Suppose $L$ is a lattice (free abelian group) and $\sigma$ is a (pointed) spanning rational cone in $L\otimes\mathbb Q$. Then $M=L\cap \sigma$ is a monoid with $M^{gp}=L$. A monoid of this form is called a *toric monoid*. Toric monoids are are precisely the finitely generated, commutative, integral (cancellative), sharp (unit-free), saturated monoids. Let $TMon$ be the category of toric monoids.

A *face* of a monoid $M$ is a submonoid $F\subseteq M$ so that $a+b\in F$ implies $a,b\in F$. The faces of a toric monoid $\sigma \cap L$ are exactly submonoids of the form $\tau \cap L$, where $\tau$ is a face of the cone $\sigma$.

Suppose $D$ is a diagram of toric monoids in which every morphism is the inclusion of a face, and so that for every pair of monoids $D_i$ and $D_j$ in the diagram, there is a unique maximal common face $D_i\cap D_j$ in the diagram. Let $M$ be the colimit of the diagram in $TMon$. Are the maps $D_i\to M$ inclusions of faces?

My feeling is that this problem should be straightforward, but I've done a good job getting myself confused.

**Problem 1:** What is $M$? Colimits exist in the category of finitely generated commutative monoids, but they're unwieldy. See, for example, the first chapter of William Gillam's notes on log geometry. The inclusion of integral saturated monoids into all commutative monoids has a left adjoint, so the colimit $M$ can be formed by taking the colimit in commutative monoids, and then "integral-and-saturifying" it. In particular, $M$ exists. This description makes it impossible to deal with $M$.

Here is another description which is a bit better, but which I'm still not sure how to handle. Let $L$ be the colimit of the induced diagram of free abelian groups $D^{gp}$. This $L$ is probably free even if you take the limit in the category of abelian groups (rather than free abelian groups). Then $M$ is the image of $\bigoplus D_i\to L$. The problem with this description is that it's hard to keep track of faces once you turn everything into groups.

**Problem 2:** The fact that every pair of monoids in the diagram has a unique maximal common face in the diagram is necessary, but I'm having trouble making use of it. Here is a counterexample where this condition fails.

Let $\def\N{\mathbb N}M_n\subseteq \N^2$ be the submonoid generated by $\{(1,0), (1,1),\dots, (1,n)\}$

Let $f_1,g_1:\N\to M_2$ be given by $f_1(1)=(1,0)$ and $g_1(1)=(1,2)$. Let $f_2,g_2:\N\to M_3$ be given by $f_2(1)=(1,0)$ and $g_2(1)=(1,3)$. Consider the diagram

$$\begin{array}{ccc} & \N & \\ {}^{f_1}\swarrow & & \searrow^{f_2}\\ M_2 & & M_3\\ {}_{g_1}\nwarrow & & \nearrow_{g_2}\\ & \N \end{array}$$

In this case, I'm pretty sure the colimit is $M_6$ with the maps $\begin{pmatrix}1&0\\ 0&3\end{pmatrix}:M_2\to M_6$ and $\begin{pmatrix}1&0\\ 0&2\end{pmatrix}:M_3\to M_6$. These maps are not inclusions of faces.