Well, I decided to include some additional information.
Definition 1. Let $\mathcal{C}$ and $D$ be categories, $T\colon\mathcal{C}\to\mathcal{D}$ and $S\colon\mathcal{D}\to\mathcal{C}$ be functors. Then the pair $(T,S)$ is called an equivalence iff $S\circ T\cong I_{\mathcal{C}}$ and $T\circ S\cong I_{\mathcal{D}}$. In this case functors $T$ and $S$ are also called equivalences, and categories $\mathcal{C}$ and $\mathcal{D}$ are called equivalent.
It is a basic definition and Tim's answer shows why we need to use equivalence even more frequently than isomorphism. Here's another important definition:
Definition 2. Let $\mathcal{C}$ be a category, $\mathcal{S}$ be a subcategory of $\mathcal{C}$. Then the category $\mathcal{S}$ is called a skeleton of $\mathcal{C}$ iff it is a full subcategory of $\mathcal{C}$ and every object of $\mathcal{C}$ is isomorphic to precisely one object of $\mathcal{S}$.
Note, that if the axiom of choice holds, then every category has a skeleton. See also nLab article. The connection between equivalences of categories and their skeletons is described in the following proposition:
Proposition 1. Let $\mathcal{C}$ and $\mathcal{D}$ be categories, $\mathcal{S}_{\mathcal{C}}$ and $\mathcal{S}_{\mathcal{D}}$ be their skeletons. Then $\mathcal{C}\simeq \mathcal{D}$ iff $\mathcal{S}_{\mathcal{C}}\cong\mathcal{S}_{\mathcal{D}}$.
The proof follows from the fact that every category is equivalent to its skeleton and if two skeletal categories are equivalent, then they are isomorphic. You can also search Mac Lane's "Categories for the working mathematician" for the details.
Thus we can use the notion of skeleton instead of the original definition of equivalence, but sometimes it is not a simplification. As it was mentioned, even an attempt to prove that a category has a skeleton may lead to the set-theoretical difficulties. Tim also gave arguments.
You write: But why go through this whole process of defining this notion of equivalent categories if we can just create a single equivalence classes of objects via the equivalence relation of being isomorphic, and make morphisms defined on the same equivalence class "the same?"
Okay, it could be a good idea if we want to define something like a skeleton. But it isn't, because the straightforward applying this idea leads to wrong definition. Let's try to do this.
Definition 3. Let $\mathcal{C}$ be a category. Then define the graph $\text{Equiv}(\mathcal{C})$ in the following way: $\text{Obj}(\text{Equiv}(C))=\text{Obj}(\mathcal{C})/\cong_{\mathcal{C}}$ and $$\text{hom}_{\text{Equiv}(\mathcal{C})}([a],[b])=(\coprod_{a'\in[a],b'\in[b]}\text{hom}_{\mathcal{C}}(a',b'))/[(f\sim g)\Leftrightarrow(\exists a,b\in \text{Iso}(\mathcal{C})|\quad g\circ a=b\circ f) ].$$
But the graph $\text{Equiv}(\mathcal{C})$ doesn't inherit the composition law from $\mathcal{C}$. The graph $\text{Equiv}(\mathcal{C})$ doesn't even coincide with graph of any skeleton of $\mathcal{C}$ in general case. For example, it may paste two morphisms with the same domain: in the category $\mathbf{Finord}$ we have $\text{end}_{\text{Equiv}(\mathbf{Finord})}([2])=\text{hom}_{\text{Equiv}(\mathbf{Finord})}([2],[2])\cong2$, but $\text{end}_{\mathbf{Finord}}(2)=2^2=4$.
There is something fundamental going wrong in your reasoning. You are defining two different relations on $S$:
- $a\sim_1 b$ if and only if $a$ and $b$ are both not squares.
- $a\sim_2 b$ if and only if $a$ and $b$ are both circles.
However, these relations are not equivalence relations yet; there are elements in $S$ that do not relate to itself, so it is not reflexive. There are many ways to modify $\sim_1$ and $\sim_2$ such that it does define an equivalence relation. For instance:
- $a\sim_1 b$ if and only if $a$ and $b$ are both squares or are both not squares.
- $a\sim_2 b$ if and only if $a$ and $b$ are both circles or $a=b$.
Now, both relations are transitive, symmetric and reflexive and therefore equivalence relations. However, each of the relations defines its own partition of $S$.
For instance, $\sim_1$ partitions $S$ into two equivalence classes: the set $A$ that consists of all the elements of $S$ that are not squares and the set $B$ that consists of all elements of $S$ that are squares. Note that $A$ and $B$ are disjoint sets! (This is what is meant).
Similarly, $\sim_2$ partitions $S$ into many classes: the class that contains all circles and each element in $S$ that is not a circle defines its own equivalence class consisting of just itself. These classes are obviously disjoint as well.
What you did instead is compare equivalence classes defined by $\sim_1$ with classes defined by $\sim_2$.
I hope this helps.
Best Answer
Here are two examples :
$1 - $ Consider the relation $\equiv$ ( an equivalent relation), then
$$a \sim b \Leftrightarrow a\equiv b \mod 2 $$
That is, $a$ and $b$ will be in the same class $\overline{a}$ if their remainders of the division by $2$ are the same. For example $4$ and $6$ belong to the same class, which we are going to choose a representant $0$, because
$$6 = 3 \cdot 2 + \color{red}{0} \ \ \text{and} \ \ 4 = 2 \cdot 2 + \color{red}{0}$$
then we say $\overline{4} = \overline{6} = \overline{0}$. If we think, there are two distinct classes: $$\overline{0} = \{x \in \mathbb Z ; x \equiv 0 \mod 2, \text{$x$ is even}\}\ \ \text{and}\ \ \overline{1} = \{x \in \mathbb Z ; x \equiv 1 \mod 2, \text{$x$ is odd}\}$$
The set of all classes is
$$\mathbb Z_2 = \{\overline{0}, \overline{1}\}$$
$2-$ Consider the relation
$$(a,b) \sim (c,d) \Leftrightarrow ac = bd $$
This equivalent relation gives us the fractions, that is the filed of fractions of $\mathbb Z$. Similarly we choose a class representant for example,
$$\frac{1}{2} = \frac{2}{4} = \frac{3}{6 } = \cdots$$
we choose $\frac{1}{2}$ to be the class representant. Notice that $\mathbb Q = \{ \frac{a}{b} ; a,b \in \mathbb Z, \ \ \text{where}\ \ b \neq 0\}$ is the set of all classes.