While reading a book about super Lie groups I came upon the claim that there is a unique morphism $\phi: G \to \mathbb{R}^{0|0}$, where $G$ is any supermanifold (I'm using the superspace definition of supermanifolds, i.e. supermanifolds here are understood to be locally super ringed spaces), but I am having some trouble with coming up with a proof of this fact. Let $\mathcal{O}_G$ be the structure sheaf on $G$, and $\mathcal{O}_{0|0}$ the structure sheaf on $\mathbb{R}^{0|0}$. Let's notice that $\mathcal{O}_{0|0}(\mathbb{R}^{0|0}) \cong \mathbb{R}$ is the only nontrivial ring of sections on $\mathbb{R}^{0|0}$. Obviously the morphism of underlying topological spaces is unique, since $\mathbb{R}^0$ is a single point. However I am struggling with proving that the ring morphism $\phi^*: \mathbb{R} \to \mathcal{O}_G(G)$, which is part of the sheaf morphism, is also unique.
From here, I'll be using real numbers to denote constant functions equal to those numbers. Without loss of generality, we can start by working on one coordinate chart, as we can later glue sections from all the charts together. As such we only really need to prove that there is only one ring morphism $\mathbb{R} \to C^\infty(U) \otimes \bigwedge(V)$, where $U$ is an open subset of $\mathbb{R}^n$, and $V$ is an $m$-dimensional vector space.
So far i managed to prove that $\phi$ is unique on $\mathbb{Q}$. This stems from the fact that as a morphism of rings $\phi^*$ has to preserve the unity, so $\phi^*(1) = 1$ is the constant section equal to 1 with no nilpotent part. Using addition and division we can conclude that $\phi^*(x)$ has to be the constant section equal to x with no nilpotent part for every $x \in \mathbb{Q}$. We also know that $\phi^*(x)(p) = x$ , since in general for a supermanifold morphism $\eta^*(s)(p) = s(|\eta|(x))$. So for an $x \in \mathbb{R} \setminus \mathbb{Q}$ we know that $\phi^*(x) = x + f_I \theta^I$, where $I = (i_1, \dotsc)$ is a multi-index such that $i_k < i_{k+1}$ and $\theta^I = \theta^{i_1}\theta^{i_2}\dotsc$.
This is where I got stuck - I do not know how to prove that the nilpotent part $f_I \theta^I$ is also $0$ for irrational $x$.
