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 if 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?



Are presheaves of constant functions sheaves?
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.


