So, I'm trying to sort my head around this notion of equivalences of categories, and so I decided to construct a short and simple example for my own reference to kind of motivate why one would create such a definition as equivalence of categories in the first place, making use of everyone's favourite example of an isomorphism, the homeomorphism between the donut and the cup. I made the following illustrative picture:
...and wrote as follows:
Arguably the most celebrated, classic example of an isomorphism is the homeomorphism between the donut and the cup, both being topological variations of the torus $S^1 \times S^1$. From these, we may construct ourselves a category which we dub $\mathcal{DnC}$. Its objects are the donut and the cup, denoted $d$ and $c$ respectively, and its morphisms are the identity morphisms on either, $e_d$ and $e_c$, the isomorphism $f_{dc}$, which maps points on the donut to points on the cup, and its inverse $f_{cd}$.
The defining feature of both these shapes, which leads to them being topologically isomorphic, is that they both possess precisely one hole. We construct the category $\mathcal{Circ}$ consisting of the circle, which we denote $a$, as its sole object and the identity morphism unto itself, $e_a$ as its sole morphism, and to this construct the functors $F, G$ which relate the two categories (see Figure).
The functor $F$ is defined by: $F(d)=F(c)=a$ and $F(e_d)=F(e_c)=F(f_{dc}) = F(f_{cd}) = e_a$, and the functor $G$ is defined by $G(a)=d$ and $G(e_a)=e_d$.
The two categories $\mathcal{DnC}$ and $\mathcal{Circ}$ clearly are not isomorphic as the difference in their object counts renders it impossible to construct a bijective functor between them. Nevertheless, there is a certain degree of "similarity" between the two categories.
If we start at the donut in the category $\mathcal{DnC}$, go down to the circle in the category $\mathcal{Circ}$ and then back up again with $G$, we end up at the donut where we started, and while it may be true that if we start at the cup and do the same that we end up at the donut, the donut is still isomorphic to the cup, and what is isomorphism is not a measurement of "similarity"? The same line of reasoning may be applied to the morphisms. While $f_{dc}$ and $f_{cd}$ may be morphisms between different objects that both map to the identity morphism $e_a$, the objects they are morphisms between are nonetheless isomorphic to each other.
This sort of made me wonder: can you define an equivalence of categories in such a way that two categories $\mathcal{A}, \mathcal{B}$ are equivalent if the categories $\mathcal{A}', \mathcal{B}'$ whose objects are the equivalence classes of objects in $\mathcal{A}$ and $\mathcal{B}$ respectively that are isomorphic to one another, are isomorphic to one another?
This seems to be to be reasonable to me, and yet I am nevertheless having a bit of a hard time to define the categories $\mathcal{A}', \mathcal{B}'$ in particularly good way, seeing the definition of a quotient category is such that any quotient category of a category $\mathcal{C}$, call it $\mathcal{Q}$ has the same objects as $\mathcal{C}$.
Specifically, is my understanding correct, and if so, given a category $\mathcal{C}$, how does one go about properly defining the category $\mathcal{C}'$ whose objects are the isomorphism-equivalence classes of $\mathcal{C}$? Particularly, how do you define the morphisms on $\mathcal{C}'$?
I look forward to your response.