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?