I am constructing two categories, $\bf\mbox{ T-Poly}$ and $\bf Poly$, and a forgetful functor $U:\bf\mbox{ T-Poly}\longrightarrow{\bf Poly}$.

The category $\bf\mbox{ T-Poly}$ has disjoint unions of triangulated, labelled polygons as objects; labelled with three colours. See the example in the first diagram. The morphisms are breaking the polygons down into smaller ones (the "cut" respects the triangulation); or gluing them together (along common edges) into larger ones. These morphisms compose, and the identity morphism leaves everything unchanged. Every morphism is an isomorphism. The idea of disjoint union is important here - if $f$ breaks down a polygon $A$ into smaller ones, then $B=f(A)$ is the disjoint union of the smaller ones. (This is the two-dimensional case; it can be generalized to $n$ dimensions. I should add, this is the non-oriented case; you can extend it to the oriented case.)

I am constructing $\bf\mbox{ T-Poly}$ because I feel it is a powerful tool to study some combinatorial questions. For example, you can define a simple boundary parity functor (parity of the boundary edges) $Par:\bf\mbox{ T-Poly}\longrightarrow \bf Z_2$. This functor gives a one-line proof of Sperner's Lemma. I will give details of that below.

The second category $\bf Poly$ has disjoint unions of labelled polygons as objects; labelled with three colours. See the example in the second diagram. The morphisms are breaking the polygons down into smaller ones (through adding corresponding edges - I have coloured these added edges in red in the diagram); or gluing them together (along common edges) into larger polygons. Again, every morphism is an isomorphism. As before, the idea of disjoint union is critical here (and you can generalize this category to $n$ dimensions and to the oriented case.)

The forgetful functor $U:\bf\mbox{ T-Poly}\longrightarrow{\bf Poly}$ is simply forgetting the triangulation inside the polygon(s), but it keeps the labelling on the boundary. In fact, whenever a polygon is broken into smaller pieces, $U(f)$ adds the same edges that $f:A\rightarrow B$ was adding. And it removes the common edges when smaller polygons are glued together along common edges into a larger one. I hope the third diagram can clarify what I mean by that. This diagram also shows that $U$ is not faithfull.

What could be the left adjoint of $U$? I seem to be unable to construct it. I have tried mapping an object from $\bf Poly$ to the disjoint union of all corresponding triangulated polygons in $\mbox{ T-Poly}$, but this construction was not functorial. I would be grateful for any suggestions or hints that come to your mind.

Maybe it is helpful for the understanding of these categories if I add the following comment: a critical property of these categories is that the morphisms preserve the "parity of the boundary". Let me make this more precise for $\mbox{ T-Poly}$:

Take any 2-coloured edge on the boundary of $A$, for example $[yg] := $ the edge with a yellow and a green vertex. Any morphism $f:A\rightarrow B$ in $\mbox{ T-Poly}$ adds or removes pairs of edges, so the parity of $[yg]$-edges on the boundary is the same for $A$ and $B$. Another observation is that the parity of the $[yg]$-edges is in fact the same as for the other 2-coloured edges.

This allows you to define the "boundary parity functor" $Par:\bf\mbox{ T-Poly}\longrightarrow \bf Z_2$ which expresses odd parity as $1$ and even parity as $0$. For $f: A\rightarrow B$ define

$$Par:\left\{ \begin{array}{ll} A\mapsto (\sum\, [yg] \mbox {-edges on boundaries of $A$}) \mbox{ mod } 2 \\ \\ f\mapsto \mbox{id}_{\bf Z_2}\\ \\ B\mapsto (\sum\, [yg] \mbox {-edges on boundaries of $B$}) \mbox{ mod } 2 \end{array}\right.$$

Note that $Par$ is only counting 2-coloured edges (ignoring edges with the same colour on both vertices).

This functoriality reduces the proof of Sperner's Lemma to one line. Recall that Sperner's labelling conditions ensure that the boundary of the triangle $A$ in his Lemma has odd parity. Now simply apply a morphism $f$ that breaks the triangle $A$ down into a disjount union $B$ of all the simplices of the triangulation: To keep odd parity, $B$ must contain a 3-coloured simplex (or any odd number of them), because the simplices with 2 colours all have even boundary parity.