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So I've been reading Mathematical Foundations of Supersymmetry by Carmeli, Caston and Fioresi, and I cannot wrap my head around the proof of lemma 4.2.3 from that book. The first part of the lemma (which is the one I have a problem with) goes as follows:

Let $M$ and $N$ be supermanifolds, and let $\phi := (|\phi|, \phi^*): M \to N$ be a super ringed space morphism. Then for any $U \subset N$ open and $f \in \mathcal{O}_N(U)$, we have $\phi^*(f)(x) = f(|\phi|(x))$.

The second part of this lemma is a simple corollary that says that any super ringed space morphism between supermanifolds is automatically a supermanifold morphism (since the lemma is equivalent to $\phi$ being local).

The authors give a proof as follows:

Suppose that $\phi^*(f)(x) \neq f(|\phi|(x))$. Define $g = f - f(|\phi|(x)) \in \mathcal{O}_N(U)$. Then $\phi^*(g)(x) \neq 0$ by assumption. Furthermore, by construction $g(|\phi|(x)) = 0$, therefore g is not invertible on any neighbourhood of $|\phi|(x)$. However, since $\phi^*$ is a superalgebra morphism and $\varphi^*(g)(x) \neq 0$, we have a contradiction.

I've thought about it for a while now, but I really cannot see where the contradiction comes from. It seems to me that the authors are trying to say that if $g$ is not invertible on any neighbourhood of $|\phi|(x)$, then its image $\phi^*(g)$ cannot be invertible either, but non units can map into units, right?

Konrad Gębik
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1 Answers1

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Let $$ \varphi: C^\infty(\mathbb{R}^n)_0[\theta^1, \dots, \theta^s] \longrightarrow C^\infty(\mathbb{R}^m)_0[\theta^1, \dots, \theta^t] $$ be an $\mathbb{R}$-superalgebra map. Here, $C^\infty(\mathbb{R}^n)_0$ denotes the germs of smooth functions around $0 \in \mathbb{R}^n$. I want to show the locality of $\varphi$.

Let $f \in C^\infty(\mathbb{R}^n)_0[\theta^1, \dots, \theta^s]$.
Define $$ \sigma(f) = \{ \lambda \in \mathbb{R} \mid f - \lambda \text{ is not invertible} \}. $$

Then we have $$ \{ f(0) \} = \sigma(f) \supset \sigma(\varphi(f)) = \{ \varphi(f)(0) \}. $$

Hence, we conclude that $$ f(0) = \varphi(f)(0). $$

hipokaba
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