Why does functor's definition care so much about preserving $id$?

Question

I do understand what category and functor are, and yet would like to ask this question just to get more confidence and build correcter "intuition". So, among other requirements, functor is not allowed to tear apart existing arrows and, in particular, composites of those: $$F(g \circ f) = F(g) \circ F(f) \tag1$$ $$F(1_A) = 1_{F(A)} \tag2$$

My question is: why does second (#2) equation make any sense? Why can't we use $F(1_A \circ m) = F(m) = F(1_A) \circ F(m)$ or even $F(1_A \circ 1_A) = F(1_A) = F(1_A) \circ F(1_A)$ to prove both left and right identity properties?

Is it because $1_A$ has a special spirit, which differs it from some possible endomorphisms and especially automorphism (which can't be defined without identity)? Just to specify explicitly the uniqueness of the identity and it's significant role comparing with other morphisms? Or am I missing some important principle, which makes second equation absolutely necessary?

Answer 1

Categories are models of an essentially algebraic theory, much like how groups and rings are models of algebraic theories—what this means is that they are sets with structure that satisfy some equational axioms. Any (essentially) algebraic theory has a natural notion of homomorphism, namely functions on the underlying sets which commute with the structure.

For instance, a group is a set $G$ equipped with a unit element $e_G$, an inverse function $i_G : G \to G$ and a multiplication function $m_G : G \times G \to G$, satisfying the usual axioms. A homomorphism of groups is then a function $f : G \to H$ such that:

• $f(e_G)=e_H$;
• $f \circ i_G = i_H \circ f$; and
• $f \circ m_G = m_H \circ (f \times f)$.

The fact that the first two equations can be deduced from the third is a convenient coincidence.

Now a category consists of a set $\mathrm{ob}(\mathcal{C})$ of objects, a set $\mathrm{mor}(\mathcal{C})$ of morphisms, domain and codomain functions $\mathrm{dom}_{\mathcal{C}},\mathrm{cod}_{\mathcal{C}} : \mathrm{mor}(\mathcal{C}) \to \mathrm{ob}(\mathcal{C})$, an 'identities' function $\mathrm{ids}_{\mathcal{C}} : \mathrm{ob}(\mathcal{C}) \to \mathrm{mor}(\mathcal{C})$, and a composition function $\mathrm{com}_{\mathcal{C}} : \mathrm{comp}(\mathcal{C}) \to \mathrm{mor}(\mathcal{C})$, where $$\mathrm{comp}(\mathcal{C}) = \{ (f,g) \in \mathrm{mor}(\mathcal{C})^2 \mid \mathrm{cod}_{\mathcal{C}}(f) = \mathrm{dom}_{\mathcal{C}}(g) \}$$

A functor, then, is nothing more than a homomorphism of categories: it's given by maps on the underlying sets (of objects and of morphisms), which respects the structure (domain, codomain, identity, composition). Explicitly, a functor is a pair of functions $F_0 : \mathrm{ob}(\mathcal{C}) \to \mathrm{ob}(\mathcal{D})$ and $F_1 : \mathrm{mor}(\mathcal{C}) \to \mathrm{mor}(\mathcal{D})$, satisfying the following equations:

• $F_0 \circ \mathrm{dom}_{\mathcal{C}} = \mathrm{dom}_{\mathcal{D}} \circ F_1$;
• $F_0 \circ \mathrm{cod}_{\mathcal{C}} = \mathrm{cod}_{\mathcal{D}} \circ F_1$;
• $F_1 \circ \mathrm{ids}_{\mathcal{C}} = \mathrm{ids}_{\mathcal{D}} \circ F_0$; and
• $F_1 \circ \mathrm{com}_{\mathcal{C}} = \mathrm{com}_{\mathcal{D}} \circ (F_1 \times F_1)$.

Unlike with groups, the identity axiom (the third listed above) cannot be deduced from the other three.

In order to deduce that $F(\mathrm{id}_A) = \mathrm{id}_{FA}$, you would need it to be the case that $F(\mathrm{id}_A) \circ g = g$ for all $g : C \to FA$ and $h \circ F(\mathrm{id}_A) = h$ for all $h : FA \to D$ in the codomain category. When the morphisms $g,h$ are in the image of $F$ these equations do hold, but in general, there will be morphisms which are not in the image of $F$.

As a concrete example, let $\mathcal{C} = \mathbf{Set}$ and let $\mathcal{D}$ be the category with one object $\star$ and one non-identity morphism $e : \star \to \star$ such that $e \circ e = e$. Define $F : \mathcal{C} \to \mathcal{D}$ by letting $F(X) = \star$ for all sets $X$ and $F(f)=e$ for all functions $f$. Then $F$ satisfies all the laws a functor is required to satisfy except for the requirement that $F(\mathrm{id}_X) = \mathrm{id}_{FX}$ for all sets $X$.

Answer 2

Nearly the entire point of category theory is to clarify when things are isomorphic, hopefully naturally so. Since isomorphisms (in any setting) are defined by equalities like $f \circ g = 1_Y$ and $g \circ f = 1_X$ then yes, the identity plays a special role that must be respected.

In order to be useful in comparing different categories, a functor should respect isomorphisms. Hence a functor should preserve both composition and identities, which is why we have it defined as such.