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.