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.)

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.

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.

EDIT: Based on very helpful comments from Alec Rhea and Qiaochu Yuan I am adding some specification on objects and morphisms, hoping that this clarifies the idea behind these categories. I have also reflected these comments in the actual question.

I am constructing two categories, $\bf\mbox{ T-Poly}$ and $\bf Poly$, and a forgetful functor $U:\bf\mbox{ T-Poly}\longrightarrow{\bf Poly}$. If size issues do not stand in the way, these categories are groupoids (I have looked up several definitions of groupoids, and it seems they all assume small categories; which I am not sure about for $\bf\mbox{ T-Poly}$ and $\bf Poly$)

The category $\bf\mbox{ T-Poly}$ has finite disjoint unions of triangulated, labelled polygons as objects; labelled with three colours. These polygons are meant to be "rubber sheet" polygons, i.e. you can stretch and change angles. See the example in the first diagram. 

The morphisms are breaking the polygons down into smaller ones (the "cut" respects the triangulation); or translating or rotating them; or gluing them together (along common edges) into larger ones. The only operation that is not admissible is "flipping" the objects, i.e. changing their orientation. I have added two diagrams at the bottom to clarify my idea behind it with two more examples $h\circ g\circ f: C\rightarrow E$ and $q\circ p: C\rightarrow E$. Maybe the last example is the most important one, as it clarifies how broad the definition of morphisms is meant to be.

These morphisms compose, and the identity morphism leaves everything unchanged. Every morphism is an isomorphism. Another way of thinking about these morphisms, they define an equivalence relation among the objects.

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 helpful tool to study some combinatorial questions. For example, you can define a simple boundary parity functor (parity of 2-coloured 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); translating or rotation them; 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.)

What could be the left adjoint of $U$? Based on the helpful comments from Alex Rhea and Qiaochu Yuan, the left adjoint might not exist at all. Because (if size issues do not stand in the way?), $\bf\mbox{ T-Poly}$ and ${\bf Poly}$ could be groupoids. A left adjoint to $U$ would require them to be equivalent, which is not the case. So I feel my questions might come down to a question of size.

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.