Why is equivalence of functors defined differently from the equivalence of categories?

Question

My way of thinking:

1. when it comes down to a regular sets, the right notion of "sameness" of any two elements is strict, regular equality;
2. for any two objects belonging to the same category - isomorphism, which is less strict version of equality (probably, it makes sense to think of "isomorphism" as "partial equality");
3. then there are functors, generating yet another $[A, B]$ functors category; by the #2, fine notion of "sameness" for them is (natural) isomorphism;
4. however, when it comes down to "sameness" of somewhat categories, which also build separate category $CAT$, somewhy isomorphism considered to be "unreasonably strict". As such, $G \circ F = 1_A$ and $F \circ G = 1_B$ becomes $G \circ F \cong 1_A$ and $F \circ G \cong 1_B$.

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QUESTION: Why #2 notion isn't good enough to reason about equivalence of categories? Are categories special? As far as I understand, $CAT$ allows reuse #2 "sameness" definition with respect to any two categories...

Answer 1

Part of the importance of equivalence of categories has to do with isomorphism of the objects in said categories.

Consider the category of all finite sets $\mathsf{FinSet}$ (and mappings between those). That's a huge category, since it's collection of objects is a proper class. However in a sense it should not be so huge, since essentially there only as many finite sets as they are are natural numbers.

Consider another category $\mathcal A$, which is only the finite sets of the form $\{1,\dots, n\}$. Now for every $n\in \mathbb N$, $\mathcal A$ has one set-representative of that size while $\mathsf{FinSet}$ has many, but in $\mathsf{FinSet}$ all these sets of the same size are isomorphic and we should not treat isomorphic sets as being different.

Hence it doesn't make any real difference if we use $\mathsf{FinSet}$ or $\mathcal{A}$ to deal with finite sets. So they ought to be the same. And they are: these categories are equivalent. But they can't be isomorphic: $\mathcal{A}$ has a countable set of objects, but $\mathsf{FinSet}$ a proper class.

Answer 2

Categories are indeed special, in that they form a 2-category: natural transformations give a good notion of 2-morphism, a morphism between morphisms. That's what leads to the possibility of using equivalence rather than isomorphism. As the other answer suggests, there are numerous examples of equivalent pairs of categories which are not isomorphic. In fact, categories encountered "in the wild", while doing mathematics, are hardly ever isomorphic to each other, while there are many very important equivalences between categories.

For a few examples, in increasing order of difficulty:

(1) The category of finite dimensional vector spaces over a field is equivalent to its opposite, by taking the dual space. (Beware that the dual space functor is famously not isomorphic to the identity!)

(2) The category of adjunctions from the category of sets to any category $C$ with coproducts is equivalent to $C$: given $c$, define the left adjoint on any set $S$ to be the coproducts of $S$ copies of $c$ and the right adjoint to send $c'$ to the set $\mathrm{Hom}_C(c,c')$.

(3) The opposite of the category of sets is equivalent to the category of complete atomic Boolean algebras. This generalizes to Stone duality, the theorem that totally disconnected spaces are equivalent to the opposite of arbitrary Boolean algebras, and then Gelfand duality, which says that arbitrary compact Hausdorff spaces are equivalent to the opposite of commutative $C^*$ algebras.

(4) Some very important conjectures center around equivalences of categories. For instance, homological mirror symmetry posits an equivalent between the "Fukaya category" of a symplectic manifold and the "derived category" of a related complex manifold, its mirror, thus leading to a unification of two very disparate fields of mathematics.

There's no question of giving an isomorphism in any case but (1), and no one would ever want to: the point is that, whatever math you can do in one category, you can transfer to an equivalent one. In category theory, we usually care about objects up to isomorphism and morphisms up to equality, and that's what an equivalence reflects.

Answer 3

Actually, #2 is a degenerate instance of the notion behind #4.

Intuitively, in a mere set, elements are either equal or not, there is no "in between". This shall be consider a tremendous lack of looseness of sets. This lack of looseness reverberates on (parallel) morphisms of a given category : given $f:A\to B$ and $g: B \to A$, the morphisms $f\circ g$ and $\mathrm{id}_B$ are either equal or not, there is no "in between"; the same goes for $g\circ f$ and $\mathrm{id}_A$. If now $A$ and $B$ are categories and $f$ and $g$ functors, there is multiple ways for $f\circ g$ and $\mathrm{id}_B$ to be "equal-ish", namely all the isos $f\circ g \simeq \mathrm{id}_B$. So the idea is that multiple ways to be equal-ish should be the norm and the either equal or not just happens to represent the particular cases when the multiple ways are reduced to just one.

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But enough with the hand waving, let's try to be (more) formal.

Define a $2$-category to be a category enriched over small categories. Technically a $2$-category $\mathcal K$ is then the data of

• a collection of objects
• for each pair of objects $A,B$, a category $\mathcal K (A,B)$
• for every objects $A,B,C$ a functor $c_{A,B,C} : \mathcal K(A,B)\times\mathcal K(B,C) \to \mathcal K(A,C)$
• for every object $A$ a functor $i_A: 1\to \mathcal K(A,A)$ (where $1$ is the final category)

such that commute the diagrams making the $c$'s associative compositions and the $i$'s neutrals for those. I let you draw these. (Hint: these are the same diagrams as for locally small categories, where the $\mathcal K(A,B)$'s are just sets). For $c_{A,B,C}(f,g)$ we write $g\circ f$ and we also write $\mathrm{id}_A$ for the object of $\mathcal K(A,A)$ selected by $i_A$.

In any $2$-category $\mathcal K$, you have a notion of equivalence as follow: an object $f$ of $\mathcal K(A,B)$ is an equivalence if there is an object $g$ of $\mathcal K(B,A)$ such that $f\circ g$ is isomorphic to $\mathrm{id}_B$ in $\mathcal K(B,B)$ and $g\circ f$ is isomorphic to $\mathrm{id}_A$ in $\mathcal K(A,A)$.

Small categories, functors and natural transformations forms a $2$-category $\mathbf{Cat}$ (objects are small categories, and $\mathbf{Cat}(A,B)$ is the category you denoted $[A,B]$; compositions and identities are the usual compositions and identities of functors). Equivalence in $\mathbf{Cat}$ gives you back #4.

Now any category $\mathcal C$ can be considered as a $2$-category $\underline{\mathcal C}$: objects are those of $\mathcal C$ and for objects $A,B$, the category $\underline{\mathcal C}(A,B)$ has

• as objects the morphisms $f:A\to B$ of $\mathcal C$
• as morphisms only the identity morphisms

An equivalence in this $2$-category $\underline{\mathcal C}$ is exactly an isomorphism in the original category $\mathcal C$. Indeed, an isomorphism of the form $f \circ g \simeq \mathrm{id}_B$ can only be an identity morphism, forcing the two-side to be equal. So it gives you back #2.

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Of course, you can now say that it is fishy to ask, in the definition of an equivalence $f$, for $f\circ g$ and $\mathrm{id}_B$ to be isomorphic, as it means there is $\varphi : f\circ g \to \mathrm{id}_B$ and $\psi: \mathrm{id}_B \to f\circ g$ such that $\varphi \circ \psi$ and $\mathrm{id}_{\mathrm{id}_B}$ are equal (and the same for $\psi \circ\varphi$ and $\mathrm{id}_{f\circ g}$). This is only due to the lack of looseness of the set of endomorphisms of $f\circ g$ and $\mathrm{id}_B$. Following the lines of before, shouldn't this only be a degenerate case of a setting where those $\varphi$ and $\psi$ are just isomorphic ? Yes, it could, if you move to $3$-categories and you view $2$-categories as degenerates $3$-categories. And you can carry on, it is turtles isomorphisms all the way up...

(Actually, people are usually more interested in the cases where even associativity and identity laws hold up to iso, leading eventually to the world of $(\infty,1)$-categories and beyond, but this is way beyond the scope of this answer.)