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I am reading Nigel Ray's "Universal Constructions in Umbral Calculus" (1998, published in "Mathematical Essays in Honor of Gian-Carlo Rota", page 344). The article reads:

We define an umbral calculus $(C, r)$ to consist of a coalgebra $C$ in $\mathcal{C}_{R}$ [the category of coalgebras over a commutative ring $R$] and an evaluation functional $r$ in $C^*$ (the dual $\mathrm{Hom}_{R}(C, R)$). For added convenience we insist that $C$ be supplemented, in the sense that it is equipped with a summand of scalars whose projection is a counit $\epsilon: C\to R$, and that $r$ acts on this summand as an identity.

I am completely lost beginning at "that $C$ be supplemented". What is meant here by scalars? Are scalars elements of the ring $R$? What is the summand of scalars? Summand with respect to what operation, what objects are being summed? Is it an operation? An object? A subset? What does it mean for a "summand" to have a "projection" that is a map from $C$ to $R$? Projection by what map? This terminology is neither explained nor used neither before nor after, and I wasn't able to google what is a "supplemented coalgebra".

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    $\begingroup$ I think “summand of scalars” means something like $C=R\oplus C’$. Think of how a polynomial algebra in its degree zero component has a copy of the coefficient field. $\endgroup$
    – Sam Hopkins
    Commented Mar 1 at 12:49
  • $\begingroup$ What would it mean, then, for this subspace to have the counit, a map from $C$ to $R$, as a projection? Something like "counit acts as identity on $R$ injected in $C$"? $\endgroup$
    – Daigaku no Baku
    Commented Mar 1 at 12:51
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    $\begingroup$ Yes, projection means it should be the identity on the $R$ component of $C$. $\endgroup$
    – Sam Hopkins
    Commented Mar 1 at 12:59
  • $\begingroup$ So, do I understand it correctly that here "supplemented coalgebra" means just "coaugmented coalgebra"? $\endgroup$
    – Daigaku no Baku
    Commented Mar 3 at 21:52
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    $\begingroup$ The coaugmentation $u:R\rightarrow C=R\oplus C'$ is the inclusion in the first direct summand since $\varepsilon\circ u=\text{id}$. $\endgroup$
    – F Zaldivar
    Commented Mar 4 at 5:12

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