I'm trying to understand a particular point in the paper The Unipotent Albanese Map and Selmer Varieties on Curves by Kim.
We fix a basepoint $b \in X(L)$ on a curve $X$ over $L$ a characteristic 0 field. We then construct the universal pointed unipotent connection (relative to the fibre functor $e_b$ taking a unipotent connection $\mathcal{V}$ to the fibre at $b$) denoted by$\mathcal{E}$. This is defined by $P^{dr}\times E/U^{dr}$, where $P^{dr}\rightarrow X$ is the canonical right $U^{dr}$ torsor with fibre at $x \in X$ the torsor of de Rham paths from $b$ to $x$ and $E$ is the universal enveloping algebra of $LieU^{dr}$
Now I'm satisfied that the fibre $x^*\mathcal{E}$ is functorially isomorphic to $Hom(e_b,e_x)$, and that with the comultiplication $\Delta$ on $x^*\mathcal{E}$ induced by the universal property of $\mathcal{E}$ we therefore identify $P^{dr}(x)$ with the grouplike elements of $x^*\mathcal{E}$.
Where I'm confused is by the statement following this immediately after. It is claimed that letting $\mathcal{P}:=\mathcal{E}^*$, then $\mathcal{P}$ is the coordinate ring of $P^{dr}$.
Looking at this pointwise, where I'm confused is as follows: the fibre of $P=Spec(\mathcal{P})$ at $x$ will be $ P(x)=Spec(x^*\mathcal{P})$. Then $L$-points of $P(x)$ correspond to $L$-algebra homorphisms $x^*\mathcal{P} \rightarrow L$. As far as I can see then $x^*\mathcal{P}$ corresponds to $L$- algebra homomorphisms $ x^*\mathcal{E}\rightarrow L$.
As to why this should give an equality pointwise of $P(x)$ with $P^{dr}(x)$ I'm unsure. The only thing I can think of is that because we have this comultiplication $\Delta$ on $x^*\mathcal{E}$ which is itself an algebra homorphism to $x^*\mathcal{E} \otimes x^*\mathcal{E}$, we might expect that the other $L$- algebra homomorphisms we are looking at respect this comultiplication. Then, if this were true, the $L$- algebra homomorphisms $x^*\mathcal{P} \rightarrow L$ corresponding to elements of $x^*\mathcal{E}$ must be induced by group like elements of $x^*\mathcal{E}$. But this is a stab in the dark.
Essentially, what's confusing me is why going up to the dual somehow excludes the elements of $Hom(e_b,e_x)$ which aren't tensor compatible. Maybe I'm missing something very obvious.
Thanks for any insight anyone can offer - I'm trying my best to nail this down.
