I am confused with morphisms of supermanifolds. Take a simple example $f:R^{01}\to R^{01}$. By (one of) definition, $f$ is a morphism of superalgebras of functions $C(R^{01})\to C(R^{01})$. Morphisms of superalgebras preserve the grading, I deduced that $f$ have the form $1\mapsto 1, \theta\mapsto x\theta$, i.e. $Hom(R^{01}\to R^{01})=R^1$ (as a set?). But I read from a paper that $Hom(R^{01},R^{01})=R^{11}$. What is going on? Thanks in advance!
2 Answers
You are right that the set of supermanifold morphisms $Hom(\mathbb R^{01},\mathbb R^{01})$ to itself is $\mathbb R^1$. However, one can define for supermanifolds $X,Y$ with $\dim X=0d$ 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 $map(\mathbb R^{01},\mathbb R^{01})=\mathbb R^{11}$.

$\begingroup$ Very interesting! I reread the paper, it says actrually interhom, maybe it your $map$. Should it be $X=R^{0d}$? $\endgroup$– Ma MingCommented Feb 10, 2011 at 1:39

1$\begingroup$ @Ma: Yes, the inner hom is the same thing as map. Which paper are you reading, by the way? $\endgroup$ Commented Feb 10, 2011 at 2:26

1$\begingroup$ As @Ma Ming said, it suffices that $X$ be $0\delta$dimensional for $\operatorname{Maps}(X,Y)$ to be finitedimensional. However, it's not good enough for $Y$ to be $0\delta$dimensional. For example, $\operatorname{Hom}(\mathbb R^{11},\mathbb R^{01}) = \operatorname{Hom}(\mathbb R[\epsilon], \mathcal C^\infty(\mathbb R)[\epsilon])$, where $\epsilon^2 = 0$ and $\epsilon$ is in odd degree; but such a map is any $\epsilon \mapsto f(x)\epsilon$, so this hom space is infinitedimensional $\mathcal C^\infty(\mathbb R)$. $\endgroup$ Commented Feb 10, 2011 at 2:55

1$\begingroup$ Thank you all! Could you give a clue how to calculate the $map$, for example, $map(R^{0d},M)$ for any supermanifold $M$? $\endgroup$– Ma MingCommented Feb 10, 2011 at 11:49
@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 wellknown (and due to Kontsevich, I think), that $map(R^{01},M)$ is the total space of the odd tangent bundle $\Pi TM$ of $M$.
If $U$ is a superdomain of dimension $pq$, then $map(R^{0d},U)=U\times R^{pr+qsps+qr}$ where $(r+1)s=2^{d1}2^{d1}$ 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 0dimensional supersymmetric field theories" by Hohnhold, Kreck, Stolz and Teichner.

$\begingroup$ Clickable arxiv:math/0307303: Kochan and Severa  Differential gorms, differential worms (not a typo, I think!). $\endgroup$– LSpiceCommented Dec 23, 2019 at 1:42