# Equality of objects in category theory

I am trying to understand the notion of equality of objects in category theory.

My understanding is that an identity functor $1_\mathcal C:\mathcal C \to \mathcal C$ acts in such a way that for every object $X$ in $\mathcal C$ we get $$1_\mathcal C(X)=X$$ but I'm not actually sure whether this is correct, given the 'evilness' of equality in category theory.

Assuming this is correct, an isomorphism of categories $$F:\mathcal C \cong \mathcal D:G$$ would yield $$G \circ F=1_\mathcal C$$ and thus we could say that $$G\circ F(X)=1_\mathcal C(X)=X$$

On the other hand, an equivalence of categories $$F:\mathcal C \simeq \mathcal D:G$$ only yields$$G \circ F\cong 1_\mathcal C$$ and thus the equality of objects fails to transition across the whole formula $$G\circ F(X) \cong 1_\mathcal C(X)=X$$

So we can only say that $G\circ F(X)$ is isomorphic to $X$, but we cannot say anything about equality between them. To me, this is an easy way to understand the difference between isomorphism and equivalence of categories.

This reasoning rests on the assumption that $1_\mathcal C(X)=X$. So my question can be phrased as: Is the object part of identity functors idempotent?

• The identity functor sends everything (objects and arrows) to itself. You could also say that it leaves everything untouched. – drhab Feb 27 '18 at 18:30
• There is a general notion of equality of objects, it's just not what category theorists are interested in (usually) – Max Feb 27 '18 at 19:00
• It's more or less impossible to talk about category theory without any notion of equality of objects whatsoever-how do you even define composition? – Kevin Carlson Feb 27 '18 at 23:14
• AFAIK, if you had a suitable language in which you couldn't distinguish between equality and isomorphism of objects, you wouldn't distinguish between isomorphism and equivalence of categories either. – Hurkyl Feb 28 '18 at 6:22
• @KevinCarlson: Composition doesn't require any sort of equality of objects; it only requires using object-valued variables multiple times. e.g. people talk about $\hom(A,B) \times \hom(B,C) \to \hom(A,C)$, not about $\hom(A,B) \times \hom(C,D) \to \hom(E,F)$ in the case that $B=C$, $A=E$, and $D=F$. Although if you did talk about the latter in a weaker sense where the equalities were simply given isomorphisms, composition does transport along those isomorphisms. – Hurkyl Mar 1 '18 at 3:59

The identity functor is an identity arrow in the (large) category of (locally small) categories, so it is idempotent like any identity arrow. To be clear, a function, $f$, being idempotent means $f\circ f = f$. Also, $1_{\mathcal C}(X)=X$ is likely a definition. In an axiomatic approach, this might be accomplished by asserting that formula as an axiom, but in a set-theoretic approach, this would usually be interpreted as the set $\{(X,X)\mid X\in\mathsf{Ob}(\mathcal C)\}$ which is then referred to with the shorthand $1_\mathcal C$. The principle of equivalence talks about properties (or sometimes operations), what property, i.e. predicate, are you referring to with $1_\mathcal C(X)=X$? Are you taking $X$ as free? Well then this predicate is just always true by definition and thus trivially satisfies the principle of equivalence.
One signature for the DFOL theory of categories could be: $$\begin{array}% \mathsf{O}\ \mathsf{type}&& () \\ \mathsf{Hom}(A,B)\ \mathsf{type}&& (A,B:\mathsf O) \\ 1_A : \mathsf{Hom}(A,A)&& (A:\mathsf O) \\ f\circ g :\mathsf{Hom}(A,C)&& (A,B,C:\mathsf O,f:\mathsf{Hom}(B,C),g:\mathsf{Hom}(A,B)) \\ f = g\quad \mathsf{formula} && (A,B:\mathsf O, f,g:\mathsf{Hom}(A,B)) \end{array}$$ to which would be added axioms making $=_{AB}$ a congruence and the usual category axioms in terms of $=_{AB}$. Models of this wouldn't be categories exactly, but categories enriched in setoids. This is just to say that we have an equivalence relation on each hom-set, the interpretation of $=_{AB}$, and the operations are only defined up to equivalence. We can, of course, quotient by those equivalence relations in most set theories to get a standard category. It's actually awkward to add "equality" to $\mathsf{O}$ since presumably we'd then want something like $\mathsf{Hom}(A,B)=\mathsf{Hom}(A',B)$ when $A=A'$, but types aren't objects of the logic, i.e. there is no sort of types. There's a solution to this, but it is not obvious.
Homomorphisms of models of the above theory will be (setoid-enriched) functors. It can be proven that the principle of equivalence holds with respect to formulas written using the above signature. If we wanted to have a DFOL theory whose models were functors, we could start with two copies of the above signature, which I'll use subscript $1$ and $2$ to differentiate, and extend that with: $$\begin{array}% \mathsf F(A):\mathsf O_2 && (A:\mathsf O_1)\\ \mathsf F_{AB}(f) : \mathsf{Hom}_2(\mathsf F(A),\mathsf F(B)) && (A,B:\mathsf O_1, f:\mathsf{Hom}_1(A,B)) \end{array}$$ to which the functor laws are added as well as $\mathsf F_{AB}(f)=\mathsf F_{AB}(f')$ when $f=f'$. To be a bit more precise, this is the theory of a pair of categories and a functor between them. The models are a pair of (setoid-enriched) categories and a (setoid-enriched) functor between them. Continuing, we could extend this theory with: $$\begin{array}% \mathsf G(A):\mathsf O_1 && (A:\mathsf O_2)\\ \mathsf G_{AB}(f) : \mathsf{Hom}_1(\mathsf G(A),\mathsf G(B)) && (A,B:\mathsf O_2, f:\mathsf{Hom}_2(A,B)) \end{array}$$ and analogous laws to have a theory with a pair of functors going in both directions. Now, not only can we not state that $\mathsf G(\mathsf F(A))= A$ simply because we don't have an equality relation on objects, but we can't even state $\mathsf G_{\mathsf F(A)\mathsf F(B)}(\mathsf F_{AB}(f))=f$ as this isn't a well-formed formula: the left hand side is a term of sort $\mathsf{Hom}_1(\mathsf G(\mathsf F(A)),\mathsf G(\mathsf F(B)))$, while the right hand side is a term of sort $\mathsf{Hom}_1(A,B)$. We can easily add as data or assert the existence of isomorphisms between $\mathsf G(\mathsf F(A))$ and $A$ and similarly for $\mathsf F(\mathsf G(B))$ and $B$.