, the gluing condition says that $\mathcal{P}(\emptyset)$ should be the equaliser of two maps between two sets. Each of those sets is the product of no terms, which is a single point. Thus $\mathcal{P}(\emptyset)$ must also be a single point. The first stage of sheafification is to force $\mathcal{P}(\emptyset)$ to be a single point, and then we are back to Ryan's answer.