Understanding a Natural Isomorphism Induced by Graphs and Categories

83 Views Asked by At

In Categories for the Working Mathematician, the author states a fundamental connection between graphs and categories:

Theorem 1. Let $G = \{ A {\rightarrow \atop \rightarrow} G\}$ be a small graph. There is a small category $C = C_G$ with $O$ as set of objects and a morphism $P : G \rightarrow UC$ of graphs from $G$ to the underlying graph $UC$ of $C$ with the following property. Given any category $B$ and any morphism $D : G \rightarrow UB$ of graphs, there is a unique functor $D' : C \rightarrow B$ with $(UD') \circ P = D$.

Within this context, the author continues:

Graphs may be used to describe diagrams. If $G$ is any graph, a diagram of the shape $G$ in the category $B$ may be defined to be a morphism $D : G \rightarrow UB$ of graphs. By the theorem, these morphisms $D$ correspond exactly to functors $D' : C_G \rightarrow B$, via the bijection $D' \mapsto D = UD' \circ P$.

Immediately after, the author states:

This bijection

$$ \mathbf{Cat}(C_G , B) \cong \mathbf{Grph}(G, UB) $$

is natural in $G$ and $B$.

Question: I am confused by this statement. What is the precise natural isomorphism being referred to here? Are the functors of the natural isomorphism from $G \times B$ to $\mathbf{Set}$?

1

There are 1 best solutions below

2
On BEST ANSWER

Well, $G$ is a graph, not a category. When 'considering it as a category', we will get exactly $C_G$, which can also be called the free category over $G$.

The functors are of the form ${\bf Grph}^{op}\times{\bf Cat}\to{\bf Set}$, acting as follows: $$\matrix{(G,B)\mapsto {\bf Cat}(C_G,\,B) &|& (G,B)\mapsto {\bf Grph}(G,\,UB) \\ (\gamma,\beta)\,\mapsto\, \big[f\mapsto \beta\circ f\circ\gamma' \big] &|& (\gamma,\beta)\,\mapsto\, \big[f\mapsto U\beta\circ f\circ\gamma\big] }$$ where $\gamma':C_G\to C_{G'}$ denotes the image of the graph morphism $\gamma:G\to G'$ under the free functor, which can be constructed by the theorem.

$\quad\quad$ by choosing $B=C_{G'}$ and $D=\gamma$