I encountered an example in a paper telling that $\underline{SM}(\mathbb{R}^{0|1},X)\cong \pi TX $, where $X$ is some fixed ordinary Riemannian manifold, $\pi TX $ is the supermanifold with base manifold $X$ and structural sheaf $\pi(\wedge^\*(TX)^\vee) $ according to my advisor, seen as a functor $\pi TX:=SM(-,\pi TX): SM^{\mathrm{op}} \to Set$, $S\mapsto SM(S,\pi TX)$ where $SM$ stands for the category of supermanifolds. $\underline{SM}(\mathbb{R}^{0|1},X)$ is the functor which acts on objects by $\underline{SM}(\mathbb{R}^{0|1},X)(S)= SM(S\times \mathbb{R}^{0|1},X) $, with the obvious action on morphisms. "$\cong$" means that there is a natural isomorphism between the two functors.


Now, by the theorem that supermanifolds are "affine" we have
$$\underline{SM}(\mathbb{R}^{0|1},X)(S) \cong SAlg\_{\mathbb{R}}(C^\infty(X), C^\infty(S)\otimes C^\infty(\mathbb{R}^{0|1})) $$


where $SAlg\_{\mathbb{R}}$ is the category of super $\mathbb{R}$-algebra in which the morphisms are parity-preserving $\mathbb{R}$-algebra homomorphisms. Let $\theta$ be the odd coordinate of $\mathbb{R}^{0|1}$ and we see that an element in $\underline{SM}(\mathbb{R}^{0|1},X)(S)$ is identified with a super $\mathbb{R}$-algebra morphism $\Phi^\*:C^\infty(X)\to C^\infty(S)\otimes C^\infty(\mathbb{R}^{0|1})\cong C^\infty(S) \oplus C^\infty(S)\cdot \theta $.


Decompose $\Phi^\*$ using the direct sum and we can write $\Phi^\*=f+\phi\theta$ where $f:C^\infty(X)\to C^\infty(S)$ is a super $\mathbb{R}$-algebra morphism and $\phi:C^\infty(X)\to C^\infty(S)$ is a parity-reversing map such that


$$\phi(ab)=\phi(a)f(b)+f(a)\phi(b)= \phi(a)f(b)+(-1)^{p(a)}f(a)\phi(b)$$
for any $a,b\in C^\infty(X)$, where $p(a)$ stands for the parity of $a$ which is always zero.


The paper refers to Deligne & Morgan's *Notes on Supersymmetry (following Joseph Bernstein)* and says that this is the standard description of $\pi TX$ in terms of its $S$-points so we get $\underline{SM}(\mathbb{R}^{0|1},X)\cong \pi TX $, but Deligne & Morgan's description is not clear to me at all. My advisor told me that $\pi TX$ is supposed to be as above and left the rest to me as an exercise.


I have no difficulty understanding $\pi TX$ for sure, but I don't have a clue how a morphism $\varphi: S\to \pi TX$, which can be identified with a super $\mathbb{R}$-algebra morphism $\varphi^\*:\pi(\wedge^\*(TX)^\vee) \to C^\infty(S) $, could be translated to some $\Phi^\*:C^\infty(X)\to C^\infty(S)\oplus C^\infty(S)\cdot \theta$ satisfying the axioms above.


For the simplest case, say $X=\mathbb{R}^1$, we have $\pi(\wedge^\*(TX)^\vee) =C^\infty(\mathbb{R})\oplus C^\infty(\mathbb{R})\cdot dx$ where $dx$ is considered to be odd. From this $\varphi^\*$ gives a parity-preserving $\mathbb{R}$-algebra $\bar f:C^\infty(X)\to C^\infty(S)$ which I guess is the candidate of $f$ (or not?). Put $s:=\varphi^\*(dx)$ and we see that $\varphi^\*$ is determined by $\bar f$ and $s$, as


$$ \varphi^\* (a+bdx)=\bar f(a)+\bar f(b)s.$$


Unless $\phi=\bar f\cdot s$, which doesn't make any sense, I cannot see where $\phi$ could possibly come from. I think I do need some help about this. Thanks in advance.



Parity reversed tangent bundle as a supermanifold
My advisor told me the answer; I was just one step away from it.


A morphism $\varphi:S\to \pi TX$ is determined by the natural transformation between the structural sheaves, which is locally determined, so it suffices to assume that $X$ has coordinates $x\_1,\cdots,x\_n$. Now $C^\infty(\pi TX)=\pi(\wedge^\*(TX)^\vee) $ is finitely generated by $d x\_1,\cdots,dx\_n$ over $C^\infty(X)$, the composition $C^\infty(X)\hookrightarrow \pi(\wedge^\*(TX)^\vee)\to C^\infty(S)$ gives our $f$ and the remaining data of $\varphi^\*:\pi(\wedge^\*(TX)^\vee)\to C^\infty(S)$ is given by the images of $d x\_1,\cdots,dx\_n$.


Let $s^i:= \varphi^\*(dx\_i)$, then $\varphi^\*$ on $dx\_i$'s can be expressed by (with abuse of notation) $\varphi^\*=\sum\_i \partial\_{x\_i}\otimes\_f s^i$ in light of $\left(\sum\_i\partial\_{x\_i}\otimes \_f s^i\right)(dx\_j)= \sum\_i dx\_j(\partial\_{x\_i})\otimes\_f s^i =s^j $.


Now, $\sum\_i \partial\_{x\_i}\otimes\_f s^i$ gives an element in $\text{Der}(C^\infty(X),C^\infty(X))\otimes\_f C^\infty(S)$, which is by definition (see [Wikipedia](https://en.wikipedia.org/wiki/Inverse_image_functor)) a global section of the pullback bundle $f^\*\pi TX$ on $S$. Finally, as $\text{Der}(C^\infty(X),C^\infty(X))\otimes\_f C^\infty(S)\cong \text{Der}\_f(C^\infty(M),C^\infty(N))$, where $\text{Der}\_f$ denotes derivations with respect to $f$, via $W\otimes s\mapsto s(f^\*\circ W)$ with inverse $V\mapsto \sum\_i \partial\_{x\_i}\otimes \_f V(x\_i) $, the above relates $\varphi$ with $(f,\phi)$ bijectively.


