Morphisms between supermanifolds R^{0|1}→R^{0|1} I am confused with morphisms of supermanifolds. Take a simple example $f:R^{0|1}\to R^{0|1}$. By (one of) definition, $f$ is a morphism of superalgebras of functions $C(R^{0|1})\to C(R^{0|1})$. Morphisms of superalgebras preserve the grading, I deduced that $f$ have the form $1\mapsto 1, \theta\mapsto x\theta$, i.e. $Hom(R^{0|1}\to R^{0|1})=R^1$ (as a set?). But I read from a paper that $Hom(R^{0|1},R^{0|1})=R^{1|1}$. What is going on? Thanks in advance!
 A: You are right that the set of supermanifold morphisms 
$Hom(\mathbb R^{0|1},\mathbb R^{0|1})$ to itself is $\mathbb R^1$. 
However, one can define for supermanifolds $X,Y$ with $\dim X=0|d$ a 
supermanifold $map(X,Y)$ of morphisms from $X$ to $Y$, by $Hom(Z,map(X,Y))=Hom(Z\times X,Y)$ for all supermanifolds $Z$. 
And $map(\mathbb R^{0|1},\mathbb R^{0|1})=\mathbb R^{1|1}$.
A: @Ma: As an answer to your following question:

Could you give a clue how to calculate the $map$, for example, $map(R^{0∣d},M)$ for any  supermanifold $M$?

Take a look at arXiv:math/0307303, where this question is discussed. 
For $d=1$, it is well-known (and due to Kontsevich, I think), that $map(R^{0|1},M)$ is the total space of the odd tangent bundle $\Pi TM$ of $M$. 
If $U$ is a superdomain of dimension $p|q$, then $map(R^{0|d},U)=U\times R^{pr+qs|ps+qr}$ where $(r+1)|s=2^{d-1}|2^{d-1}$ is the graded dimension of $\bigwedge R^d$. This you can check using the definition of $map$ and the characterisation of morphisms of supermanifolds as given in Leites.
Another good source on this subject (for $d=1$), is the paper "Differential forms and 0-dimensional supersymmetric field theories" by Hohnhold, Kreck, Stolz and Teichner.
