has a particular geometric form, can, under a mild additional condition, be embedded into an ${{\ensuremath{\widehat F}}}$-coalgebra whose carrier is free and finitely generated.

\[ExEx\] Let $(X,c)$ be an ${{\ensuremath{\widehat F}}}$-coalgebra whose carrier $X$ is a compact subset of a euclidean space $\mathbb R^n$ with $\Delta^n\subseteq X\subseteq\mathbb R_+^n$. Assume that the output map ${{c}_{\text{\rm o}}}$ does not vanish on invariant coordinate hyperplanes in the sense that ($e_j$ denotes again the $j$-th canonical basis vector in $\mathbb R^n$) $$\label{nv}
        \begin{aligned}
        & \nexists\,I\subseteq\{1,\ldots,n\}{{.\kern3pt}}\\
        & I\neq\emptyset,\quad 
            {{c}_{\text{\rm o}}}(e_j)=0,j\in I,
            \quad
            {{c}_{a}}(e_j)\subseteq\operatorname{span}\{e_i\mid i\in I\},a\in A,j\in I.
        \end{aligned}$$ Then there exists an ${{\ensuremath{\widehat F}}}$-coalgebra $(Y,d)$, such that $X\subseteq Y\subseteq\mathbb R_+^n$, the inclusion map $\iota\colon X\to Y$ is a ${{\sf Coalg}({{{\ensuremath{\widehat F}}}})}$-morphism, and $Y$ is the subconvex hull of $n$ linearly independent vectors (in particular, $Y$ is free with $n$ generators).

The idea of the proof can be explained by geometric intuition. Say, we have an ${{\ensuremath{\widehat F}}}$-coalgebra $(X,c)$ of the stated form, and let $\tilde c\colon \mathbb R^n\to\mathbb R\times(\mathbb R^n)^A$ be the linear extension of $c$ to all of $\mathbb R^n$, cf. Lemma \[LinExt\].

(0,0) – (0,2) to \[out=20,in=0\] (3,0) – (0,0); (0,0) – (12,0); (0,0) – (0,5); at (0,1.5) [[$\bullet$]{}]{}; at (-0.5,1.5) [[$e_2$]{}]{}; at 