# reperiendi

## Functors as shadows

Posted in Category theory, Math by Mike Stay on 2007 September 15

The last example in the previous post said that the collection of all algebraic gadgets of a given kind and structure-preserving maps between them forms a category. The example given was the category of rings. It’s also true that a category itself is an algebraic gadget with structure (the ability to compose morphisms); a structure-preserving map between categories is called a functor. A functor $F:C\to D$ between categories $C$ and $D$ maps

• objects of $C$ to objects of $D$
• morphisms of $C$ to morphisms of $D$

such that identities and composition are preserved.

One way of thinking about a functor $F:C\to D$ from the category $C$ to the category $D$ is as a “copy” or a “shadow” of $C$ in $D.$ For example, recall that graphs and manifolds both give rise to categories. If we let $C$ be the graph of a cube, where

and $D=\mathbb{R}^2$ be the plane, where

• objects are points of the plane and
• morphisms are paths in the plane,

then a functor $F:C\to D$ is a two dimensional picture of a cube, a shadow cast by the cube. The functor $F$ takes

• each vertex to a point in the plane, and
• each path on the cube to a path in the plane

such that

• composing paths is associative and
• identity paths on the cube map to constant paths in the plane.

These last two requirements imply that the paths in the shadow of the cube are generated by the shadows of each edge. Illustrated are the images of four functors from the cube $C$ to the plane $D$. Figure (a) maps each vertex to a distinct point and each edge to a distinct path. Figure (b) maps the front face of the cube and the back face of the cube to the same square, and maps the edges running from the front to the back to the constant paths at the corners. This is a degenerate functor, i.e. a functor that maps multiple vertices to the same point in $\mathbb{R}^2.$ Figure (c) maps all the points of the cube to the same point and all the edges to the constant path at that point. This functor is totally degenerate, since there’s no way to map to a smaller subset of points in $\mathbb{R}^2.$

Exercise: define a functor $F:C\to D$ such that figure d) is the image of $F$.

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