Franklin's Notes


If $\mathsf{C}$ and $\mathsf{D}$ are categories , then a (covariant) functor $F:\mathsf{C}\to\mathsf{D}$ assigns an object $Fc\in\mathsf{D}$ for each object $c\in\mathsf{C}$, and a morphism $Ff:Fc\to Fc'$ in $\mathsf{D}$ for each morphism $f:c\to c'$ in $\mathsf{C}$. A functor must also satisfy the following two functoriality axioms:

1. If $f,g$ are composable morphisms of $\mathsf{C}$, then $Ff,Fg$ are composable in $\mathsf{D}$, and $F(gf) = (Fg)(Ff)$.
2. For each $c\in\mathsf{C}$, $F1_c = 1_{Fc}$.

In other words, functors preserve domains, codomains, identity morphisms and composition of morphisms. There is another type of functor, called a contravariant functor, which swaps domains and codomains and reverses the order of composition of morphisms. That is, if $F$ is a contravariant functor, then it maps each morphism $f:c\to c'$ of $\mathsf{C}$ to a morphism $Ff:Fc'\to Fc$ of $\mathsf{D}$, and satisfies $(Ff)(Fg)=F(gf)$. A contravariant functor can also be thought of in terms of the opposite category as a covariant functor $F:\mathsf{C}^\text{op}\to \mathsf{D}$.

Here are some of my favorite examples of covariant functors taken from Riehl:

The following fact has a straightforward proof:

Proposition 1. Functors preserve isomorphisms. That is, if $F:\mathsf{C}\to\mathsf{D}$ is a functor and $f$ is an isomorphism of $\mathsf{C}$, then $Ff$ is an isomorphism of $\mathsf{D}$.

However, although functors preserve isomorphisms, they need not reflect isomorphisms. That is, if $f$ is an isomorphism, $Ff$ is guaranteed to be one; but if $Ff$ is an isomorphism, $f$ might not be.

Exercise 1. (Riehl 1.3.iii) Find an example to show that the objects and morphisms in the image of a functor $F:\mathsf{C}\to\mathsf{D}$ do not necessarily define a subcategory of $\mathsf{D}$.

Proof. Let $\mathsf{C}$ be the discrete category with one object $X$ and the required identity morphism $1_X$, and let $\mathsf{D}$ be the category whose only object is an object $Y$ with an identity morphism $1_Y:Y\to Y$ and one additional idempotent endomorphism $g:Y\to Y$ such that $gg=g$. Then we may construct a functor $F:\mathsf{C}\to\mathsf{D}$ mapping $X\mapsto Y$ and $1_X\to g$. However, the object $Y$ and the morphism $g:Y\to Y$ do not constitute a subcategory of $\mathsf{D}$, since they exclude the required identity morphism $1_Y$. $\blacksquare$



back to home page