Finite product of supermanifold.

23 Views Asked by At

I am trying to prove that finite product exist in $sMan$ (the category of supermanifolds). the product of supermanifolds that mention is defined as:

Take $M,N$ supermanifolds, $F:\operatorname{sMan}\longrightarrow \operatorname{Set}$ the contravariant functor defined by $F(B):=\operatorname{Hom}(B,M)\times \operatorname{Hom}(B,N)$ and $F(\phi)(\alpha,\beta):=(\alpha\circ \phi,\beta\circ \phi)$ for every $\phi \in \operatorname{Hom} (B_1,B_2)$ and every $(\alpha,\beta)\in F(B_2)$. A representation of $F$ in $sMan$ is called a product of $M$ and $N$, and is denoted by $(M\times N, \theta)$.

For that I try to use the special case for Construction of supermanifolds by Gluing. and for that I try use claim A: If $(Y,\mathcal{O}_Y)$ is a ringed space, then the map $F:Hom((X,\mathcal{O}),(Y,\mathcal{O}_Y))\longrightarrow \left\{(\phi_U)_{U\in \mathcal{U}}\in \prod_{U\in \mathcal{U}}Hom((U,\mathcal{O}_U),(Y,\mathcal{O}_Y)):\phi_U\circ \lambda_{UU'}=\phi_{U'}\right.$ whenever \$U' \subset U \left.\right\}$ given by $F(\phi):=(\phi_U)_{U\in \mathcal{U}}$, where $\phi_U:=\phi_{|(X_{|U})}\circ \lambda_U^{-1}$, is a bijection with $\lambda_{UU'}:(U',\mathcal{O}_{U'})\longrightarrow U_{|U'}$ and $\phi_U:U_{|U'}\longrightarrow (Y,\mathcal{O}_Y)$, so that $\phi_{U'}:(U',\mathcal{O}_{U'})\longrightarrow (Y,\mathcal{O}_Y)$.

If any body has an idea please share it. Or if any body has a good reference for finite product of ringed space share it. Thank.